Lecture Notes in
Mathematics
Edited by A Dold and B Eckmann
758
C. Nàstàsescu
F. Van Oystaeyen
Graded and Filtered
Rings and Modules
Springer-Verlag
Berlin Heidelberg New York 1979

Authors
C. Nàstàsescu
Faculty of Mathematics
University of Bucharest
Str. Academiei 14
R-70109 Bucharest 1
Romania
F. Van Oystaeyen
Department of Mathematics
University of Antwerp UIA
Universiteitsplein 1
B - 2610 Wilrijk
Belgium
AMS Subject Classifications (1980): Primary: 16A03
Secondary: 16A05,16A33,16A45,16A50,16A55,16A63,16A66
ISBN 3-540-09708-2 Springer-Verlag Berlin Heidelberg New York
ISBN 0-387-09708-2 Springer-Verlag New York Heidelberg Berlin
Library of Congress Cataloging in Publication Data
Nàstàsescu, Constantin.
Graded and filtered rings and modules.
(Lecture notes in mathematics; 758)
Bibliography: p.
Includes index.
1. Associative rings. 2. Graded rings. 3. Graded modules. 4. Filtered rings. 5. Filtered
modules. I. Oystaeyen, F. van, 1947- joint author. II. Title. III. Series- Lecture notes in
mathematics (Berlin); 758.
QA3.L28 no. 758 [QA251.5] 510'.8s [512'.4] 79-26585
ISBN 0-387-09708-2
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PREFACE
The aim of these lecture notes is to present coherently the latest and less known
results concerning the theory of filtered and graded rings and modules. With an eye
to projective algebraic geometry, the theory of commutative graded rings and
Noetherian graded modules over them lias been expounded in the large, e.g. using cohomo-
logical methods and Serre's theorem, which states that the category of coherent
sheaves over projective n-space is equivalent to a category of graded Noetherian
modules over a graded polynomial ring, it is possible to prove generalizations of
;iax Noether's and Bezout's theorem within the latter category, cf. [15 ] .
In writing these notes we have adopted a completely different point of view, i.e.
a purely ring theoretical one, and the only reference to projective geometry will be
in 11.14, 11.15 and 11.16 but then over a non-commutative ring. The material
presented falls naturally into three parts.
Chapter I is devoted to the study of the relation between a filtered ring R and the
category of filtered R-modules on one hand, and the graded ring and module category
associated to R on the other hand. In particular we focussed attention on : Krull
dimension in Gabriel's and Rentschler's sense; free, projective and finitely
generated objects, homological dimension i.e. projective and flat dimension etc.». Most
of the results included in this chapter have been developed after those included in
[35].
In Chapter II we introduce several topics in, what may be called, "Graded Ring Theory".
The general problem faced is the following : what information about R-mod, the
category of left R-modules, can be drawn from certain properties holding in R-gr, the
category of graded left R-modules, i.e. the feed-back from the graded theory into
ungraded ring theory. Of course in investigating this feed-back problem it will also
be necessary to develop some graded techniques paralleling the ungraded theory. In
this perspective we study : finitness conditions, Krull dimension, homological
dimension, primary decompositions, graded rings and modules of quotients, etc.—
Although the graded version of A. Goldie's theorems does not yield a necessary and
sufficient condition for the existence of a graded semisimple gr. Artinian ring of

IV
fractions, under some conditions on the gradation of the ring, which may be rather
easily verified in real life circumstances, our graded Goldie rings do have the
desired rings of fractions. Part of the results in Chapter II have been obtained by
us in recent papers, cf. [28] , [29] , [31] , but we have also included several
generalizations and new results.
Chapter III contains some new results in the theory of Noetherian graded commutative
rings on a problem put forward by Nagata in [26] . This problem has been solved by
several authors amongst whom we quote [ 8] , [23] , [24] . However, the presentation of
the problem given in this chapter is by far more simple than the one given in the
papers cited above; moreover our solution to the problem allows a number of
generalizations as well as some new results. Actually part of Chapter III may be generalized
to the case of a non-commutative ring, using the localization techniques of Chapter
II, but we thought that full generality everywhere would have been disturbingly in-
congruent with respect to the aim of the chapter and so we presented only the
commutative theory.
A certain knowledge of the elements of Ring Theory and of Homological Algebra, on the
level of graduate courses, is necessary for the understanding of these notes; in
other words, we assume some familiarity with concepts like : Ext , the Jacobson
radical of a ring, Grothendieck categories,but the indispensable background material on
filtrations and gradations has been included in the notes.
C. Majstâiucu
F. Van Oyitae.ye.yi

Acknowledgement
Each of the authors expresses gratitude towards the department of mathematics
of the co-author's institution for its hospitality while these notes were
being conceived and written.
We thank Prof. Dr. P.M. Cohn for reading the manuscript and for making some
useful suggestions.
Our typist, Ludwig Callaerts, nearly got sick turning our illegible notes
into this nicely typed book; we really appreciate his work and thank him a lot
F. Van Oystaeyen thanks Danielle for her patience and support, while he was
working at home, doing things that could have been done during normal working
hours.

CONTENTS
I. FILTERED RINGS AND MODULES 1
1.1. Preliminaries 1
1.2. The Category of Filtered Modules 2
2.1. Remarks 3
2.2. Functors on R-filt. 3
1.3. Graded Rings and Modules 5
3.1. Examples 5
3.2. The Category of Graded Modules 5
3.3. Some Properties of Graded Modules 7
3.4. Tensor Product of Graded Modules 12
1.4. Filtration and Associated Gradation 13
1.5. Free, Projective and Finitely Generated Objects of R-filt 16
1.6. The Functor HOM^(-,-). 2k
1.7. Projective Modules and Homological Dimension of Rings. ?i
1.8. Weak (Flat) Dimension of Filtered Modules. 30
II. TOPICS IN GRADED RING THEORY 36
11.1. Homogenization 36
11.2. The Structure of Principal Graded Rings 38
11.3. Noetherian Objects h]
11.4. Krull Dimension of Graded Rings ^7
11.5. The Krull Dimension of Some Classes of Rings 53
5.1. Polynomial Rings 53
5.2. Rings of Formal Power Series 55
5.3. The Enveloping Algebra of a Lie Algebra 56
5.4. Weyl Algebras 56
11.6. Graded Division Rings 58
11.7. The Structure of Simple Objects in R-gr. 59
11.8. The Jacobson Radical of Graded Rings. 6s
11.9. Semisimple Graded Rings. Goldie's Theorems for Graded Rings •?;
9.1. Graded Matrix Rings ''7
9.2. Rings of Fractions and Goldie's Theorems 70
IT.10. Primary Decompositions 7h
11.11. External Homogenization 78
11.12. Graded Rings and Modules of Quotients 83
12.1. Torsion Theories over Graded Rings 83
12.2. Examples of Torsion Theories in R-gr. 89
12.3. Injective Objects and Torsion Theories 91

VIII
12.4. Graded Rings and Modules of Quotients 95
12.5. More Properties of Graded Localization 100
11.13. Graded Prime Ideals and the Ore Condition 107
11.14. The Presheaves on Proj R. 110
11.15. Graded Zariski Central Rings H6
15.1. Basic Facts about Zariski Central Rings 116
15.2. GZ- and ZG-Rings 118
11.16. Scheme Structure of Proj over a Graded Zariski Central Ring 121
III. LOCAL CONDITIONS FOR NOETHERIAN GRADED RINGS 127
III. 1. Injective Dimension of Graded Rings 127
111.2. Regular, Gorenstein and Cohen-Macaulay Rings 133
111.3. Graded Rings and M-sequences. 137
REFERENCES Ihh
SUBJECT INDEX 11+6

SOME NOTATIONS AND CONVENTIONS
All rings have identity elements. All R-modules are left R-modules unless otherwise
stated. Ideal of R will mean two-sided ideal of R.
References without chapter indication refer within the chapter containing the
reference.
IN
TL
R-mod
R-gr
L (R)
g
EM
M
T
n
Ext£
R+
G
R+
M~
X~
rad
g
M(d,k)
Bg(M)
J(R)
the natural numbers {0,1,2,...}
the integers
the category of left R-modules
the category of graded left R-modules
if Me R-gr then M is the underlying R-module
the lattice of graded left ideals of R
if MeR-filt then FM is the filtration {F.M, iei}
if MeR-filt then M is the completion of M
the suspension functor on R-gr
the n-th derived functor of Hom„(—,—)
R.
i>o
R. =R^
i >o
the ,.associated gradation" functor : R-filt->G(R)-gr
R+ = ® R. =R.
i>o x >0
M = © M. =M^
i«o x <0
the graded module generated by homogeneous elements of maximal degree in
the decompositions of elements of X.
the graded module generated by homogeneous elements of minimal degree in
the decompositions of elements of X.
the prime radical
If M is an R-submodule contained in a graded module N then (M) is the
graded submodule of N maximal with the property of being contained in M.
if (d,k) e TL1 such that d>1, o«k«d-1 then M^d'k^
the graded bi-endomorphism ring of M
the Jacobson radical of R
M
ies
'id+k

X
J°(R)
AnnJ-1
Mn(R)
ht(P)
Ass M
*
x
£00
of
\
G(P)
CM.
Proj R
p. dim
inj . dim M
w.dim M
gl.dim R
gl.w.dim R
the graded Jacobson radical of R
the annihilator of M in R.
the n by n matrices over R
the height of P
the set of prime ideals associated to M
the external homogenization in M[T] of x£M
the filter associated to the kernel functor k
the graded localization functor associated to k
the localization functor in R-mod associated to k_
Goldie's multiplicative set associated to the prime ideal P
Cohen-Macaulay
the projective spectrum of R i.e. the set of graded prime ideals not
containing R+.
the projective dimension of M
the injective dimension of M
the weak (flat) dimension of M
the left global dimension of R
the left weak global dimension of R
All these dimensions have analogues in R-gr denoted by gr.p.dim M,
gr.inj.dim M, gr.w.dim M, gr.gl.dim R, gr.gl.w.dim R.

CHAPTER I. FILTERED RINGS AND MODULES.
1.1. PRELIMINARIES.
1.1. Definition. An associative ring R with unit is said to be a filtered ring if
there is an ascending chain {F R,ne2Z ,} of additive subgroups of R satisfying the
following conditions : 1 eF R, F R.F RCF + R for any (n,m) e 7L2. The family of these
subgroups, denoted FR is called the filtration of R.
Note that the definition implies that F R is a subring of R.
1.2. Examples.
1°. Any ring R can be made into a filtered ring by means of the trivial filtration,
i.e. F R = o if n<o, F R = R if n>o.
n n
2°. Let I be an ideal of R. The I-adic filtration is obtained by putting F R = R for
n>o while F R=I for n<o.
3°. Let R be any ring and let t : R-*-R be an injective ring homomorphism, 5 ;R->Ra
if -derivation i.e. a group homomorphism for the additive structure of R such that
8 (xy) =8 (x)y+ xl,58 (y) for x,y£R. The ring of skew polynomials R[X,y>,8] is obtained
from R by adjoining a variable X and defining multiplication according to :
Xb = r X+8(r), for reR, in such a way that R becomes a subring of R[X,y>,8] . (Note
that the action of endomorphisms is written exponent-wise). The fact that <p is
injective allows to introduce the degree function (denoted by : deg) in the usual way
and so we may define the degree-filtration by : F R[X,y>,8] = {P £R [X,y>,5} ,deg P<n};
hence F R[X,v,8] =R.
1.3. Definition. Let R be a filtered ring. A left R-module M is called a
filtered module if there exists an ascending chain {F M,n eZ} of additive subgroups of
M such that FnR«FmMCFn+mM for ^y (n>m) e ^■ Tne family (FnfI>n £Z} is the
filtration of M. Obviously, if R is a filtered ring then DR (resp. RD) is a filtered left
K K
(resp. right) R-module.
If M is a filtered left R-module then its filtration F may have one of the

2
following properties
(E) F is exhaustive if M = U FM.
neZn
(D) F is discrete if there is an n e TL such that F.M = o for all i<n„.
/-\ ° -1
(S) F is separated if I I F M = o.
n e Z n
(C) F is complete if M = lim M/F M.
n
If the filtration of R is trivial and MeR-mod, then any ascending chain of submodules
of M defines a filtration for M; in this case the trivial filtration of M is given by
F M = o if n<o, F M=M if n>o.
n ' n
1.4. Definition. Let R and S be filtered rings an MeR-mod-S is said to be a
filtered R-S-module if there exists an ascending chain {F31,n e Z} of additive
subgroups such that : F R.F MCF M for all n,me Z, and F M.F_SCF A.M for all m,t e Z.
b r n m m+n ' ' m t m+t
1.2. THE CATEGORY OF FILTERED MODULES.
Let R be filtered ring, M and N filtered left R-modules. An R-homomorphism
f eHomR(M,N) is said to have degree p if f(F.M) CF.+ N for all ie TL. Homomorphisms
of finite degree from a subgroup HONL(M,N) of HomR(M,N), homomorphisms of degree p
form a subgroup F HOMR(M,N) of KOvL(M,N),
One easily checks :
1. If p=Sq then : F HCW^CM.N) CF HOf-L (M,N).
2. HOMD(M,N) = U FHCMuCM.N).
R p ê2 P R
3. If f : M-*-N has degree p and g : N-»-P has degree q then gof has degree p+q.
These properties allow us to introduce the category R-filt of filtered left R-modules
where the morphisms are the homomorphisms in H0M_(—,—) of degree o. If M,Ne R-filt
then FH(îi(M,N) will simply be denoted by HomFR(M,N). Clearly R-filt is an additive
category and furthermore if feHomFR(}l,N) then Kerf and Coker f are in R-filt.
Indeed, Kerf has the induced filtration F.Ker f = Ker f nF-M, whereas Coker f is
filtered by putting F. Coker f=(Imf+F.N)/Imf. In particular it follows that monomor-
phisms and epimorphisms in R-filt coincide with the injective resp. surjective mor-

3
phisms of R-filt.
Obviously, arbitrary direct sums as well as direct products exist in R-filt. Moreover,
if Qi-,<p ■ ■), with i,j in some index set I, is an inductive (projective) system in R-
filt then its inductive (projective) limit exists in R-filt; it is the R-module lim M.
(lim M.) with filtration F lim M. = lim F M. , (F lim M. =lim FM.).
< i p —► i —>• p i' p < i < p i
i i
2.1. Remarks.
1. Let {M. ,iel} be a family of objects from R-filt such that their filtrations FM.,
iel, are exhaustive, then the filtration of © M. is exhaustive. This property fails
I 1
for the direct product n M-! The similar result however does hold for inductive or
I
projective limits.
2. R-filt is preabelian (cf. [13 ]) but not abelian. Indeed, take M/o in R-mod and
put F.M = o for all i eZ. Let F'M be another filtration on M and denote G, resp. H,
the filtered modules obtained from M using FM resp. F'M. The identity morphism of M
is in HompR(G,H) but not in HomFR(H,G) and thus the identity of M is a bijective
mapping which fails to be an isomorphism.
3. In a similar way as before one defines the category R-filt-S for a pair of filtered
rings R and S.
2.2. Functors on R-filt.
Before studying some special functors in R-filt let us point out that, if R is
a filtered ring with filtration FR, we have a functor D : R-mod-*R-filt which associates
to Me R-mod the R-module M with filtration F M = F R.M. This filtration of M is called
n n
the deduced filtration. Note that F D(M) -F R.M=M for any n>o.
n n '
The suspension functor. For ne % we define the n-th suspension functor T :
R-filt -* R-filt, which associates to Me R-filt with filtration FM the filtered module
M(n) with filtration F.M(n) =F.+ M hence F.T (M) =F. (M) . Suspension is
characterized by : TnoTm = Tn+m> Tn = Id. (the identity functor). In particular, every T is an
equivalence of categories which commutes with direct sums, products, inductive and
projective limits. Consequently MeR-filt is complete (i.e. FM is complete) if and
only if T (M) is complete. It is easily verified that the following holds : for p eZ

4
and for M.NeR-filt :
FpHOMR(M,N) =HomFR(M(-p),N) = HomFR(M,N(p)) .
The exhaustion functor. Let R have filtration FR. If we put R' = U F R then,
ne Z n
R' is a filtered subring of R. If MeR-filt then, clearly, M' = U FM is a filtered
neZ
R'-module and the filtration FM' given by FM' =F M for all neZ is an exhaustive
filtration. In this way we obtain a functor : R-filt—►R'-filt carrying M into the
corresponding M', which is called the exhaustion functor.
The completion functor. Let MeR-filt. A sequence (m^). e „, of elements of M is
said to be a Cauchy sequence if for every pSK there is an N(p) e ]N such that
m -m eF_ M for all s,t>N(p). A sequence (m.)- elN converges to meM if for every
p e ]N there exists N(p) e ]N such that m -me F M whenever s>N(p). It is well known,
s -p
cf.[ 3 ] , that M is complete if and only if FM is separated and all Cauchy sequences
converge in M. Using this completeness criterion it is straightforward to check that
any finite direct sum of complete modules is again a complete module.
Now let MeR-filt and consider the projective system :
{M/F M, it :M/F M->M/F M p>q}. Then M = lim M/F M is a filtered Z-module be-
p pq -p -q F H ^ *— -p
cause of the results in 1.2. Moreover, M is in this way a complete abelian group and
there is a natural isomorphism between M/F M and M/F M for every pez. If the filtra-
tion of R is exhaustive then M is a filtered R-module, in particular R is a complete
filtered ring. If FM is exhaustive too then M has the structure of a filtered
R-module. Thus we obtain a functor ç, called the completion functor, from the
category of filtered R-modules with exhaustive filtration to the category of complete
filtered R-modules associating M to M.
Taking the limit of the system of the canonical maps it :M-*-M/F M yields an R-linear
map q. :M-»-M and M is complete if and only if c,, is an isomorphism i.e. if Msc(M) =M-
Note that the assumption that the filtration considered are exhaustive is not really
restricting since we may always apply the exhaustion functor if this were not the
case.

5
1.3. GRADED RINGS AND MODULES.
A ring R is said to be a graded ring of type Z if there is a family of
additive subgroups {R ,nez} of R such that R= ® R and R.R. cr. . for i,j eZ. It
follows from this that 1 eR and that R is a subring of R. An MeR-mod is said to be
a graded left R-module if there is a family {M ,neZ} of additive subgroups of M with
the properties : M= ® M and R.M. cm. . for all i.jez.
™ n in i+n
neZ J J
Although for an abelian group G it is very well possible to study gradations of type
G_ we will only consider gradations of type Z in this paper.
The elements of h(R) = U R and h(M) = W M are called homogeneous elements of R
nez n nez n
and M resp. If m/o, meM., then m is called an homogeneous element of degree i, we
write : deg m = i. Any nonzero meM may be written, in a unique way, as a finite sum
m- + ._ +m, where none of the terms is zero and such that deg m. <deg m, < — <deg m^ ;
the elements m.,... ,m, in such an expression are the homogeneous components of m.
3.1. Examples.
1°. Let R be a ring, if : R-*R an injective ring homomorphism. The ring of twisted
polynomials S=R[X,¥'] is a graded ring with grading : S^ =o if i<o, S. = {aX ,aeR} if
i>o.
2°. Any ring R may be considered as a graded ring with trivial gradation : R =Rand
R. =o if i/o.
i
3°. Let k be a field, X and Y variables and consider the ring k(X,Y), consisting of
rational functions rv'vi wnere f and >P are homogeneous polynomials in X and Y. This
is a graded ring with grading induced by the degree function : deg(\\„,J.) = deg y>(X,Y)
deg 4/ (X,Y). Commutative graded rings of this type are well known, they arise as
homogeneous coordinate rings and function fields of projective curves in Algebraic
Geometry.
3.2. The Category of Graded Modules
Let R = e R- be a Z-graded ring and M a Z-graded R-module (we omit Z in what
iez
follows).
A submodule N of M is a graded submodule if N = © (NHM.) or, equivalently, if for
4 i-"7! 1

any xeN the homogeneous components of x are again in N.
Now consider graded left R-modules M and N. An R-linear f :M-»-N is said to
be a morphism of degree p if : f(M.) CN.+ for any iez. Morphisms of degree p from
an additive subgroup of Hom_(M,N), which we will denote by HOMn(M,N) and it is clear
that U HQML(M,N) =HQML(M,N) is a graded abelian group. Composition of a morphism
p ez P
f : M-*-N, of degree m, with a morphism g : N-*-P, of degree n, yields a morphism gof of
degree m+n.
If N is a graded submodule of M then M/N= ® (M.+N)/N is graded and the canonical
ieZ 1
projection M-*M/N is a graded morphism of degree o.
Now we introduce the category R-gr of graded left R-modules where the morphisms in
this category are taken to be the graded morphisms of degree o. This category
possesses direct sums, products, inductive and projective limits. One easily verifies
this statement for sums and products, furthermore, if (M , f J is some inductive sys-
a ap
tern (resp. projective system) of objects in R-gr, then the left module lim M (resp.
a
lim M ) may be equipped with the grading (lim M ) =lim (M ) , (resp. (lim M ) =
a a a a
lim(M ) ). If M,N£R-gr then Homp_ (M,N) denotes the graded morphisms of degree o
a
from M to N. If f eHonL _ (M,N), then both Kerf and Coker f are in R-gr. Indeed,
since Im f is a graded submodule of N, it follows that Coker £= N/Imf= ® (N. +Imf)/
ieZ
Im f, is the cokernel of f in R-gr. It is not hard to check that Im f and Coim f are
isomorphic in R-gr, therefore R-gr is an abelian category which satisfies Grothendieck's
axioms Ab3, Ab4, Ab3 and Ab4 , cf. [13 ]. The axiom Ab5 may equally easily be deduced.
Obviously we have a functor U : R-gr-*■ R-mod (U for ungrading) which associates to a
graded Me R-gr the corresponding ungraded R-module.
Notation. We will write U(M) = M, but we omit this for R since it will be clear from
the context whether R is being considered as a graded ring or not. One of the main
problems concerning R-gr that we will be dealing with may be phrased as follows : if
MeR-gr has property P, then it is true that M has this property too. The converse is
usually true and far more easier to deal with.
Similar to the definitions in 2.2 we define the n -suspension M(n) of MeR-gr

to be the module M together with the gradation M(n)^ =M +.. This defines a functor
T : R-gr-► R-gr which may be characterized by
1 . T„ o T = T
n m n+m
T oT =Id.
n -n
3 . UoT =U.
n
In particular T is an equivalence of categories, for all neZ. It is
straightforward to prove that the family {R(n),nez> is a family of generators for R-gr,
consequently © R(n) is a generator of R-gr and R-gr is a Grothendieck category with
neZ
enough injective objects, cf. [ 13] . An F eR-gr is said to be free if it has a basis
consisting of homogeneous elements, equivalently, if there is a family {n^,iel} of
integers such that F= © R(n.). Since any M eR-gr is isomorphic to a quotient of a
iel 1
free object of R-gr, whereas free objects are certainly projective in R-gr, it follows
that R-gr has enough projective objects.
3.3. Some Properties of Graded Modules.
An M eR-gr is left limited (resp. right limited) if there is an n ez such
that M. =o for all i<n , (resp. i>n ). If i<o yields M. =o then M is said to be
positively graded, (negatively graded is defined in the same way).
3.3.1. Lemma. Let R be a left limited graded ring and M eR-gr, then the following
statements hold :
a. If M is finitely generated then M is left limited.
b. If M is left limited then there exists a free graded module F, left limited, and
an epimorphism F->M-»-o.
c. M is finitely presented in R-gr if and only if M is a finitely presented R-module.
Proof : a. Since M is finitely generated there exists a finitely generated free
graded module F such that we have an exact sequence in R-gr : F-*-M-*-o. Since R is
left limited and F being finitely generated, it follows that F is left limited and so
is M.

8
b. Let n e Z be such that R. =o for i<n and let m e Z be such that M. =o for
o 100 i
all Km . If x/o is in h(M) then deg x>m . To an xehQf) there corresponds a
morphism of degree o, <p : R(-iO -+M, given by <p (1) =x, where n = deg x. Note that
for i<n +m we have R(-ri ) - =o. Putting <p = ® t yields an exact sequence :
00 ^ x xeh(M) x
<P
® R(-nJ—>M—»-o
xeh(M) x
where © R(-n ) is clearly a free, left limited graded module.
xeh(M) X
c. Obvious. □
3.5.2. Lemma. Suppose that M,N£R-gr and that M is finitely generated, then :
HOMp(M,N) = HomR(M,N).
Proof : Obviously, graded morphisms are R-linear. Conversely, let f :M-»-N be
R-linear and let x.,._,x be homogeneous generators for M. Express f(x. ) ,j =1...n,
J;
as 2 y.. with deg y.. <.„<deg y., . Each element y.. defines a graded morphism g.-,
g.. :M-*-N, in the following way : g.. (x, ) =o if k/j, g.-(x.) =y.. (it is easily
checked that this is we 11 -defined). Now nutting g= 2^ g.. yields f = g and there-
i=1 J1
fore feHOMR(M,N). n
3.3.3. Remark. If M is not finitely generated then it can happen that HOM„(M,N) f
Hom_(M,N). For example let R have a non-limited grading, then there exists an element
(aj,) e n RJ1 which is not in ® R . Put M=R® and define f eHon^CM.R) by
nez n eZ
putting f((xn)n) = 2 a^. Clearly, f ÉHOl^fM.R) !
The derived functors of the left exact functor HQM_(.,.) will be denoted by
EXT^(.,.), then we have
3.3.4. Corollary. If R is a graded left Noetherian ring and M is a finitely
generated graded R-module then, for any n>o : EXT?[M,N) =Ext„(M,N) for any graded R-mo-
dule N.
Proof. Since R is left Noetherian, M has a finite resolution

9
...—F2-* F1— F0— M — o
where the F. are free graded R-modules and also finitely generated. Now Lemma 3.3.2.
finishes the proof. □
Now consider a positively graded ring R and consider the graded ideal
R = ® R. of R.
+ i>o x
3.3.5. Lemma. Let R be a positively graded ring, let Me R-gr be left limited,
then RM =M if and only if M = o.
Proof. Suppose M/o and pick n eZ such that M. =o for all i<n , whereas M /o.
rr r o i o n
o
Clearly N^ nR+M = o and therefore R+M/M. n
o
3.3.6. Lemma. Let R be a graded ring, M,N,PGR-gr, and consider the following
commutative diagram of R-morphisms :
h
-*■ N
f
P'
where f has degree o. If g (resp. h) has degree o then there exists a graded mor-
phism h' (resp. g') having degree o such that f = goh' (resp. h = g' oh).
Proof. Let us prove the case where g has degree o. Take xeM and put h'(x) =h(x) .
It is not hard to see that f = goh'. □
3.3.7. Corollary. Let PeR-gr, then P is a projective object in R-gr if and only
if £ is a projective R-module.
Proof. If P is projective in R-gr then it is a direct summand of a free graded
object F but then it is also true that £ is projective in R-mod. Conversely, suppose
that £ is projective in R-mod. Since PeR-gr there exists a free graded R-module F
such that F—►P-^o is exact in R-gr. Projectivity of P_ entails that there is an
R-morphism g : P_-*-F_ such that fog = 1p. According to Lemma 3.3.6. there exists a

10
graded morphism g' : P-*-F, of degree o and such that fog' =1 . Consequently P is a
projective object in R-gr. □
3.3.8. Corollary. Let Me R-gr and N a graded submodule of M. Then N is a direct
summand of M if and only if N^ is a direct summand of M.
A graded left R-submodule I of R is called an homogeneous or graded left
ideal of R, if I is moreover two-sided then it is called an homogeneous or graded
ideal of R.
3.3.9. Lemma. Let R be a graded ring, EeR-gr. The following assertions are
equivalent :
1. E is an injective object of R-gr.
2. The functor H(M,(-,E) is exact.
3. For every homogeneous left ideal I of R, the morphism :
HCMj^iJg) iHOM^R.E) ->H0MR(I,E) deriving from the canonical inclusion i : I-*-R,
is surjective.
Proof. 1 »2. Obviously E is an injective object if and only if the functor
HomR_ (— ,E) =H0MR(—,E) is exact. Moreover E is injective if and only if E(n) is
injective for every nez.
2 =► 3. Obvious.
3 =► 1. The proof is similar to the proof of Baer's theorem in the ungraded case. □
3.3.10. Corollary. If EeR-gr is such that E_ is an injective R-module then E is
injective in R-gr.
3.3.11. Remarks.
1°. If EeR-gr is injective then E_ need not be injective. For example, let K be any
field and consider R = K[X,X ] with gradation R. ={aX1,aeK}, iez. it may be checked
that R is injective in R-gr but not in R-mod. More about this kind of objects will
be in
2°. If Le R-gr is such that L is a free left R-module then L need not be free in R-gr.

11
For example, let R be Z xZ with trivial grading. Let L be R endowed with the
following grading : L = Zx{o}, L, = {o} xZ and L. =o for i/o,1.
Obviously, L is not free in R-gr.
The projective dimension of an R-module M will be denoted by p.dim„M. If
Me R-gr, then the projective dimension of M in the category R-gr will be denoted by
gr.p.dinuM. One easily deduces from 3.3.7. that :
3.3.12. Corollary. If MeR-gr then gr.p.dinuM = p.dimpM.
For further use we state :
3.3.13. Lemma. Let MeR-gr and let N be a graded submodule of M. If N is
essential in M in the graded sense, then N_ is essential in M.
Proof. Recall that a subobject of M is said to be essential if its intersection with
all non-zero subobjects of M is nonzero. Hence, the fact that N is R-gr-essential
in M means that,for each xeh(M)-{o} there exists aeh(R) such that axeN-{o}. If
x/o is in M then we may write x=x- + .„ + x. , where the x. are the homogeneous
1 n k
components of x of degree i, . The proof goes by induction on n, establishing that
for each xeM there is an aeR such that ax£N. For n = 1, this is obvious from
graded essentiality of N in M. Define y = x. + ._ +x. . Since x. /o there is an
"""I n-1 "Si
aeh(R) such that ax. £N and a x. /o. Hence ax-ax. = ay where ay has at most n-1
n n n
homogeneous components. So if ay = o then axeN and we are done. If ay/o then by
induction, we may take some beR such that bayeN and bay/o. Thus, bax = bax. +
n
bayeN. Now bax = o yields bay = bax. =o, contradicting the choice of b, therefore
n
bax/o. □
3.3.14. Remark. Let MeR-gr and let N be a small graded submodule of M (small is
the notion dual to essential), then N need not be small in M. For example, consider
the graded ring R = K[X] over a field K. Then the homogeneous ideal (X) is small in
the graded sense but not in R-mod.

12
3.4. Tensor Products of Graded Modules.
When talking about Z-gr we assume that Z is equipped with the trivial grading
(it is the only possible grading on Z).
3.4.1. Definition. Let R be a graded ring, Megr-R and NeR-gr. Consider the
abelian group M®N_ and define its grading by putting (M®N) equal to the additive
R R n
subgroup generated by elements x®y with xeM.jeN., such that i+j =n. In this way
we obtain an object M®N of Z-gr, called the tensor-product of the graded modules M
R
and N.
Let R° be the opposite ring of R i.e. the same abelian group but with
multiplication reversed. From the above definition it is clear that we obtain a functor
® : R°-gr xR-gr-*-Z-gr. Fixing MeR-gr, then the functor ® M : R°-gr-*-Z-gr is right
R R
exact. M is said to be gr-flat if ® M is exact. Let us mention some elementary pro-
R
perties, the proof of which is similar to the corresponding proof in the ungraded
3.4.2. For any m,neZ : M(m) ®N(n) = (M®N) (m+n). If R and S are graded rings,
R R
Megr-R, NeR-gr-S, then M®N is a graded right S-module.
R
3.4.3. If Megr-R, Pegr-S, NeR-gr-S, then there is a natural isomorphism :
IBM (M ® N, P) = KOL (M, BOM (N, P) ).
3.4.4. If Pegr-R, Nes-gr, Mes-gr-R, then there exists a canonical morphism :
<P : P ®HOMs(M,N) -HOMg (H0MR(P,M) ,N),
defined by 0(p®f)(g) = (fog) (p) forpeP, feHCMs(M,N) and g eiîDMR(P,^ .
3.4.5. Proposition. An MeR-gr is gr-flat if and only if M is flat in R-mod.
Proof. If M is flat in R-mod then M is obviously gr-flat. Conversely, as in [21 ] ,
it is easy to show that M is the inductive limit in R-gr of free graded modules,
consequently M will be flat in R-mod. □

13
3.4.6. Corollary. The gr-flat dimension of M, denoted by : gr-w dinuM, is defined
as in the ungraded case, (in R-mod, the flat dimension of M is denoted by w.dim_M).
The above proposition implies : gr-w.dim^M = w.dimRM.
Let o-*-M'-*-M-*-M"-*-o be exact in R-gr. This sequence is said to be gr-pure
if for any Negr-R, o -*-N®M' -*-N®M-*-N®M" -*-o is exact in Z-gr .
1 R R R
3.4.7. Corollary. An exact sequence o-*-M' -*-M-*-M"-*-o in R-gr is gr-pure if and
only if o-*-M' -»-M-»-M" ->-o is pure in R-mod.
Proof. The idea of the proof for ungraded modules together with Theorem 2.3. of[21 ]
allow to deduce that if o-*-M' -»-M-»-M"-»-o is gr-pure, then it is an inductive limit of
a filtered set of exact splitting sequences of the form ;
o—«-M1—«-M1 ®pW —>pM—-+o ,
where each P^ ' is of finite presentation. Consequently the sequence is pure in
R-mod and so is o-*-M' ->M-»-M"-»-o. The converse is obvious. □
1.4. Filtration and Associated Gradation.
Let R be a filtered ring, MeR-filt. Consider the abelian groups :
G(R) = © F.R/F. -R, G(M) = ® F.M/F. 1M. If x£F M then x denotes the image of
ieZ 1 1-1 iez 1 1_l P p
x in G(M) =F M/F M. If aeF.R, xeF.M then define a..x. = fax). - and extend it
P P P-1 i J i J 1+J
to a Z-bilinear mapping m : G(R) xG(M) -*G(M). Taking M=R, m makes G(R) into a graded
ring and in general G(M) is thus made into a graded G(R)-module. Let f eHompR(M,N)
for some M,NeR-filt; then f induces canonical mappings f- : F.M/F._-M-»-F-N/F._1 N,
given by f.(x.) = (f(x)). for xeF.M, each ieZ. Putting G(f) = ® f. defines a mor-
111 iez 1
phism of G(R)-modules.
All of this amounts to the statement that G : R-filt ->G(R)-gr is a functor. We have
the following :
4.1. Proposition. The functor G has the following properties :
1. If MeR-filt and FM is exhaustive and separated then M = o if and only if G(M) =o

14
2. If neZ and MeR-filt, then G(T M) =T (G(M)).
3. If MeR-filt and FM is discrete then G(M] is left limited.
4. If MeR-filt and both FR and FM are exhaustive then G(M) =G(M).
5. The functor G commutes with direct sums, products and inductive limits.
Proof. Assertions 1., 2., 3., are clear. Statement 4. is well known, cf. [ 3 ]
chapter III. In order to prove statement 5, let (M ) , be an arbitrary family of
objects of R-filt, and put M= ® M . We have :
atJ
G(M) =F (M)/F .,00- e F (in/ 9 F ,01)
v Jn n^ -" n-1 *• ' =7 n^ a." _, n-1 ^ aJ
a e J a e J
= ® F M /F . M = e Gp4 )
_ 7 n a n-1 a _ , crn
a e j a e j
Hence G0Q = ® G01 )• In a similar way we may establish that G commutes with
aeJ a
direct products and inductive limits. □
4.2. Remark. If R is a graded ring and MeR-gr then we can define an exhaustive
and separated filtration on R (resp. M) by means of the subgroups F R = ® R.,
p i<p x
(resp. F M= ® M.). Let us denote the obtained filtered ring (resp. filtered
p i=Sp x
module) by R' (resp. M'). It is straightforward to prove that G(R') =R, GQ4') all.
Similarly, the subgroups F'R= ® R. (resp. F'M= ® M.) define a filtration on
P i>-p 1 P i>-p 1
R (resp. M). Denoting by R" (resp. M") the obtained filtered ring (resp. module)
then again G(R") =R and GOl'O -M-
In studying the effect of G on morphisms we need :
4.3. Definition. Let M,NeR-filt. A filtered morphism f :M->N is said to be strict
if f(F M) = Imf nF N for each pez.
A sequence L—►M-^N in R-filt is strict exact if it is an exact sequence in R-mod
such that both f and g are strict morphisms in R-filt.
4.4. Theorem. Let R be a filtered ring and consider the following o-sequence in
R-filt :
f e
(*) L—►M-^N ,

15
as well as the sequence G(*) in G(R)-gr
G(£) G(g)
G(L) ► GfM) >G(N).
Then :
1. If (*) is strict exact then G(*) is exact.
2. If G(*) is exact and FM is exhaustive then g is strict.
3. If G(*) is exact while FL is complete and FM is separated then f is strict.
4. If G(*) is exact and FM is discrete then f is strict.
5. If FK is complete, FM is exhaustive and separated, or if FM is exhaustive and
discrete then (*) is strict exact if and only if G(*) is exact.
Proof. 1. Clearly G(g) oG(f) =o. Look at an x£F M which is such that G(g)x =o.
This says (g(x)) =o and therefore g(x) £F 1N. Now since g is strict there must
exist an x' eF _.M such that g(x) =g(x') i.e. x-x' =f(y) with yeF L, but then
G(f)(y ) =xp i.e. Im G(f) =Ker G(g).
2. Let yeF N n Im g and y^F ,N. There is an x£M such that g(x) =y and since FM
is exhaustive we may assume that xeF M for some s>o. If s = o then we are done.
If s>o then G(g) (x ) =o and the fact that G(*) is exact implies that x =
p s p s
G(f)(zp+s) for some z£Fp+sL. Thus x-f(z) eF ^M and y = g(x) =g(x-f(z)) =g(x')
with x' eF _.M. Repeating this procedure we finally get that there is an ueF M
p s- i p
such that y = g(u).
3. Take yeFMnim f. Exactness of G(*) yields : G(g) (y ) =g(y ) =o, therefore
y =G(f) (xl-p-)) for some x^PJ ep l. Hence y-ffx^) e Imf nF .M. By induction we
7p v •" v p ' p / v. -i p_i /
find a sequence x^ ,x^p"1^ ,.„ ,x^p_s^ with x^p_s^ eF _ L such that :
y-ffx^) - .„ -ffx(p-s)) eimf nF .M.
' ' ^ ' p-s-1
Completeness of FL allows to define x = £°° x'-p-s-' eF L. Hence
s=o p
y-f(x) =y-lim f(x^ +xl-p" > + ._ +x'-p_s-)) =o, the latter since FM is separated,
o -*■«>
and it follows from this that yef(F L). That f(F L) CF Mnimf is obvious.

16
4. Along the lines of 3.
5. Strict exactness of (*) implies exactness of G(*) because of 1. Conversely,
suppose that G(*) is exact and let yeM, y/o be such that g(y) =o. Since FM is
exhaustive we have y£F M and y^F . for some peZ. We obtain thus : G(g)(y ) =o
and so : y = G(f) (x^) =£{x^) for some x^ eF L. Consequently y-f(x^) eF _1M.
By induction we obtain x^,.„,x^ s-^ with x^ ^ eF L and such that :
' ' ' p-s
y-f(xM+..+x(p"s:))eF -M .
' l ' p-s-1
If FM is discrete then for some index s, F ,M = o ; hence y = f(x^ + ... + x'-P ')
' p-s-1 ' '
and thus (*) is exact. If FM is complete then we may take x= 2°° x^ ' and get
s=o
y = f(x). The fact that f and g are strict morphisms will follow from 2. and 3..
4.5. Corollary. Let f : M-»-N be a morphism in R-filt and suppose that FM and FN
are separated and exhaustive, while FM is also complete. If G(f) is bijective then
f is bijective too.
4.6. Remark. The restrictions on the filtrations i.e. being exhaustive and
complete may be keyed down a lot in most situations because the functor G commutes with
the exhaustion and the completion functor so that we have a commutative diagram of
functors : (G(R) = G'(R-) = G(R'))
G(R)-gr
1.5. Free, Projective and Finitely Generated Objects of R-filt.
Let R be a filtered ring, then Le R-filt is filt-free if it is free in R-mod
and has a basis (x^). . consisting of elements with the property that there exists a
family (A:). ej of integers such that :

17
F L= £ F R .x. = $ F R.x. .
P iej P"ni x iej P"ni
Note that for any iej x. eF L and x. £F .L. We say that (x.,n-)- tT is a filt-
' 'in- l n--1 ' l i' riEJ
basis for L. Next lemma follows easily from this definition and Theorem 4.4.
0-
5.1. Lemma. Let R be a filtered ring, LeR-filt, then :
1°. L is filt-free with filt-basis (x.,n.)- ,-, if and only if L= ® R(-n-'
— i' riej ' iej '
2°. If L is filt-free with filt-basis (x.,n.). T then G(L) is a free graded module
in G(R)-gr with homogeneous basis {(x.) ,iej}.
i
3°. If G(L) is a free object in G(R)-gr with homogeneous basis {(x.) ,iej} and if
i
FL is discrete, then L is filt-free in R-filt with filt-basis (x. ,n.)- ,- ■,•
4°. If MeG(R)-gr is free graded then there exists a filt-free LeR-filt such that
G(L) sM.
5°. Let LeR-filt be filt-free with filt-basis (x.,n-)- ,, let Me R-filt. If
l' l;lEj'
f : {x.,ieJ}-*-M is a function such that f(x.) eF M, then there is a unique filtered
l i s+n. '
i
morphism of degree s, g : L-»-M which extends f.
£_. Let Me R-filt and suppose that L is filt-free. If g : G(L) -*-G(M] is a graded
morphism of degree s then there is a filtered morphism f : L -+M of degree s such that
G(f) -g.
7°. Let LeR-filt be filt-free, then FL is exhaustive (separated) if and only if FR
is exhaustive (separated).
If FR is discrete and {nez, iej} is bounded below then FL is discrete. If J is
i
finite and FR is complete then FL is complete.
8°. Let Me R-filt be such that FM is exhaustive. Then there is a free resolution
of M in R-filt :
f, f, f
(*) ..._l2_L,Li_Ulo—£Um—o
where every L- is filt-free and every f. is a strict morphism. Moreover, if FR and
FM are discrete then we may assume that every FL-, j >o, is also discrete.
9°. If FR is exhaustive and complete and if G(R) is left Noetherian and G(M) being

18
finitely generated as a left graded R-module then we may select the L. in the
resolution (*) to be finitely generated too.
Note that an MeR-filt is said to be finitely generated or filt-f.g., if
there is a finite family (x.,n.)- ,_ T with x. eM and n. £Z, such that FM= ï F R.x.
; i' riej i i ' P i=l Pi
5.2. Remarks. 1°. If LeR-filt is filt-free as well as finitely generated in
R-mod then R is filt-f.g.
2°. If MeR-filt is such that FM is exhaustive then M is filt-f.g. if and only if
there exists a filt-free L which is finitely generated, and a strict epimorphism
it : L-*-M.
5.3. Proposition. Let R be a filtered complete ring and let MeR-filt be such
that FM is separated and exhaustive. Then M is filt-f.g. if and only if G(M) is
finitely generated as a graded G(R)-module. Furthermore, if G(M) may be generated
by n homogeneous elements, then M may be generated as an R-module by less than n
generators.
Proof. From the remark in 5.2.2° it follows that, if M is filt-f.g., then G(M) is
finitely generated. Conversely let G(M) be generated by homogeneous generators xj- ',
Pi
1 <i<n.
Consider L eR-filt, L filt-free with filt-basis (y. ,p.).. Define f : L->M by
f(y.) =x'-1-'. Since G(f) : G-(L) -* G (M) is an epimorphism, Theorem 4.4.5 yields that f
is a strict epimorphism and then Remark 5.2.2° entails that M is filt-f.g. and
generated by x1-1-1 as an R-module. □
Proposition 5.3 has some consequences for the determination of the Krull
dimension of certain modules. In [34] Gabriel and Rentschler defined the Krull
dimension of ordered sets for finite ordinal numbers, in [16 ] this definition has
been extended by G. Krause for some other ordinal numbers. Let (E,<) be an ordered
set. If a,beE we put : [a,b] ={xeE, a«x<b}, r(E) = {(a,b) ,a<b}. By transfinite
recurrence, we define on r(E) the following filtration : r (E) = {(a,b), a=b},
r0(E) ={(a,b) er(E), [a,b] is Artinian}, supposing r (E) has been defined for all

19
g<a, put :
r (E)={(a,b) er(E),Vb>b1 >._»b >._>a there is anneH such that [b-. ! ,b.]6ro(E)
a I n l+1 1 p
for all i>n}.
We obtain an ascending chain : r_. (E) cr. (E) c ._ cr (E) c ._ . There exists an
ordinal I such that r (E) =r.+. (E) = ._ . In case there is an ordinal a such that
r(E) »r (E), E is said to have Krull dimension. The smallest ordinal with the
property that r (E) =r(E) will be called the Krull dimension of E and we denote it by
K-dim E.
5.4. Lemma. Let E,F be two ordered sets and let f : E-»-F be a strictly increasing
mapping. If F has Krull dimension then E has Krull dimension and K dim E<K dim F.
(cf. [34])
5.5. Lemma. Let E,F be ordered sets with Krull dimension, then ExF has Krull
dimension and K dim E xF = sup(K.dim E,K dim F). (cf. Note that ExF has the product
ordering.). If A is an arbitrary abelian category, MeA, then we consider the set E
of all subobjects of M ordered by inclusion. If E has Krull dimension then M is said
to have Krull dimension and we denote it by : K.dim.M or simply K.dim M if no
confusion is possible. Let MeA have K.dim M = a ; then M is said to be ct-critical if
K.dim M/M' <a for every nonzero subobject M' of M. For example M is o-critical if
and only if M is a simple object in the category A. Also it is obvious that if MeA
is a-critical then any nonzero subobject of M is a-critical.
If A is R-mod, Me R-mod, and if M has Krull dimension in the above sense then we
denote it by K-diiOl. The Krull dimension of Re R-mod is called the left dimension
of R; it is denoted by K-dim R.
Returning to the consequences of Proposition 5.3 we have :
5.6. Corollary. Let MeR-fiit, such that FM is separated and exhaustive and
suppose that FR is complete. If G(M) is left Noetherian as an object of G(R)-gr then
M is left Noetherian too. Moreover, we have then that : K-dimJKK-dimr(-R-.G(M).
If K-dimJ^l = K-diny,,jv.G(M) =a and G(M) is a-critical, then M is an a-critical R-module.

20
Proof. Let N be a submodule of M equipped with the induced filtration F N = F M^N.
Since G(N) CG(M), G(N) is finitely generated. Proposition 5.3. entails that N is
finitely generated, so M is left Noetherian. Consider submodules N and P of M, NCP,
both equipped with the induced filtration, and assume that G(N) = G(P). If
{x1-1^, 1 <i<n} is a set of homogeneous generators for G(N), where x^ ' eN for all i,
1 <i<n, then, again from Proposition 5.3., we infer that {x^ ' ,1 <i<n} generates N
as well as P. Thus N = P. Lemma 5.4 applies and yields K-dim_M<K-diiiy, ,R,G(M). In
order to establish our last claim, let N/o, NCM and filter M/N by putting :
F (M/N) = (N+F M)/N. Clearly, we obtain an exact sequence in G(R)-gr
o >G(N) >G(M) 'G(M/N) >o .
Since G(N) fo, the Krull dimension of G(M/N) is less than a, hence :
K-dimpM/N^K-dùig^GOVN) <a .
This shows exactly that M is an a-critical module. □
5.7. Corollary. Let R be a complete filtered ring such that G(R) is a left
Noetherian ring, then :
1°. R is left Noetherian and K-dim R<K-dim G(R)
2°. F R is left Noetherian and K-dim F R< K-dim G(R) +1.
o o
Proof. 1. follows from the foregoing corollary.
2. F R is a subring of R and it is clearly complete; Corollary 5.6. together with
Proposition 3.2 and Corollary 4.12 of Chapter II will finish the proof. □
5.8. Proposition. Let MeR-filt with FM being exhaustive. Assume that submodules
of M are closed, i.e. if NCM then N= n (N + F M). Consider submodules NcpcM,
peZ p
then :
1. If G(N) =G(P) then N = P. In particular if the Krull dimension of G(M) is well
defined then the Krull dimension of M is defined and K-dimRM<K-dimr ,R-.G(M).
2. If K-dinuM = K-dimf,(.R..G(M) =a and if G(M) is a-critical then M is an a-critical

21
module.
3. If G (M) may be generated by n homogeneous generators then M may be generated by
n generators.
Proof. 1. Take x^P, then xef.M for some i, x£F. -M. Because :
xieG(P)i=G(N)i= (PnF.M) + F-^M/F^.M = (NnFiQ + F._1M/Fi_1M, there is a
y. eNnF.M such that x-y. e F..M. Hence x£y +PnF. JL Repeating this procès we
end up with elements y..,y7,...,y EN such that x-(y. +y? + ._ +y ) is in F._ Mnp, s>o.
Thus xEN + F. M for s>o. It follows that xe C) (N + F M) =N, whence N = P follows.
1_s p£Z p
2. Proceed as in the proof of Corollary 5.4.
3. Let {x^1 ,1 <i<n} be a family of homogeneous generators for G CM). Write M' for
the submodule of M generated by the elements x'-1 ,1 <i<n. Obviously, x*-1J eG(M') =
^i Pi
CM' nF M)/(M' nF ^M). Therefore G CM1) =GQ1) and by 1. this yields M' =M.
Pi Pi
5.9. Remark. The above proposition is generally applied in case FM is exhaustive
and discrete.
Let R be a ring, I an ideal of R, MeR-mod. Using the I-adic filtration, as
introduced in 1.2.2°, we get the I-adic completion M of M by M = lim M/rVl. Let
i :M->M be the canonical filtered morphism. In studying I-adic completion the
following property is very relevant. I is said to satisfy the left Artin-Rees property
if for any finitely generated left R-module M, any submodule N of M and any natural
number n, there exists an integer h(n) >o such that I ™MnNCInN. In other words,
I satisfies the left Artin-Rees property if the I-adic topology of N coincides with
the topology induced in N by the I-adic topology of M.
5.10. Proposition. Let R be a left Noetherian ring and I an ideal satisfying the
left Artin-Rees property.
1. The functor carrying M to M is exact in the category of left R-modules of finite
type.
2. R is a right flat R-module.
3. If MER-mod is finitely generated then M=R®M.
R

22
4. If MeR-mod is finitely generated then Ker i= n InM={xeM, (1-r)x = o for
n>1
some r e I}.
In particular if I is contained in the Jacobson radical of R then n inM=o.
, , , n>1
5. If MeR-mod is finitely generated then M = R.i(M) and M is a finitely generated
R-module.
Proof, cf. [ 3 ] , chapter 3, Theorem 3 p.68; Proposition 5 p.65. □
5.11. Corollary. If R is a left Noetherian ring, I an ideal of R satisfying the
left Artin-Rees property, then the following statements are equivalent :
1. I is contained in the Jacobson radical of R.
2. If MeR-mod is finitely generated then submodules of M are closed in the I-adic
topology.
An ideal I of R is said to be generated by a central system if I =Rx- + ... +R
n
where x. eZ(R) and where the image of x. in R/(x.,... .x.,) is in the center of the
latter ring.
5.12. Proposition. Let R be a left Noetherian ring, I an ideal generated by a
central system, then I satisfies the left Artin-Rees property.
Proof- Let N be a submodule of a finitely generated MeR-mod. Let sen and
consider a submodule M' of M maximal with the property that M' nN = IsN. This choice
makes M/M1 into an essential extension of N/ISN, where Is (N/ISN) =o. If we are
able to establish that under these circumstances M/M' may be annihilated by some I
then I1 (M/M1) =o yields 1*1.1 on cisN. Note also that it is sufficient to do this for
s-1
s = 1 because for arbitrary s it will follow from this case applied to IN,...I N
instead of N and combination of the inclusions thus obtained. If I = Re. + .„ + Re ,
where (c.,._,c ) is a central system, let f :M-»-M be the map given by m-^c-m for all
meM. Since M is left Noetherian, there exists r e flj such that Ker vT = Ker m for
all ke]N i.e. Im p nKer m =o. However N is contained in Ker n and N is an
essential submodule of M, therefore Im m -o. Let t be the smallest natural number such
that m =o, then n gives rise to an inject ive R-morphism M/Ker m ->Ker m .To prove

23
our original claim it will therefore be sufficient to prove that both Ker m and
Ker it can be annihilated by large powers of I. Now c.j being central, Ker m
is an R/Rc.-module and we apply induction on n to deduce that 1^ Ker m =o for some
peH. Since there is an injective morphism : Ker m /Ker m -*Ker m for 1 <m<t,
the fact that I4 Ker
t-1
:o for some q e Is( will follow by induction on t. □
5.13. Example. If ^ is a nilpotent Lie algebra, finite dimensional over a field
of characteristic zero, then any ideal of the enveloping algebra U(^) is generated
by a central system.
PeR-filt is said to be filt-projective if, the diagram in R-filt where f is
a strict morphism
P
m" fhf--
may be completed by a filtered morphism h such that the diagram is commutative o
5.14. Proposition. Let R be a filtered ring, PeR-filt such that FP is exhaustive,
then the following statements are true :
1. P is filt-projective if and only if P is a direct summand in R-filt of a filt-free
object.
2. If P is filt-projective, then G(P) is projective in G(R)-gr.
3. If FR is complete and if P is finitely generated filt-projective then FP is
complete.
Proof. 1. Let L be filt-free with filt-basis (x'-1-',n.). £,. Consider the
following diagram in R-filt :
L

24
where f is a strict epimorphism. Since x1-1-' eF L, it follows that gfx1-1-1) eF M" =
f(F M). Hence there exists y^ eF M such that gCx^1-1) = f(yM). Lemma 5.1.5°
ni ni m
yields the existence of a filtered morphism of degree o, h : L->M such that h(xl J) =
y . Clearly g=f oh, hence L is fi It-projective. If P is any filt-projective ob-
f
ject then there exists a filt-free object L and a strict epimorphism L -*-P-*-o (see
Lemma 5.1.8°). Projectivity of P implies the existence of a morphism g' : P-*-L of
degree o, with fog' =1 . We have g'(F-P) cF.Lnim g' while conversely xeF.Lnim g'
implies x.g'(y)eF.L i.e. f(x) =fog'(y) =y£F.P. Therefore xeg'(F.P) or g' (F.P) =
F.Lnim g' follows and the latter means that g' is strict. Note first that in R-mod
we have : L = Im g' ®Ker f. Now we claim that :
F.L =F. Img' ®F. Ker f = (Im g' nF.L) ® (Ker fnF.L).
Indeed, if xeF.L then x=y + z with y elm g' and zeKer f. Hence f(y) =f(x) eF.P and,
while y elm g', we get that g' o f (y) =y or y eg' (F.P) = F.L n Im g'. On the other hand,
z =x-yeF.LnKer f and thus L - Im g®Ker f in R-filt. Finally since g' is a strict
monomorphism and since we have Im g' =P in R-filt it follows that P is a direct sum-
mand of a filt-free object.
The converse implication is easy.
2. Directly from 1 and the properties of the functor G.
3. The construction of L in 1 shows that, in case P is finitely generated in R-filt,
we may choose L to be finitely generated too and we have that L =P©Q for some QeR-
filt. Since R is complete, L is complete and so a Cauchy-sequence (x ) e ~. in P
converges to some xeL. Write x=p + q with pep, qeQ. Since FP is exhaustive,
peF.P for some k ez and it is clear that x^"P will be in F.L for t large enough i.e.
q = lim (x -p) =o and xep or P is complete follows. □
n
1.6. The Functor HCMp (-,-)•
The properties of HCMp mentioned in 1.2 make it into a functor R-filt x
R-filt ->Z-filt; and it is clear that, by definition of the filtration in HOM^(-,-),

25
F HCMpCM.N) is exhaustive for any M,NeR-filt. First let us claim the analogies of
Lemma 3.3.2 and Remark 3.3.3.
6.1. Lemma. Let R be a filtered ring and let M,N£R-filt be such that M is a
finitely generated object while FN is exhaustive, then : HONL(M,N) =Hom_(M,N).
Proof. Let {x^ ,n.} be a system of generators for M and take f eHom_(M,N). There
exists sel such that f(x^) eF N for any 1 <i<n. It easily derives from this
i
that f eF HOL(M,N) and hence f eHOL(M,N). Note that, if FN is not exhaustive and
if N' is the exhaustion of N then HOL(M,N) =HCMR(M,N') =HomR[M,N').
6.2. Remark. In general HCMpCM.N) /Hom_(M,N). For example let R be a discretely
and exhaustively filtered ring such that R = F.R for all i when F. (R) /o. Take M to
be filt-free with filt-basis {x^.o} with 1 <i<oo, and take N = R. Define f :M->N
by ffx'-1-') =a'-1-) where a^ £F.R. If the degree of f is p, let i eZ be such that
F.N = o for any i<i . Consider i<i -p, hence f(x^ ') =a ' =o, contradiction.
6.3. Proposition. Let M,NeR-filt, then :
1. If FM is exhaustive and FN is separated, then FHCMpfMjN) is separated.
2. If M is a finitely generated R-module, if FM is exhaustive and FN is discrete
then FHOM^.N) is discrete.
3. If FM is exhaustive and FN is complete then FHOMR(M,N) is complete.
Proof. 1. Let f ef>F HOMn[M,N). Then f(F M) cH F N = o. Therefore f(M) =
p p K P s P+s
fP(UFM) =o.
P P
2. Let x1- \.„ ,x^ ' be generators for M. Since FM is exhaustive, there is an r ez
such that x^ eF M for all ie{1,..,n}. Since F.N = o for all i<n for some
o ,.,
n eZ, it follows that for any f eF.HOMR(M,N) where i<n -r we have f(x^ ') =o,
hence F^OM^.N) =o if i<n -r .
3. Let q<pez and consider the projective system :
HOM^M.N)
HCMR(M,N)/F_ BOMj^CM.N) HOMR(M,N)/F_ HOM^M.N) .

26
We claim that HCM(M,N) = lim HOL(M/N)/F_ HDNL(M,N) =X. Indeed, let (f^p\peZ) be
an element of X such that Pir (f(p)) =f^ for p>q. Let fM =<p (gM), with
g^ eHOIv^CM.N). If q<p then g^ -g^ GF_ HCM^(M,N) .and therefore
(g -g )(F M) CF=_ N f°r any seZ. Since FM is exhaustive, for any x£M there
exists some teZ such that xeF M and thus g P^(x) - g (x) eFt_ N, consequently the
sequence {g^(x),peZ} is a Cauchy sequence in N. Completeness of N entails that
the mapping f :M->N defined by f(x) = lim g P^x, is well defined. If xeF M,
knowing that g ™ (x) eg^(x) +F N for p>q, we may conclude that f(x) eg^(x) +
F N. Therefore (f-g(q))(FM] cf N or f-g^ eF_ HO^CM.N). Hence f eHO^CM.N)
and <p (f) =f^ for the selected q. n
If M,Ne R-filt, we introduce a natural map t = y(M,N),
* : G (HOI^CM.N) ) ► HOMç [R) (G (M) ,G(N) ) ,
as follows : for f eFpHCM^i.N), xeF M put * (f ) (xq) =f(x)p+q.
6.4. Lemma. y(M,N) is a monomorphism. Moreover, y(M,N) is an isomorphism if M
is filt-projective.
Proof. Fixing M,Ne R-filt we write <p instead of y(M,N). If <f (f ) =o, then
ffx) ^ =o for every xeF M, qeZ. It follows that f(F Ml cf .N for any q ez i.e.
p+q ' q ' n q P+q-l
feF HOL(M,N]. Consequently f =o. In order to prove the second statement, first
assume that M is filt-free and let (x^1 ,p.)- ^, be a filt-basis for M. In this case
'*! i ej
{x^.ieJ} is a homogeneous basis for GfM). Letting ge H0M„,m (Gpl) ,G(N)) , we ob-
*i . . ..'■-' P
tain g(xi^) =y^-. Define f :M->N by putting ffx^1-') =y , with this definition
^i P Pi
it is clear that feF H0M„(M,N) and also that <p(£ ) =g, proving surjectivity of if.
If M is filt-projective then there is a filt-free L such that L =M®M' in R-filt. We
know that HOMp commutes with finite direct sums, therefore we may use the previous
part of the proof. □
6.5. Definition. Let FR be an exhaustive filtration of R. A filt-injective
obj ect is an obj ect MeR-filt such that, if I is a left ideal of R equipped with the

27
induced filtration then any filtered morphism f : I ->M extends to a filtered morphism
R->M, equivalently, the canonical mapping HCM(i,1 ,) : HO>L(R,M)->HCMR(I,M) is an epi-
morphism.
6.6. Remarks. 1. Let FR be exhaustive and complete and suppose that G(R) is left
Noetherian. If M is R-filt injective then it is infective in R-mod, because of Lemma
6.1.
2. Let FR be exhaustive. If Qe R-filt is injective in R-mod then it is injective in
R-filt.
6.7. Theorem. Let FR be exhaustive and let Qe R-filt be complete. If G(Q] is
injective in G(R)-gr then Q is filt-injective.
Proof. Equip I, left ideal of R, with the filtration F I =F Rni. Since i : I-»R
is a strict morphism we obtain the following commutative diagram :
G HCM^I.Iq)
GCHOM^R.Q)) ► GCHC^CI.Q))
*>(R,Q)
HCMG[R)(G(i),G(1Q))
*(I,Q)
HOMG[R)(G(R),G(Q]) ► HOMG[R)(G(I),G(Q)) .
Since G(i) is a monomorphism, injectivity of G(Q) entails that HOM,,™ (G(i) ,G(I„)) is
an epimorphism. Hence y>(I,Q) is an epimorphism too, therefore an isomorphism. Thus
GH0Mn(i,1 J is an epimorphism, while F H»L(R,Q) and F HQMj^I.Q) are exhaustive and
complete, therefore by Theorem 4.4.3., H0t>L(i,1 ) is epimorphic. □
1.7. Projective Modules and Homological Dimension of Rings.
We will say that xeR is topologically nilpotent if the sequence {xn,ne]N}
is a Cauchy sequence converging to o. Recall the following lemma about topological
rings :
7.1. Lemma. Let R be a complete topological ring and consider a fundamental set

28
of neighborhoods of zero, consisting of additive subgroups. If t :R-»-S is a ring
homomorphism such that every xeKery is topological!/ nilpotent, then every idem-
potent element of Im <t may be lifted to R.
2
Proof. Let e e Inv be idempotent, e=y>(x) for some xeR. Since $ (x -x) = o,
2 2
a = x -xeKer t ■ Put b = x + X(1-2x) with XeR, and determine X such that b =b while
2
X commutes with a. We get : (X -X)(1 + 4a) +a = o.
Put : (*) X =1(1 - (1 + 4a)Y) =1 2 (-1)k_1 ck ak, where ck is the coefficient
2 2 Kk<» 2k ' 2k
k k 2k 1 k
of 1 (-1) in the binomial expansion of (1-1) =o i.e. j c?, is an integer.
Therefore X may be viewed as a power series with integer coefficients which is easily seer
to converge. Hence (*) determines X and it commutes with a and consequently b is
idempotent. Furthermore, aeKer <p yields X s Ker y and thus v> (b) =<f (x) =e. □
7.2. Lemma. Let R be a filtered ring, Me R-filt such that FM is exhaustive and
2
complete. Suppose that f :M-*-M is a filtered morphism such that G(f) =G(f). Then
2
there exists a strict morphism g :M-*-M in R-filt satisfying : g -g and G(g) =G(f).
The proof follows from the foregoing plus the fact that every idempotent f in
HOMR(M,M) is a strict morphism. Indeed, let f(x) eF Mnlm f, then f(f(x)) =f(x)
yields f(x) ef(F M) hence f is strict. □
7.3. Remark. If Me R-filt is such that FM is exhaustive and complete then
F_1HOMR(M,M) CJOCM^.M)) (J =Jacobson radical). Indeed if f 6 F_..H»L(M,M) then f
is topologically nilpotent and so 1-f has an inverse 2 f .
o <n <°°
7.4. Corollary. If M is as in the foregoing lemma then any countable set of
orthogonal idempotent elements in Im v can be lifted to Horn- f., (M,M).
7.5. Theorem. Let R be a filtered ring such that FR is exhaustive. Let P be
° g
projective in G(R)-gr and suppose that, either FR is discrete and FP is left-limited
or P is finitely generated while FR is complete. Then there is a fiIt-projective
module PeR-filt such that G(P) -P . If Me R-filt, then for any morphism g : P ->-G(M)
of degree p there is a filtered morphism f : P-*-M of degree p such that g = G(f).

29
Proof. The hypotheses imply that there is a free object L in G(R)-gr and a mor-
phism h : L ->L of degree zero, such that h =h and Imh =P . If P is finitely
generated then L may be chosen to be finitely generated too. Assume that L =G(L),
where LeR-filt is filt-free. If P has left limited grading then L may be chosen
so that it has left-limited grading too and we may suppose FL to be discrete since
FR is discrete. In the case where P is finitely generated and FR is complete, we
obtain that FL is complete. By Lemma 7.2, there exists a strict filtered morphism
f : L->L with f = f and G(f) =h. Put P = Im f. Since f is strict it follows that
G(P) =P and that P is filt-projective as required. Moreover if P is finitely
generated then FP is complete. To prove the remaining statement we may write g : G(P) -*
G(M) and up to taking a suitable suspension we may assume deg g = o. We have L = P©Q
and hence there is a graded morphism h : G(L) -*-G(M) such that h|G(P) =g. However
h = G(k) for some filtered morphism k :L-»-M. Putting f =h|P we arrive at g=G(f). □
7.6. Corollary. Let R be a filtered ring such that FR is exhaustive and discrete
and let MeR-filt be such that FM too is exhaustive and discrete. Then we have :
p.dim M<p.dimGf-R. G(M) •
Proof. By Lemma 5.1.8°, there exists a free resolution for M in R-filt
f 1 fi f
._—*-L , —S^-^L -—+...—-L, '-+L °-*M—>o
n-1 n-2 1 o
where FL. are discrete for all j. Write Q for Ker f _.., where n =p.dimr(-R-.G(M) > then
G(Q) is a projective G(R)-module with left-limited grading. By the theorem we may
select a projective object QeR-filt such that G(Q) =G(Q'), let g : G(Q') ->G(Q)
denote this isomorphism. There exists a filtered morphism f : Q' -*-Q such that G(f)=g.
Now we end up with the situation where Q' is filt-projective, FR is discrete (note
that FQ' turns out to be discrete just as well) while G(f) is an isomorphism,
therefore we may apply Corollary 4.5 and deduce that f is an isomorphism, whence it follows
that Q is filt-projective and therefore projective in R-mod. □

30
7.7. Corollary. Let R be a filtered ring with exhaustive and discrete filtration,
then gl.dim R<gr.gl.dim G(R).
Proof. Equip a left R-module M with the filtration F,M = F.R.M, then FM is
exhaustive and discrete. By the foregoing corollary we have : p.dim M<p.dim G(M) <gr.gl.
dim G(R). □
7.8. Corollary. Let R be a graded ring, then : gr.gl.dim R<gl.dim R. If the
gradation of R is left limited then : gr.gl.dim R = gl.dim R.
Proof. The second statement follows from the foregoing corollary applied to the
ring R filtered with the associated filtration. The first statement follows directly
from Corollary 3.3.12. □
7.9. Corollary. Let R be a filtered ring such that FR is exhaustive and complete,
let MeR-filt be such that FM is exhaustive. Suppose that G(R) and G(M) are left
Noetherian, then: p.dimRM<p.dimr ,R,G(M) •
Proof. Apply Lemma 5.1 and Theorem 7.5 and use the argumentation of the proof of
Corollary 7.6. □
7.10. Corollary. Let R be a filtered ring such that FR is exhaustive and complete.
Suppose that G(R) is left Noetherian. We have : gl.dim R<gr.gl.dim G(R).
Proof. Let M be a finitely generated left R-module and take Xi,.-,x to be a set of
generators. Filter M with the filtration : F.M= 2 F.R.x, . In doing this we obtain
1 k=1 x K
a finitely generated filtered module with an exhaustive filtration. Corollary 7.8
finishes the proof. □
1.8. Weak (flat) Dimension of Filtered Modules.
Let us start this section with a basic lemma on tensor products of filtered
modules. If MeR-filt, Nefilt-R then the Z-module N8*I has filtration : F (N® M)
is the Z-submodule of N® M generated by all elements of type n®m where neF N,

31
me F M and p = s+t. This construction yields a functor
® : filt-R xR-filt^Z-filt
R
8.1. Lenma. With notations as above, the following statements are true :
1°. If EM and FN are exhaustive then so is F(N®M).
R
2°. If FM and FN are discrete then so is F(N®M).
R
3°. For all m,nsz : N(n) ®M(m) = (N®M) (m+n).
R R
4°. In the category R-filt we have the following isomorphism R8M2M; similarly
R
N® RsN in filt-R.
R
5°. The functor ® commutes with direct sums and inductive limits.
R
6°. For Me filt-R, NeR-filt-S, Pefilt-S, we have the following isomorphism in
Z-filt : HOMs(M® N,P) aHO^Ql.HOMgOl.P)).
Proof_. Statements 1°,2°,3° are clear.
4°. Consider the R-module morphism, <p : N ® R N, y(x®r) =xr. Its inverse is given
R
by, ■// :N-*-N® R, <P(x) = x®1. Now if x®reF (N® R) with xeF.M, reF.R, i+j =p,
R PR 1 ' j » J r >
then y(x®r) =xr6F.N.F.RCF N. In a similar way one establishes that <l> is a mor-
1 j p '
phism in filt-R.
5°. Let N= ® N , M= « M , then « N 8LM = N„®M. Indeed, let
«eA a ges 3 (a,g) eAxB a K K
t :N®DM —► « (N ® MJ
R (a,g)eAxB a R 3
be the R-module morphism which is given by :
*(.(. z x )®( 2 y )) = 2 x ®y .
aeA a geB p (a.g)eAxB a p
Then it is well known that ^ is an isomorphism of abelian groups. Let x®yeF (N® M)
P R
where xeF.N, yeF.N and i+j =p. If x= 2 x , y = 2 y„ then x eF-N and
1 ^ «eA a geB 3 a 1 a
y.eF.M . It follows immediately that x ®y £F (N 8lM.1, hence
gjg ' a ' g pv a R gJ
¥>(x®y) eF ( O N ®M ). In a similar way one establishes that <t is actu-
p (a,g) eAxB a S
ally a morphism in Z-filt. The assertion for inductive limits may be proved in an

32
equally straightforward way.
6°. Define
<P :HOMs(M® N,P) «-HO^CM.HOIgCN.P))
by : (y> (f) (x) ) (y) = f (x ®y), and define
* rHOIv^CM.HaigCN.P))—»HOMs(M» N,P)
by : <Hg)(x®y) =(g(x))(y). If feF HOM„(M® N,P), xeF M, y£F N then it follows
U o T? X. u
from x®yeF^ M® N that f(x®y) eF ^ P. Hence <f (f) (x) eF it H»L(N,P) and thus
/ ^+r ^ ^ ;> p+t+r *• JK p+t S
V (f) eF HCMjjOl.HOMS(N,P)). This expresses that «> is a morphism in Z-filt. Along
the same lines one verifies that <l> is a morphism of degree o and ^ is the inverse of
ip. □
Let NeR-filt, Mefilt-R. Define a graded morphism
¥> =¥>(N,M) :G(M) ® G(N)—*G(M®N),
G(R)
by putting : t (m ®n ) = (m®n) . Obviously this is well defined and it is clear
that $ is surjective
8.2. Lemma. If either M or N is a free object in filt-R or R-filt respectively
then t (N,M) is an isomorphism.
Proof. Knowing that ® and G commute with direct sums, the proof may be reduced to
R
the case where M = R(n)R. But then Lemma 8.1 allows us to reduce the proof further to
the case M = R_. In this case we have the following commutative diagram of graded mor
phisms, vertical arrows representing isomorphisms.:
G(RJ ® G(N) > G(R® N)
R G(R) R
I G(N) I
G(N) y G(N)

33
So, y(M,R) is also an isomorphism. □
Lemma.
Let R be a filtered ring and let MeR-filt be such that FR and FM
are discrete and exhaustive. If G(M) is a flat object in G(R)-gr then M is a flat
R-module.
Proof. If J is a right ideal of R, equip it with the filtration F J=F Rnj. Let
i : J-»-R be the canonical inclusion morphism; note that i is a strict morphism.
Application of Theorem 4.4 yields that G(i) is infective, and so we obtain the following
commutative diagram :
G (J) ® G(M)
G(R)
<P (M,J)
G(J»RM)
G(i) ®1
G[M)
-* G(R) ® G(M)
G(R)
¥>(M,R)
G(i®V
-»• g(r® m)
R
The fact that y(M,R) is an isomorphism, whereas G(i) ®1p,,„ is a monomorphism, entails
that t (M,J) is a monomorphism, hence an isomorphism. Then G(i®1 ) has to be monomor-
phic and repeated application of Theorem 4.4. allows to derive from this that F(J®„M)
is discrete. Hence, i®1w is a strict monomorphism and it follows that M is flat. □
8.4. Corollary. Let R be a filtered ring, MeR-filt, such that FR and FM are
discrete and exhaustive. Then we have :
w. dirndl < gr.w. dinu ,R, G (M)
Proof. Put n =gr.w.dim_,mG(M). Consider an exact sequence :
G(.kj
n-1
fi f
►L. —U-l —°-*M-
1 o
^o ,
where L ,.-,L _. are free objects, while FL., i=o,... ,n-1, and FL are exhaustive and
discrete. By exactness of the functor G we derive from (*) an exact sequence :

34
o—-G(L)—GCL^.,) —.-—G(Lo)—-G(M)—o .
Our assumptions imply that G(L) is a flat object of G(R)-gr and the foregoing lemma
entails that L is flat. Therefore w.dimRM<n. □
8.5. Corollary. For a filtered ring R with FR being exhaustive and discrete we
have :
gl.w.dim R<gr.gl.w.dim G(R) .
Proof. Put n = gr.gl.w.dim G(R). Any MeR-mod may be filtered by the filtration
F M = F R.M which is easily seen to be exhaustive and discrete. The inequality of
Corollary 8.4. thus holds for any M, filtered this way, an this proves our claim. C
8.6. Corollary. Let R be a graded ring, then :
gr.gl.w.dim.R<gl.w.dim R
If moreover R is left limited then :
gr.gl.w.dim R = gl.w.dim R .
Proof. Easy combination of foregoing results.
8.7. Corollary. Let R be a positively graded ring and assume that R is a von
Neumann regular ring : gl.w.dim R= left-w.dim_R = right-w.dinuR .
Proof. Take MeR-mod and equip it with the trivial filtration F M = o for p<o,
F M=M for p>o. Endow R with the filtration F R= ® R.. In this setting M is a
P P i<p x
filtered R-module and clearly G(M) = o for all p^o. Therefore G(M) is annihilated
by the ideal R = ® R . Corrolary 8.4 together with [ 2 ,ex.5 p.360] we get :
n>1 n
w.dimRM<w.dimRG(M) <l.w.dimJR + l.w.dimR G(M) and the latter is just l.w.dimRR0.
It follows that gl.w.dim R=l.w.dimJR . The other equality may be established in
exactly the same way. □

35
8.8. Corollary. Let R be a filtered ring such that FR is discrete and exhaustive.
Let NeR-filt, Mefilt-R be such that FM and FN are discrete and exhaustive, then
Tor^R:)(G(M),G(N)) =o implies : TorR(M,N) =o.
Proof. Consider a free resolution of N
f
? fi f
L.,—^-*L. '—*L —°— N—*o ,
2 1 o '
where F.L is discrete and exhaustive, for all i, and where f ,f.,,f.,,... are strict
l ' ' o 1 2
morphism. Consider the following commutative diagram, the vertical arrows of which
represent isomorphisms :
.-—>G(M) ® G(L ) >G(M)
G(R)
® G(L ) ► G(M) ® G(LJ
G(R) ' G(R) °
G(M®L9)
R L
G(M®L.)
R
G(M®L )
R °
J.n
G(R)
We have : G.'J1(M®LA) =HnG(M®LA) =Hn(G(M) ® G(LJ) = Tor^w (G(M) ,G(N))
-G(R),
G(R)
>nr
If Tor v V(G(M),G(N)) =o then G ' (M®LJ =o and, since F(M®LJ is exhaustive and
n R * R *
discrete, Tor (M,N) =H (M®LJ =o. □
Special References for Chapter I.
K.S. Brown, F. Dror [4]
R. Fossum, H.B. Foxby [8]
P. Gabriel, Y. Nouaze [27]
C. Nastasescu [28]
G. Sjôding [35]

CHAPTER II. TOPICS IN GRADED RING THEORY.
I1.1. HOMOGENIZATION.
Let R= ® R. be a graded ring. If MeR-gr then we put M+ = © M., M~ = ® M..
ieZ 1 i>o 1 i<o
It is obvious that R and R are graded subrings of R, whereas M is a graded R -module,
M is a graded R -module. If N^M are graded R-modules then we have (M/N) =M /N
and (M/N)" =M_/N_.
Let MeR-gr and let XCM be an R-submodule. Any xeX may be, in a unique way,
written as x. + ... + x with deg x. <... <deg x . By X (resp. X_) we will denote the
submodule of M generated by the x (resp. x.) i.e. by the homogeneous components of
highest (resp. lowest) degree of elements of X. With these notations we have :
1.1. Lemma. 1. X and X__ are graded submodules of M.
2. X = X if and only if X is a graded submodule of M.
3. If XCY are R-submodules of M then X~cy~, X_cy_.
4. Equivalently : 1°. X = o, 2°. X~ = o, 3°. X_ = o.
5. If M is left-or right-limited and if X (resp. XJ) is generated by r homogeneous
elements then X may be generated by less than r elements.
6. If I is a left-ideal of R and N an R-submodule of M, then : I~N~<= (IN)~ and
I_N_c (IN)„.
7. If I is an ideal of R then by rad I we mean n{P, p a prime ideal of R, P3I}.
If either R is commutative or left Noetherian, we have : (rad I) Crad(I ) and
(rad I)„crad(IJ.
Proof. Assertions 1,2,3,4 are easy to prove.
5. Filter X as follows F X=Xn ® M.. It is easily verified that X is isomor-
i>-n
phic to the graded module G(X) associated to the filtered module X. Similarly X_ is
isomorphic to G(X) when X is filtered by putting F X = Xn e M.. The statement now
follows directly from Proposition 1.5.3.
6. Let aei , x£N be homogeneous elements. Then there exist ael.xSN such that
a,x are the homogeneous components of highest degree of a,x resp. If axf o then ax

37
is the homogeneous component of highest degree of axe IN, hence axe (IN) .
7. The commutative case is easy and, as a matter of fact, it may be proven exactly
in the same way as in the following proof for left Noetherian R. Take xe (rad I) .
First note that for any ideal J of R, J and J^ are ideals of R. Suppose that x is
the component of heighest degree of xerad I. Since R is left Noetherian xerad I
is equivalent to R. xll x... R. xR. ci, for all (A.,.-,A ) eZn, for some nez.
Taking components of highest degree yields : R. xR. x... R. xR. ci (note that
12 n-1 n
there are no problems arising from zero products), for all (A.,...,A ) ez11.
1.2. Proposition. Let XCY be submodules of M, then the following conditions are
equivalent :
1. X=Y.
2. X~=Y~ and XnM~ =YnM~.
3. X_=Y_ and XnM+=YnM+.
Proof. In order to prove that 1 ■»• 2 let yeY, y=y1+.-+y the homogeneous
decomposition of y, and let deg y =t. We shall show that yex by induction on t.
If t<o then yeM and thus yex. If t>o then y eY =X yields that there is an
xex such that x = x.+._+x ., +y , with deg x. < ... <deg x , <t. It follows that
1 n-1 'm b 1 s n-1
y-xeY has a homogeneous decomposition in which the highest degree appearing is
less than t. The induction hypothesis yields that y-xex and hence yex.
The implications 1 *>Z are established in a similar way. □
1.3. Corollary. With notations as above : if M is left-limited (resp. right-
limited) then X=Y if and only if X~=Y~ (resp. X_=YJ.
If R is a graded ring and MeR-mod then it is not always possible to
consider maximal graded submodules within M; for special M however we have the following:
1.4. Lemma. Let N e R-gr and let M be a submodule of N. There is a unique obj ect
in R-gr which is maximal amongst objects of R-gr which are submodules of M and
graded submodules of N, this object will be denoted by (M) .
Proof, cf. T30 ] .

38
1.5. Lemma. Let R be a graded ring. If I is an ideal of R then (I) is a
graded ideal of R. If P is any prime ideal of R then (P) is a graded prime ideal
of R. If J is a graded ideal of R then radJ = n{(P) ,p a prime ideal containing J},
hence radJ is a graded ideal of R too.
1.6. Note. (M) as defined in 1.4 depends on the structure of N, however we
will only use this lemma in cases where all modules considered are submodules of
some fixed graded module, so there is no need to take this dependence into account
in the notation; e.g. in Lemma 1.5. all gradations considered are induced by the
gradation of the ring R.
II.2. THE STRUCTURE OF PRINCIPAL GRADED RINGS.
In this section we reduce the study of principal graded rings to the study
of twisted polynomial rings, a class of rings which will also reappear in II
Throughout this section R = R $R. ©._ will be a positively graded domain.
2.1. Lemma. Let ICR be a left principal graded ideal, then there exists a
homogeneous element which generates it.
Proof. Suppose, I =Rx and write x=x. + ... +x with deg x. <... <deg x . Since x. el
for all j we have that x? =rx with reR, rj'o. Write r = r. + ._ +r with
deg r, <._<deg r . Since r.x. ^o, x7=r-x1 follows. Repeating this argumentation
we find that x. = r.x, for some homogeneous r. e R, therefore x= (1 + r, + .„ +r .)x.
andRx = Rx1 follows. □
2.2. Theorem. In case the gradation is nontrivial the following statements are
equivalent :
1. R is a left principal ring.
2.a. R is a left principal ring.
b. There is an injective morphism ^:R -+R mapping nonzero elements of R to
units of R
o
c. R = R [X,ip] as graded rings, where degx>o.

39
Proof. 1*2. If I CR is a left ideal then RI is graded and left principal,
therefore, by the foregoing lemma there is a homogeneous aeR such that RI =Ra. If
I fo then the fact that the gradation is positive yields that aeR and
(RI ) =1 = R a; thus R is left principal. Nov/ to prove b), consider the ideal
R = © R. which is nonzero as the gradation is nontrivial. By our hypothesis
i>1 x
R+ = Rt for some homogeneous teR say deg t=s>1 i.e. AeR . From R = Rt it
follows that R =R t, Rt =R t2 ._ , while R. =o if s i. Pick aeR ; then taeRt
s o ' 2s o ' i o
since Rt is two-sided, hence there exists a unique element $ (a) eR such that
ta = v>(a)t, clearly c(a)eR . It is easy to verify that <p, thus defined, is an
inject ive ring homomorphism R -+R . Again consider aeR ,a^o. The left ideal
R
: a + R+ is principal, hence there exists a b e R , b fo, such that R a + R+ =Rb. In
this case : R a = R b,._,R =R b. Substituting R =R t we get : R t = R tb =
o o''nsns 6so £ o o
R f (b)t and so there exists AeR such that t=Xp(b)t, hence \f (b) =1. From R a =
R b we derive : R^> (a) = R t (b) = R , thus inp (a) = 1 for some /aeR. Applying <f yields
<p(ti)v (a) =1. But <p(ji) has a left-inverse m', obtained as before, so mV(m) =1
2 2
yields m' = <P (a) or <p (a)^(fi) =1. Since y> is injective it results from this that
f (ai)it = 1, consequently <f (a) is a unit of R .
For c) it suffices to check that the map R [x,<p] ->R defined by : a-*-a if aeR
X-»-t, actually defines an isomorphism of graded rings.
2*1. It is clearly sufficient to prove that in R [x.y] ; any graded left ideal is
left principal and without loss of generality we may assume that deg X =1. Every
left graded ideal I of R [X,<p] has the form 1 = 1 ®I X®I~-X2 ®..., where I ,I..,I7,...,
are left ideals of R such that I c^ (I ) for each n>o. Let r£K be minimal
o n "• n+1
such that I to, then, since I c^ (I *), I +1 contains units of R or I , =R .
We reached the situation I =R for all n>r. If I =R b then I = Rb follows, con-
n o r o
sequently R is a left principal ring. □
2.3. Corollary. The following conditions are equivalent :
1. Ris left and right principal.
2.a. L is a skewfield.
b. There is an automorphism t : R -+R such that R = R [X,y] as graded rings, where

40
deg x>o.
Proof. 2*1. Trivial.
1 =*2. Since R is left principal, R = R [X,y>] , where ? is a monomorphism R -+R . The
right ideal R «v> (R )X%2(R )X2«._ has the form aR for some aeR. It follows
that R =aR , </> ÇB. )X = aR1 = aR X and so we obtain that f (R ) = aR =R i.e. f is an
oo^o 1o ^ooo
automorphism. Since t maps nonzero elements of R onto units of R , it follows that
R is a skewfield.
2.4. Remark. In the situation of 2.3. and by construction of t it is clear that
<f is inner if and only if there exists b/o in R such that bt commutes with R .
Let K be the center of R and let kcK be the fixed field for <t. If the center C
o oo
of R properly contains k then C = k[btn] for some ne IN, beR -{o}, and in this case
f is an inner automorphism. Conversely if y is inner (n the smallest integer as
such) then C=k[btn] for some beR -{o}. If <pn is not inner for any nelN then
C=k and all ideals of R are generated by powers of t. If <p is inner for some ne]N
then all ideals of R not in R may be generated by a central element. Note also that
R is a noncommutative graded Dedekind domain.
2.5. Remark. If K is a field then the only possible ZZ-gradation on K is the
trivial gradation. Indeed if x^oeK. with i>o then expressing y= (1+x) = 2 y.
1 j£Z J
yields conditions : 1 =y +xy ., o=y. +xy. - for all j fo. So if n is minimal such
' 'O '-1* 'j 'J-1 J O
that y to and if n fo then the second relation yields y + xy . =o i.e. y =o,
'n„ o 7 7n 'n -l 'n
2 ° ° n
contradiction. If n =o then one calculates y=1-x + x -._ but as x fo for all n
this is impossible. However we shall introduce the notion of graded division ring,
(some say:graded field), this is a graded ring such that each nonzero homogeneous
element is invertible. For example, let k be a field and consider K = k[X,X ] for
some variable X, then K, equipped with the obvious grading, is a graded division
ring. Now consider K[Y] =R and grade it as follows R= 2 K.Y-1. An feR is
k i+j=n X n
written : f = a + a,Y + ._ + a, Y where a e K^,... ,a,e K , . If geR andg^o, then
there exist homogeneous elements q,r in R such that f =gq + r with degY r <deg„g.
Having a division algorithm on R it follows that graded ideals of R are generated by

41
one homogeneous element of R. However R is not a (left) principal ring as is well-
known (since K is not a field!), this shows that the condition "R is positively
graded" cannot be dropped from II.2.
II.3. NOETHERIAN OBJECTS.
Let MeR-gr. Then M is said to be a left Noetherian (Artinian) object, and
in this case we say that M is gr.l.Noetherian (gr.l.Artinian) if M satisfies the
ascending chain condition (descending chain condition) for graded submodules of M.
It is straightforward to verify that M is gr.£.Noetherian if and only if each
graded submodule of M is finitely generated , or if and only if each non-empty
family of graded submodules of M has a maximal element. Dually, M is gr.£.Artinian
if and only if each non-empty family of graded submodules of M has a minimal element,
or if and only if each intersection of graded submodules may be reduced to a finite
intersection.
3.1. Proposition. Let MeR-gr have left-(right-) limited grading. Then M is
gr.£.Noetherian, resp.gr.£.Artinian, if and only if M is I.Noetherian, resp.
I.Artinian, in R-mod.
Proof. See Corollary 1.3.
Without the left-limitedness condition the statement remains valid in the
Noetherian case as we will see later, but it fails for the Artinian case. Indeed,
let K be a graded division ring, then K is clearly gr.£.Artinian but clearly K_ need
not be I.Artinian as the example K = k[X,X ] shows.
3.2. Proposition. Let R be a graded ring, MeR-gr a left Noetherian object,
then :
1. For all ieZ, M. is a left Noetherian R -module.
1 O
2. M is a left Noetherian R -module.
3. M is a left Noetherian R -module.
Conversely if M ,M are left Noetherian objects of R -mod, R -mod resp. then M is

42
gr.£.Noetherian.
Proof. 1. Let N. be any R -submodule in M.. Then RN. is a graded submodule in M
and hence it is finitely generated by homogeneous elements x. , ._,x,, which may be
supposed to be taken from N-. Pick y£N.; then there exist homogeneous X ,._ ,X, er
J;
such that y= 2 A. x.. Comparision of degrees yields deg X. =o, j = 1,... ,k and thus
i=1 x 1 J
N. is generated as an R -module by the elements x..,._,x, .
2. By Proposition 3.1. it is sufficient to show that M is gr.£.Noetherian in R -gr.
Let N = © N. be a graded submodule of M . As in 1. we may assume that RN is gene-
i>o x
rated by homogeneous elements x..,._,x, of N. Put t=max{deg x. deg x, } and let
y EN be a homogeneous element with deg y>t. There exist X. X, eh(R) such that
y=X.x- + ._ X x, . Now, deg y>t implies that deg X.>o for all i=1._k, consequently
X. ER ,i = 1 —k. Because of 1., M ®... $J1 . is a left Noetherian R -module and
l o t-1 o
therefore N ®._ ®N 1 is a finitely generated R -module. Let y.,.-,y generate
N «... ©N i over R . Then clearly {x.,._ ,x, ,y1,._ ,y } generates N over R ,
therefore M is a left Noetherian R -module.
3. Similar to 2.
Supposing that M and M~ are both left Noetherian objects in R -mod, R -mod resp.,
let N1 cn„ c ._ cn c ... be an ascending chain of graded submodules of M. Then we
also obtain ascending chains in M and M :
1 2 p
N~cnZc._cn~c._cM~ .
1 2 p
Our assumptions amount to : N,=N,„+1 and N, =N,+1, but then N, =N,+, = ._ . □
3.3. Theorem. Let MER-gr. then the following assertions are equivalent :
1. M is gr.£.Noetherian.
2. M is a I.Noetherian R-module.
Proof. Since 2 ==■ 1 is obvious let us check 1=>2. Let X.cx,c.. ex c ._ be an
1 2 n
ascending chain of submodules of M. By 3.2 there is an n e ]N such that M nX- =
M~nXi-M andxT=X^+1 for each i>nQ. By Proposition 1.2. : X. =X. + , = ._ , i>nQ. □

43
3.4. Proposition. Let R be a graded ring, MeR-gr. Suppose there exist elements
aeR„ b£R, such that :
1. Ra = aR andRb=bR.
00 oo
2. M. =a1M and M . =b1M for all i>o.
10 -10
If M is a left Noetherian R -module then M is a left Noetherian R-module.
o o —
Proof. By Proposition 3.2. and 3.1. we may reduce the proof to the case of positively
graded R and M. Let N = ® N. be a graded submodule of M. For each i>o put :
i>0 x
P. ={xeM , a'ScEN.}. Obviously, each P. is an R -submodule of M and P =N ,
101 ' 1 O 00 0'
P. cp. . for all i. Also, a1?. = N. for all i>o. Now, M being a left-Noetherian
1 1+1 ' 1 1 ' 0 a
R -module, there is an ne IN such that : P =P . = .... Choose {z*- -*,... ,zj- °-)} to
o ' n n+1 o o
be a system of generators for P=N and let {z> ' z> x'} generate P., i=1,._,n.
Note that the set :
{z^,._,ZCro\ œ,...,az}rl) aV1\..,aV™}
oo ' 1 ' ' 1 ' n n
generates N as a left R-module. □
3.5. Corollary Let R be a graded ring and suppose there exist elements aeR.,
b eR_i such that :
1. R. =Ra=aR and R . = R b = bR .
1 o o -1 o o
2. R. R. =R.^ and R -R . =R . . for all i>o.
1 l i+1 -1 -l -1-1
If M is a left-Noetherian R -module then R ® M is a left Noetherian R-module.
0 R
o
Proof.
R ® M= ® R. ® Ma $ (a1® M) «M®( © b1 ® M) ^ ® a^ ® M e b1 M
R ieZ 1 R i>o R i<o R i>o i<o
ooo o
Now apply Proposition 3.4. □
3.6. Corollary, Let R be a graded ring as in 3.5. If R is a left Noetherian
ring then R is left Noetherian.

44
3.7. Corollary. Let R be a left Noetlierian ring. Let y:R-»-R be a ring
automorphism and let S : R->R be a ^-derivation of R. Then S=R[X,¥>,6] is left Noetherian.
Proof. Filter S by F.S = {PeS,deg P<i}; this filtration is exhaustive and discrete.
The associated graded ring G(R) is isomorphic to R[X,y] . Because of Corollary 5.6.
it will be sufficient to show that R[X,^] is left Noetherian, but this follows
directly from Corollary 3.6. □
Let R and S be any two rings and let MeR-mod-S, NeS-mod-R be bimodules.
Consider the matrix ring T-l =■ [r m ] , re R,me M,ne N,se SV with usual addition,
and multiplication defined by : ( r m\ I r', m,' U ( rr| , ™'+ms,' \ .
F ' \n s ) \n' s' J \ nr'+sn' ss' I
This ring has gradation T^ = (° °) , Tq = ( |J °) and ^ = [° Jj) , T. = o if | i | > 1. The
left graded ideals of T are of the form L,, T ] where I and J are left ideals in. R
and S resp., while M' is an R-submodule of M, N' an S-submodule of N,such that MJCM',
NICN'.
3.8. Corollary. T is left Noetherian (Artinian) if and only if R and S are left
Noetherian (Artinian), M is a finitely generated R-module, N a finitely generated
S-module.
Let R= © R. be a graded ring, M- © M. a graded R-module. For any
2 ieZ 1 iez 1
(d,k) eZ such that d>1, o«k<d-1 we put :
R^ = © R., and M^d'k) = © U. ,. .
ieZ ld iez ld+k
Clearly, R^ is a graded subring of R and, M^d,k^ is a graded R^-module. If N is
a graded submodule of M then N*- ' '•' is a graded submodule of M*- ' '■'. We shall write
M1- ' instead of M^ '0-) for every d>1. For fixed d, it is clear that M is the
direct sum of the R^-*-modules M*- >k) with o=Sk=Sd-1.
3.9. Proposition. With notations as above, the following statements are equivalent:
1. Misa left Noetherian R-module.
2. W- ' ' is a left Noetherian graded R*- -module for every 1 <k<d-1.

45
Proof. 1 =>2. Let PCM1- ' ' be a graded submodule, then RP is a graded submodule
of M and hence it is finitely generated over R, let x1(._,x be homogeneous elements
of R which generate RP. If yeh(P) then there exist homogeneous X.,... ,X £R such
that y = X1x1 +._ +X x . Put : deg x. = i.d+k,._ ,deg x =i d+k, degy=id+k. It follows
' 1 1 nn & 1 1 ' ' & n n '
that : deg X = (i-i )d,._ ,deg X = (i-i )d, hence X ... ,X eR^. So P is generated by
x1,.- ,x. over R1- '.
2 ==■ 1. Let N be a graded submodule of M. Since N*- ' '■' is then a graded R1- -^-sub-
module of M*- ' '■' it may be generated by : xj ',._ ,x^ ', over R*- . However, as
nk
N= « N^'^, it follows that u {x.W x® } generates R over R. □
o«k<d-1 o<k<d-1 ] nk
3.10. Remark. Let P be a graded R^-submodule of M^d,k:), then RPnM(d,k:) =P.
3.11. Corollary. If R is a left Noetherian graded ring then the ring R^ J (d>1) is
left Noetherian.
Another type of graded left Noetherian rings is given by the construction
of the Rees ring associated to certain ideals. Let R be a ring, I an ideal of R.
The Rees ring associated to I is the ring :
R(I) = R + IX + I2X2 + ... + InXn + ... c R [X] ,
which is obviously a graded subring of R[X] isomorphic to R®I®I ®._©I ©.- .
Moreover R(I)/I® I2 «.- «In«._ =G(I) = R/I «I/I2 « ._ «In/In+1 « ... .
3.12. Proposition. Let R be a left Noetherian ring and let the ideal I of R be
generated on the left by a central system (cf. I.5.12) : {a-,...,a }, such that the
a., i=1._n, commute amongst themselves, then R(I) is a left Noetherian ring.
3.13. Sublemma. Let S be a left Noetherian subring of an arbitrary ring R. Suppose
that R is generated as a ring by S and a single element x such that sx-xs £S for all
ses, then R is a left Noetherian ring.

46
Proof of the sublemma : Clearly R = S[x]={ax +.. *a.x + a ,a.eS,i=o...n}. We
equip R with a discrete filtration FR as follows : for teH. F.R={a
n r t n o
a.es i = o,... ,n , n<t}, in particular F R = S. For any ne IN , aeS we have
axn-xna = b 1 x +._+b for some b.eS. j=o,._,n-1. Hence ax -x aeF 1R. Now
n-1 o j n-I
let G(R) be the graded ring associated to the filtered ring R, G(R) =FoR*F.]R/FoRe .-
... ®F R/F _.R®._ . In this ring, the image x of x in F.R/F R is a central element,
thus there exists a ring epimorphism ^ : S [Y] -*G(R), ■/'(Y) =x. Since S [Y] is left
Noetherian, G(R) is left Noetherian too, but then R is also left Noetherian because
of Corollary 1.5.7. □
Proof of the proposition. Put I, = (a.. a,) for k<n. Since a.eZ(R), R(I.,) is left
Noetherian. We proceed by induction. Since R(I ) is generated as a ring by R(I ,)
and the single element x = a X (this is readily checked) we try to apply the sublemma.
This will be possible because for all zeR(I .) we have that a Xz-za X£R(I .);
indeed, this follows immediately from the fact that the image of a in R/(a..,...,a ,)
is in the center of the latter ring (also making further use of the fact that a
commutes with a.,.»,a ,). □
1' ' n-1
3.14. Remark. The conditions of the proposition are fulfilled if I is generated by
a central system (a-,a7) or in case I is generated by central elements of R.
Let I be an ideal of a ring R, and consider R as a filtered ring with the
I-adic filtration. Consider an MeR-filt. We say that the filtration FM is
I-fitting if there exists apeZ such that IF M = F _. for all n<p. As an example
F_iM=I1M, iez, defines an I-fitting filtration. (Note that F M = M for all n>o).
If FM is an I-fitting filtration then the quotient filtration on M/N, for any sub-
module N of M, is also I-fitting. If FM is I-fitting then R(M) =MQ«M_ «M_2®.-
._ ©M ©... is the Rees module associated to M. It is clear that R(M) is a graded
R(I)-module. The I-fitting filtrations, for certain I, are characterized by the
following,

47
3.15. Proposition. Let R be a left Noetherian ring and let I be an ideal of R
generated by a central system. Let R have the I-adic filtration and let MeR-filt
be finitely generated. Then, the following statements are equivalent : 1°. FM is
I-fitting, 2°. R(M) is an R(I)-module of finite type.
Proof. : If FM is I-fitting then there exists apeZ such that for all n<p : IM =
M _-. Consequently R(M) will be generated as a left R(I)-module by the set
M ©M. ©._ ©M , which is however finitely generated as an R-module. Conversely,
suppose that R(M) may be generated by elements x.,... ,x of degree d. = deg x., i = 1,.-,n.
Putting p=max{d. , i = 1,.~,n} one easily establishes that M . =IM . □
l ' p-1 p
3.16. Proposition. Let R be a left Noetherian ring and let I be an ideal of R
generated by a commuting central system. Let R have the I-adic filtration and MeR-
filt be finitely generated. If FM is I-fitting then, for every submodule N of M, the
filtration FN given by F.N = F.MnN is I-fitting.
Proof. A straightforward combination of Proposition 3.12 and Proposition 3.15. □
3.17. Remark. Proposition 3.16 may be considered as a generalization of the Artin-
Rees property (however only ideals generated by commuting central systems have been
considered here!).
II.4. KRULL DIMENSION OF GRADED RINGS
In Section 1.5 we have introduced the notion of Krull dimension in R-Mod as
well as in R-gr. Let us first state some easy lemmas and then turn to relating
K.dimR(M) and K.dinu (M) for any MeR-gr.
4.1. Lemma. Let R be a graded ring, MeR-gr. If N is a graded submodule of M then
M has Krull dimension if and only if N and M/N have Krull dimension. In this case
K. dim M = sup (K. dim N , K. dim M/N).
R-gr R-gr R-gr

48
Proof. Similar to the ungraded case, cf. Lemma 1.1. in [121 . □
4.2. Lemma. Let MeR-gr be a left Noetherian object then M has Krull dimension.
Proof. Similar to Proposition 1.3. in [12 ] . □
4.3. Lemma. Let R be a graded ring, MeR-gr. Suppose that M has Krull dimension
and let a be equal to sup{1+K.dim M/N, N an essential graded submodule of N}. Then
R-gr
we have : a > K.dim M.
R-gr
Proof. Similar to Corollary 1.5. in [12] .
4.4. Lemma. Let R be a graded ring and let MeR-gr be left- or right- limited, then:
1. M has Krull dimension if and only if M has Krull dimension, in case this happens
we have that K.dim M = K.dim M.
R-gr R
2. If M is a-critical then M is a-critical.
Proof. 1. Follows from Corollary 1.3.
2. Suppose M= © M. where M. =o for all i <n . Let X^o be a submodule of M, then
iez 1 1
K.dim M/X <a. By Proposition 1.2., Corollary 1.3. and also Lemma 1.5.4., we have
R-gr
K.dim M/X<K.dim M/X . Therefore M is also a-critical.
R R-gr
4.5. Proposition. Let R be a positively graded ring and let MeR-gr. Then the
following properties hold :
1. M has Krull dimension if and only if M has Krull dimension, and if so, we have
K.dim M = K.dimM.
R-gr R
2. If M is a-critical then M is a-critical.
Proof. 1. Let M= ® M.. For each peZ put M. = ® M . Obviously M. is a
. _™ l f f >p n i >p
lez ^ n>p v
graded submodule of M and also (M/M^ ). =o for every i>p. The foregoing lemma
yields K.dim M. =K.dim M^ and K.dim M/M. =K.dimM/M. .
R-gr >P r_>£ R-gr >P R-->E
Application of Lemma 4.1 enables us to deduce 1.

49
2. Let X^o be a submodule of M. For some peZ, XnM f o . Since M is a-critical
and the foregoing lemma then implies that K.dim M /(XnM ) is smaller than a.
R-gr p p
Clearly, there exists a strictly increasing mapping from the lattice of submodules of
M/X into the set-product of the lattices of submodules of M> /XnM and M/U>
respectively; namely :
Y/X —(YnM>p)/(XnM>p),(Y+M>p)/M>p) .
Now K.dim M/M <a, hence from Lemma 1.5.4. we derive that K.dim M/X<a or M is
R p R
a-critical. □
4.6. Lemma. Let R be a left Noetherian graded ring and let Me R-gr be a uniform
object. Suppose either that M is limited or that the gradation of R is positive.
Then M is a uniform R-module.
Proof. Since R is left Noetherian, M contains a nonzero graded submodule N which is
a-critical (cf. Theorem 2.7.,[12]). Uniformity of M yields that M is an essential
extension of N in R-gr; but by Lemma 1.3.3.13. it follows then that M is an essential
extension of N. Lemma 4.4 and Proposition 4.5, now yield that N is an a-critical R-
module, hence uniform and therefore M is uniform too. □
4.7. Theorem. Let R be a graded ring, Me R-gr a uniform graded module. Then M is
a uniform R-module.
Proof. For p ez denote ® M. by M>p. Let Xcm be a nonzero submodule such that
i>p
XnM^ =o for some pez. Then X^nM =o; indeed if x is homogeneous in X_nM>
then there is a yeX, y = y + ... +y , with y. =x and deg y. <... <deg y . Since
deg y1 >p, yeM> , hence yeXnM^ =o. So we have (XJ . =o for all i>p. Because
X_/o and X_ being uniform in R-gr, Lemma 4.6 may be used to derive that X_ is a
uniform R-module, hence M is a uniform R-module. Let us suppose now that XnM /o
for any nonzero submodule X of M and for all pez. in this case we proceed to show
that M is a uniform R -module. By Lemma 4.6 it will be sufficient to establish

50
uniformity of M in R -gr. Let x,yeh(M ) be nonzero and put n = deg x, m = deg y with
m<n. Since M is uniform RxnRy^o thus RxnRynM> ^o and we may pick a nonzero
homogeneous z in this set. We have z = Xx =^y with X ,m eh(R) and by the choice of z
it also follows that X ,n e R ( hence zeR xnR y. Consequently M is uniform in R -gr.
+ +
Further, if X,Y are nonzero submodules of M then X^M/o and Yn,M ^o yields
XnYnM+^o hence XnY^o.
4.8. Corollary. Let R be a graded ring which is left Noetherian. Let Me R-gr.
The Goldie dimension of M in R-gr is equal to the Goldie dimension of the R-module M.
Proof. From the above theorem and Lemma 1.3.3.13. □
Lemma 4.9. Let M be left Noetherian in R-gr, then M has a composition series
M DM., d ._ dm =0, where M. -/M- is a critical module for each 1 <i<n.
1 n i-1 l
Proof. As in the ungraded case, cf. [12] . □
Lemma 4.10. Let Me R-gr have Krull dimension, K.dim M = a. Then K.dim M. <a for
any i £ Z.
R-gr R
o
Proof. For any graded submodule N. of M. we have N- =RN. nM., then apply Lemma 1.5.4.
□
Lemma 4.11. Let M be an a-critical object in R-gr. Then K.dim M <a+1 and
R+
K.dim M"<a+1.
R"
Proof. Let x£h(M ) be nonzero. We shall show by transfinite induction on a that
K.dim M /R x<a. For a =o, M is a simple object in R-gr. From the structure of
R+
simple objects in R-gr (see Theorem 6.3 in this chapter) the assertion follows. In
case a/o, the fact that Rx/o implies that K.dim M/Rx < a and from the induction
R-gr
hypothesis combined with Lemma 4.1, Lemma 4.9. it results that K.dim (M/Rx) <a.
R+
Obviously : (M/Rx) =M /(Rx) and R xc (Rx) . So we have the exact sequence :
o—* (Rx)+/R+x—>M+/R+x -^M+/(Rx) + —""o

51
If deg x=k, then (Rx) /R x = R ,x®... ®R ,x (isomorphism of R -modules). By Lemma
4.10 we obtain K.dim (Rx) /R x<a. Finally Proposition 4.3 applied to the above ex-
R+
act sequence yields : K.dim M <a+1. In the similar way it is proven that
R+
K.dim M~<a+1.
R"
4.12. Corollary. Let MeR-gr be left Noetherian with K.dim M-a. Then :
R-Sr
K.dim M <a+1, K.dim M «a+1.
R R"
Proof. Directly from 4.11, 4.9 and 4.1. □
4.13. Theorem. Let M be left Noetherian in R-gr with K.dim M = a, then :
R-gr
a <K.dimn M<a+1.
Proof. By Lemma 1.5.4 : a<K.dim M. By Lemma 1.5.5 and Proposition 1.2 we obtain:
R
K.dim M<sup(K.dim M, K.dim M_)
R R-gr R"
So Corollary 4.12 finishes the proof. □
4.14. Corollary. Let Me R-gr be left Noetherian and a-critical. Suppose K.dim M-
R
a+1, then :
1. M is an (a+1)-critical R-module.
2. M+(M~) is an (a+1)-critical R+-(R~-) module.
Proof. Let XCM be a nonzero R-submodule. Obviously : K.dim M = K.dim M =a+1.
~ R R"
If XnM< =o then X r>M< =o. Considering that : K.dim M(n) =a for all n£Z,
^P P -_ R-Sr
we may assume that XnM /o. Then K.dim M/X <a and K.dim M~«a (see the proof
R-gr R"
of Lemma 4.11). By Proposition 1.2 and Corollary 1.3 we obtain : K.dim M<a, hence
R _
M is (a+1)-critical. The second assertion results from foregoing results. □
If Me R-gr is an a-critical left Noetherian object, then one expects that if
K.dinu M = a, then M is a-critical. For the commutative case this is indeed the case,
it follows from :
4.15. Proposition. Let R be a commutative graded ring and let Me R-gr be a-critical

52
Then M is an a-critical or an (a+1)-critical R-module.
Proof. It is clear that M is a-critical in R-gr if and only if there exists a graded
ideal I such that for any x£h(M) , x/o we have Ann x = I and K.dim R/I = a. Let
R-gr
xEM be a nonzero element with x =x. + ._ +x with deg x., < ._ <deg x . It is clear
that IcAnn x. If X eAnn x,X =X + ... +X with deg X < ._ <deg X then Xx = o yields
X..X., =o i.e. X.el. Then X.x = ... =X.x =o, hence X.eAnnx. Further, we obtain
11 1 1 2 1 n ' 1 '
X2,.„,X, ei , hence Ann x = I. Since R/I is a domain, R/I is a-critical or (a+1)-cri-
tical as an R-module, i.e., M is an a-critical or (a+1)-critical R-module. □
4.16. Proposition. Let R be a graded ring, Me R-gr. M has Krull dimension if and
only if for each k, o<k<d, M*- ' ' has Krull dimension over R^ . In this case,
K.dim M= sup K.dim ,,, M^d'k^
R o<k<d Rla)
Proof. From Remark 3.10 it follows that K.dim ,„ M^d'k^ <K.dim M for all o<k<d-1,
hence : sup K.dim f „ M1- ' J <K.dim M. On the other hand, the lattice of graded
o<k<d Rl) R
submodules of M maps into the product of lattices of graded submodules of Ml ' with
o<k<d, as follows :
N_[N(d,o))N(d,1)j__)N(d,d-1)) _
This mapping is strictly increasing so we may apply lemma 1.5.5. □
4.17. Corollary. Let K be a graded ring having Krull dimension. For every d>1,
R*- ■* has Krull dimension and
K.dim R- K.dim R^ .
R-gr R-gr
In the sequel of this section let R be a commutative graded ring. Denote by
Spec R (Spec R) the set of graded prime ideals (prime ideals) of R. By transfinite
induction we define on Spec R the filtration : (Spec R) ={graded maximal ideals}
(Spec R) ={p€Spec R, pCq and qeSpec R implies qe KJ (Spec R)„.
g « g f g B<a g B

53
If R is a Noetherian graded ring, there exists a smallest ordinal a (resp. g) such
that Spec R= (Spec R) (Spec R= (Spec R) ). The ordinal a,(g), is denoted by
gel.dim R (resp. cl.dim R) and will be called the classical graded Krull dimension
of R (resp. classical Krull dimension of R). It is well-known, cf.[ 9 ] p. 425,
that K.dim R = gel.dim R, K.dim R = cl.dim R. From Theorem 4.13 we derive :
R-gr
4.18. Corollary. If R is a Noetherian graded ring then gel.dim R<cl.dim R^gcl.dim R + 1.
4.19. Remark. If K is a graded division ring then we have gel.dim K = o, cl.dim K= 1.
If R = k [X,Y] , k a field and XY = YX = o, put R+ = k [X] , R~ = k [Y] , then K.dim R =
R
K.dim R = K.dim R =1.
II.5. THE KRULL DIMENSION OF SOME CLASSES OF RINGS.
Here we include explicit formula for the Krull dimension of certain rings, in
terms of the data used in the construction.
5.1. Polynomial Rings.
Let R be any ring, <f a ring automorphism R-»-R, and consider the ring of
twisted polynomials R [ X,<f] . If M is a left R-module, then M[X,y] stands for the
left R[X,v]-module R[X,y] ®M. Then M[X,y] becomes a graded R [X,v]-module if we
R
equip it with the grading : M[X,y] . = {X1 ®m,meM}, for i>o. In fact, M[X,^] may
2
be identified with the module of polynomials ny. +Xm, +Xm? + .., m. £M, with scalar
multiplication given by : aXp.Xqm = Xp+q ^"p"q(a)m, a£R, m£M.
5.1.1. Theorem. 1. M[X,y] has Krull dimension if and only if M is left Noetherian,
in this case :
K.dim M[X,v>] =1 + K.dim M.
R[X,y>] R
2. If M is left Noetherian and a-critical then M[X,y>] is (a+1)-critical.
Proof. Put S=R[X,v] , N=M[X,y] . If M is left Noetherian then N is left Noetherian

54
in S-mod, hence its Krull dimension is defined. Conversely, assume that N has Krull
dimension and that M is not left Noetherian, i.e. there exists a strictly increasing
sequence of submodules of M :
M CM, CM, c._ CM c._
o £ 1 *= 2 * n*
Set :
A= 1 «M. + X®M + ._ + Xk®Mk+1 + .-
o 1
; 1 ®M_ + X ®Ui + .- + I. ®H
Clearly Bca are graded submodules of N and XAcB. Let it. be the canonical map
M. -<-M./M- h, i = 1,2,... . Note that M./M. ,, i = 1,..., is an S-module with scalar
i i i-l li-l
multiplication :
(a +a..X+—) .ir. (m) =ir. 0 -1(a )m) for every meM. .
Define the mapping 8 :A/B->M.,/M «M./M. ®... as follows, if f = 1 ®œ, +X®m + ... ^A, pi
8 (f mod B) = (it- (m.) ,-n (m_),.-). It is straightforward to check that 8 is an
S-isomorphism. Since N has Krull dimension, so does A/B, cf. Lemma 4.1., and
therefore it is of finite Goldie dimension, contradiction.
Because of Lemma 4.9. we may assume that M is a-critical. Proposition 4.5 and the
fact that N is a graded S-module yield that we only have to prove that N is (a+1)-
critical in S-gr then the equality in 1 and also 2 will follow. Consider the
following infinite, strictly decreasing sequence :
N £ XN 3 X2N 3 .- 3 XXN 3 ...,
where xVxl+1 N^N/XN = M.
Obviously : K.dim N>a+1 .
k S
Let z =X ®m be a nonzero element of degree k in h(N) and consider the following
sequence of graded modules :
(*) N 3 XN 3 ._ 3 XkE 3 Sz ,
4iere K.dim X^/X14" N = a and K.dim X^N/Sz =Rm[X,v] . Since M is a-critical it
S S

55
follows that K.dimR(M/Rm) [X,v] <a Lemma 4.1 applied to (*) yields that
K.dim- N/Sz<a and therefore N is (a+1)-critical. □
5.1.2. Corollary. In case we also consider a ^-derivation S then we have :
If M^R-mod is left Noetherian then
K.dim-, M< K.dim M[X,y>,S] «1 +K.dimD M ,
K RR>,«]
where M [X,<f,S] = R [X,v,q ®M.
R
Proof. R[X,y>,6] is a filtered ring with associated graded ring equal to R[X,y>]
whereas M[X,y>,6] is a filtered module with associated graded module M[X,y] . The
foregoing theorem and Proposition 1.5.8 finish the proof. □
5.2. Rings of Formal Power Series.
Let R[X,y] be as in 5.1. If MeR-mod then M[[X,y]] is the "module of formal
power series" consisting of elements m +xm1 + x m_ + ..., m- <=M and with scalar
multiplication given by : (aXp). (xV) =Xp+q Y>~p~q(a)m, a£R, men. R[[X,v]] is the ring
of formal power series andM[[X,Y>]] is a left R[ [X,y>] ] -module, generally not
isomorphic to R [ [X,v] ] ®M !
R
5.2.1. Theorem. 1. M[[X,y>]] has Krull dimension if and only if M is left Noetherian
and in that case :
K.dim M[[X,v]] = 1 + K.dinu M .
R[[X,y>]] lL
2. If M is left Noetherian and a-critical, then M[[X,v>]] is (a+1)-critical.
Proof. The module M[[X,y>]] endowed with the X-adic topology is complete and its
associated graded module is M[X,y] . The proof for the first part of statement 1. is
only a slight modification of the proof of Theorem 5.1.1. (1), while the equality in 1.
and statement 2. are direct consequences of Theorem 5.1.1. and Corollary 1.5.6. □

56
5.2.2. Corollary. If R is left Noetherian and f : R->R a ring automorphism, then :
K.dim R[ [X,v] ] = 1+K.dim R.
5.3. The enveloping Algebra of a Lie Algebra.
5.3.1. Theorem. Let K be a commutative field, g a finite dimensional Lie K-algebra.
The enveloping algebra of g is a left and right Noetherian ring and its Krull
dimension is at most [g : k] =n.
Proof. Let U be the enveloping algebra of g and let U be the K-subspace of U
generated by 1 and products of the form g. >&2>'" >%m with m<n, g. eg. In this way U
is a filtered ring which is discrete, and its associated graded ring is the ring of
polynomials KQCp.-.X ] . So application of Proposition 1.5.8 finishes the proof. □
5.4. Weyl Algebras.
Let k be an (algebraically closed ) field of characteristic zero. Let A.(k)
be the k-algebra generated by the elements p and q satisfying [p,q] =pq-qp = 1; AnM
will be the k-algebra generated by the 2n elements {pi,q- ,1 <i,j <n} subjected to
the relations [p.,q.] =1 [p-,q.]= [p. ,p-] = [q.,q-l =o if i/j. So A (k) =A , (k) ®A (k
J.1 1J1J _L _J 11 lllj.1
Let us also define A (k) =k. Now A (k) may be characterized as the algebra generated
over A _.(k) by elements p,q commuting with A ,(k) and satisfying qp-pq = 1• Therefore
x£A (k) may be written as 2 x „ paq , with x £A - (k). By means of the total de-
n ' ag n ag n-1 '
gree in p and q we may equip A (k) with a discrete filtration such that the associated
graded ring G (A (k)) =A _. [X,Y] . Consequently A (k) is a left and right Noetherian
domain.
5.4.1. Proposition. 1. A.(k) is simple, i.e. there are no nonzero proper ideals.
2. For every nonzero left ideal L of A,, A../L is of finite length. In particular
K.dim A1 = 1.
Proof. 1. The following relations are easi ly verified : [p,paq3] = gpaq3"1 and
[q,paqg] =-apa_1 q3. Let I be a nonzero ideal of Ay u/o, uei, say u=2ua paq3.
Since [p,u] and [q,u] e I, the relations established imply lnk/(o), therefore I=A1.

57
2. Let L be a nonzero left ideal of A- and consider the strictly decreasing sequence
of left ideals containing L :
A, =L 3L3... 3L D ._ .
1 o 1 n
Apply the functor G; we get :
k[X,Y] =G(Lo)DG(L1)3...3G(Ln)3._ .
Since G(L) ?o we have that K.dim k[X,Y]/G(L) <1. It follows that, for n sufficiently
large, G(L )/G(L ^ is a k [X,Y] /G(L) -module of finite length. Therefore it has to
be finite dimensional over k (indeed, if M is a maximal ideal of k[X,Y] then k [X,Y]/M
is an algebraic extension of k hence finite dimensional). So, for large n, L /L -
is a finite dimensional k-space. Because of the density theorem it now follows from
the above that A./L has to be left Artinian and thus of finite length i.e. K.dim A- = 1.
□
Note that, since A- is a quasi-simple ring, the proof of 5.4.1. 1. also may be
extended to prove that A is simple for all n.
5.4.2. Lemma. Let R be a ring and let S.,S, be multiplicatively closed sets such
that R satisfies the left Ore conditions with respect to S. and S.. Suppose that for
any s.^S-, S?GS2 we have that Rs1 +Rs2 = R' then for left ideals ICJ of R such that
1 I= S1 and S? I= S?
S1 I=S1 and S9 I=S9 J, we have I =J
_1 _i
Proof. If MeR-mod is such that S M = S M = o then for any x£M there exist s. eS2,
s?eS9 such that s1x = s9x=o, hence (Rs1 +Rs7)x = o or x=o. Applying this to J/I
yields the statement. □
5.4.3. Theorem. For any n>1, K.dim A (k) =n.
Proof. Put S1 =k[p ] -{o}, S2=k[a ] -{oh Obviously A satisfies the left and
right Ore conditions with respect to these sets. Since A s (A - [X] ) [Y,1 ,S ] , where S
is the derivation of A , [X] given by common derivation in X : ^— , it follows easily

58
that : S~1 An^An_1 (k(X)) [Y.1,8] .
Hence, the induction hypothesis on n and Theorem 5.1.1. imply that K.dim S. A <n.
Similarly one establishes K.dim S~1 A (k) <n. Put s =P(p ) e S., s2 =Q(qn) eS2,
L=A^s1 +A^s2. Clearly Lnk[p ] ?o, Lnkl^] ?o, so it follows that L=An(k)
and we may apply the lemma to get that K.dim A (k) <n. It remains to establish
n<K.dim A (k). Consider the A- (k)-module A- (k)/A- (k)a which is simple and has
commutator exactly k. It is known that the A _1(k) «>A,-sutamodules of
A iffl 8A-/A. (k)q may be represented as I ®A. (k)/A- (k)q where I is some left ideal
of A .j. Then K.dim A^jj = K.dim (A^jj 8A.. (k)/A (k)q. Now consider the sequence :
A1
An(k) = An-1 (k) •A1 (k) 3An-1 (k) ®A1 (k)q DAn-1 (k) «A^2 3 .-
The Krull dimension of each factor is K.dim A _. (k), hence K.dim A (k) >1+K.dim A , (k)
i.e. K.dim A (k) >n. □
n
II.6. GRADED DIVISION RINGS.
6.1. Lemma. If R is a graded domain then left-invertible elements are homogeneous.
Proof. Let x= 21 x. £R, suppose 2r y. s1 x. = 1 then yM x^/o and y x /o yields
i=n j=m -1 i=n 1
M + N = o, m+n = o but then M>m, N>m yields N =n, M =m. □
Note that if R is not a domain, the lemma does not hold anymore, as may be
seen in case R=k[X]/(X) , where 1-X is invertible. If every nonzero homogeneous
element of a graded ring R is invertible then R is said to be a graded division ring.
Note that this definition implies that a graded division ring R is a domain and that
R is a field.
6.2. Lemma. Let R be a graded ring. The following properties of R are equivalent:
1. R has no nonzero proper graded left ideals.
2. R has no nonzero proper graded right ideals.
3. R is a graded division ring.

59
Proof. Straightforward.
6.3. Theorem. If R is a graded division ring then R = R or R=R [X,X ,<p] where
t :R -+R is an automorphism, X an indeterminate of degree t>o such that Xa=y(a)X
and X"1a=Y>"1(a)X~1.
Proof. If R/R then there exists a homogeneous element a with smallest positive
degree, t say. Since R = Ra, R =R a, and for all i£Z, i £ (t), R. =o; moreover a /o
yields R = R„a for all n£Z, Let xeR be arbitrary, then there is a unique element
t (a) £R such that aX =y>(A)a. It is easily seen that <f :R -+R is a field isomorphism
O 00
and we have equalities iPï, = ¥>n(X)an, a X =<p (X)a~ . The graded ring homomorphism
_i _i _i
a :R [X,X ,<p]—>R defined by a (X) = a, a (X ) =a is an isomorphism (note that we
put deg X = t). □
6.4. Corollary. Graded division rings are fields or two-sided principal ideal rings.
Proof. If the graded division ring R is not a field then it is isomorphic to the
ring of fractions with respect to the multiplicative system {1,X,X ,...} of R [X,y>] .
The latter is left- and right-principal because of our results in Section II.2., hence
R is left- and right-principal too. □
II.7. THE STRUCTURE OF SIMPLE OBJECTS IN R-gr.
7.1. Definition. Let R be a graded ring. An SeR-gr is said to be simple if 0 and
S are its only graded submodules.
7.2. Lemma. Let SeR-gr be simple, then for each ieZ, S. =o or S. is a simple
R -module.
o
Proof. Suppose S. /o. If x/o, x<=S. then Rx = S, hence R x = S.. Consequently S. is
a simple R -module. □
o
7.3. Remarks. 1. If R is positively graded and SeR-gr is simple then there is a
j S7L such that S = S.. Indeed if S. and S. are both nonzero, select x., x. /o in S.,S.

60
resp, then x. =rx-, x- =r'xi for some r,r' <=h(R) and depending whether i>j or j>i,
one of these relations is impossible, hence i=j.
2. If S is simple in R-gr and S is left-or right-limited, then S is a simple R-module.
This follows directly from Corollary 1.3.
7.4. Lemma. Let R be a graded ring. S a simple object of R-gr. Then D=END (S) is
a graded division ring and if D /D then S^ is a 1-critical R-module.
Proof. Let f : S-*-S be a nonzero graded morphism of degree n. Then f may be
represented as a morphism of degree o : S-*S(n) but as S and S(n) are simple objects in
R-gr, f has to be an isomorphism in R-gr, i.e. invertible.
To prove the second statement we first prove that for any nonzero R-submodule X of S_,
S/X has finite length. Since X/o, X_/o hence X_ = S. We may assume Xns /o,
because if Xns =o then X_ns =o hence S =o i.e. S is right-limited and î> is simple
in R-mod by Remark 7.3.2. Now let X. 3X. D...DX 3..., be a descending chain of sub-
modules of S containing X. Then we obtain the descending chain of R -submodules in
S+ :
X, n s+ 3 X n s+ 3 ... 3 X n s+ 3 ... ,
12 n '
and also the descending chain of graded submodules of S :
(X1 n S+)~ 3 (X2 n S+)~ 3 ... 3 (X n S+)~ d ... d (Xn S+)~
where the operation ~ is defined in S . Now the only graded R -submodules of S are
SL with p>o,(the proof of this claim is part 1. of the following Theorem 7.5) thus
we may write : (Xns+) =S> and (X,ns+) =S>p , where p1 <p, <—<p. <—<p. It
follows that for some j :p. =p- + 1 = — , whence (X.ns ) = (X.+, ns )
By Proposition 1.2 and Corollary 1.3 we obtain that X- =X. + 1 = .-. Therefore S/X is
an Artinian object of R-mod, whereas by Theorem 3.3. S is a left Noetherian R-module,
so S/X has finite length.
Supposing that D has nontrivial grading, then we have D=D [X,X ,<p] because of
Theorem 6.3. The endomorphism of S represented by 1-X is an injective morphism which

61
is not invertible. We obtain a strictly descending chain of R-submodules of 55 :
S3 (1-X)S=> (1-X)2§3 ... , hence i> is not left Artinian. Therefore K.dimRS = 1 and the
above implies that S^ is 1-critical. □
7.5. Theorem. Let S be simple in R-gr, then :
1. The only graded R -submodules of S (resp. R-submodules of S ) are S^ (resp.
s>.p) for p>o.
2. Any R -submodule of S and R -submodule of S is principal.
3. Any R-submodule of S_ is principal.
4. î> is either simple in R-mod or a 1-critical R-module.
5. The intersection of all maximal R-submodules of ï> is zero.
Proof. 1. Suppose that M = © M. is a graded submodule of S . Let p be the
i>o x
smallest natural number such that M /o, then, by Lemma 7.2., we have that M =S .
Further, the fact that RS = S then yields R+S = ® S. =S> and thus M=>R+M^ =S>
follows.
2. Easy.
3. Let Xcî> be a nonzero R-submodule. If Xns+=o, then (XnS+)_ = o but also X_ns =o.
Indeed, xeh(X_ns ) and x/o means that there is a nonzero y£X such that y =y-i + —
+ y with y1 =x and deg y, <.~ <deg y . Since deg x>o, ye S and hence yeXnS ,
contradiction. X__ is not S since X^nS =o but this contradicts simplicity of S,
hence we may assume that Xns /o and then we have that (XnS ) =S> for some peK
(~ operates in S ). Pick x /o in S . There is a yexns such that y = y-t + •- +y
with y =x and deS y-i <— <deg y . But this means that :
(XnsV3(R+y)~3Rxp = S>p ,
hence (Xns+)~= (R+y)~ and therefore Xns+ =R+y. On the other hand X ns+3Ry ns+3R+y»
whence XnS =Ryns follows. Because Ry/o, (Ry)_ = S=X_ and then Corollary 1.3
entails that X = Ry. □
4 & 5. To prove these statements we need some preliminary facts, which we will
present here as sublemmas.

62
Let M be a graded R-module. M is a left graded module over S = END„(M) = HCM_(M,M) and
the graded ring B^(M) = END„(M) is called the graded bi-endomorphism ring of M.
The canonical morphism p : R-*-B^(M) defined by p (r) (x) =rx, is a morphism of graded
rings.
Sublemma 1. (The Density Theorem).
Let M be a semi-simple object in R-gr (i.e. direct sum of simple objects), let
x^.-.x eh(M) and a€B^(M) homogeneous. There exists reh(R) such that a(x.) = rx.,
1 <i<n.
Sublemma 2. Let L be a minimal graded left ideal of R (i.e. L is simple in R-gr).
2
Then either L =o or L=Re where e is an idempotent homogeneous element of R.
Moreover the ring eRe is a graded division ring. Conversely if R is a semi-prime ring,
e an idempotent homogeneous element of R such that eRe is a graded division ring then
L=Re is minimal graded in R. If eRe is a field (i.e. eRe is trivially graded) then
I is a minimal left ideal of R.
Both sublemmas may be proved in a way similar to the method of proof in the
ungraded case. Now let us return to the proof of the theorem.
4. Since S is a finitely generated R-module we have D=END S= End^S which is a
graded division ring. If p_ is not a field then Lemma 7.4. yields that S^ is 1-critical.
Suppose D=D is a field and let P be the annihilator in R of S. Clearly P is a
graded prime ideal and since S is an R/P-module we may reduce the proof to the case
where P=o.
According to sublemma 1, R is dense in Bg(S) =T. Obviously S is simple in T-gr and S
is isomorphic to a graded left ideal L of T. Since L =L and END (L) ^ENDpfS) =D,
the second sublemma may be applied, yielding minimality of L_ in T, hence S is a simple
T-module. Choose x f o, x e S and y£S; put x = x- + ._ + x with deg x- < ... < deg x .
Now 5; being simple in T-mod, there is an a£T such that y = ax. Let a, a be the
homogeneous components of a. Sublemma 1 provides the existence of homogeneous r. eR
m -1
such that a.x. =r.x., 1 <j =Sn. Then a.x = r-x and hence y = ax = ( £ r-)x, proving
J J ii ^_1 i
that S is a simple R-module.

63
5. By 4. it suffices to consider the case where D = ENDR(S) is a graded division ring
with nontrivial grading. Using sublemma 1 again, it results that any maximal T-sub-
module of ^ is a maximal R-submodule, therefore it will be sufficient to show that
the Jacobson radical J(î>) of ï> over the ring T is zero. Now S is a projective T-module
of finite type and it is well known that in this case :
ENDR (S/J (S) ) = ENDij, (S) /J (ENDp (S) ) = D/J (D),
where J(D) is the Jacobson radical of the ring D. It is easy to calculate that for
any graded division ring D, J(D) =o. So, if J{S) were nonzero then S/J(S) is a
T-module of finite length and together with END (S/J(S)) =D this states that D is a
field i.e. trivially graded, contradiction! □
For modules of polynomials over simple objects we have :
7.6. Theorem. Let R be an arbitrary ring, S a simple left R-module. Consider the
left module S [X] over the ring of polynomials R [X] ; then we have the following
properties :
1. Any R [X] -submodule of S [X] is principal.
2. S DO is 1-critical.
3. The Jacobson radical of S [X] (over the ring R [X] ) is zero.
Proof. 1. Consider S [X] and R [X] as graded objects with the obvious grading. The
graded submodules of S[X] are of the form ® SX1, hence principal. By Proposition
i>p
1.5.8 (3), statement 1 follows.
2. It is easy to show that S [X] is 1-critical in R [X ] -gr. From Proposition 4.5 we
then deduce that S[X] is 1-critical in R[X]-mod.
3. Put D=ENDR(S), Bg(S) =ENDD(S). The ring Bg(S) is a regular ring in the sense
of von Neumann. As a Bg(S) module S is isomorphic to a minimal left ideal L of Bë(S).
The canonical morphism p : R->Bg(S), p (r) =<f where <t (x) =rx, extends to a morphism
R[X] -+Bg(S) [X] . Now S [X] is Bg(S) [X]-module isomorphic to L [X] , which is a direct
summand in Bg(S) [X] . Furthermore, Bg(S) being regular, a classical result of Amitsur

64
yields that the Jacobson radical of Bg(S)[X] is zero, hence the Jacobson radical of
R [X] is zero, The proof of 3 may now be finished by showing that maximal Bg (S) [X] -
submodules M of S [X] are also maximal R [X]-submodules. Pick
s(X) =sq + s1 X+._ +s. X3" eS[X]-M . Then we have : M + Bg (S). s (X) = S [X] . Let
b+b.,X+._+b XmeBg(S) [X] , write (b +b1X+._+b Xm) (s + s, X + .- + s -X^) =
o1 m v. -i i j » *- o 1 m-^ol j J
s b + (b s. +b,s )X+ (bs, +b1s1 t^s )X + .„ +b s.Xm+J. Because of the density
oovo1lovo2 112o m j '
theorem there exists a .a.,,....a £R, such that a s. =b s-, a-s. =b1s.,...,a s. =b s.
o' T ' m ' 01 o 1' 1 1 1 1 m 1 mi
for each o < i < j . Thus Bb (S). s (X) = R [X] . s (X), hence M + R [X] . s (X) = S [X] , proving that
M is a maximal R [X] -submodule of S [X] . □
7.7. Corollary. If MeR-mod has finite length n then every R [X] -submodule of M [X]
may be generated by less than n elements.
Proof : We proceed by induction on n. If n =1 then M is simple and the assertion
follows from the foregoing theorem. Suppose now that the statement is true for n
and let M have length n+1. Consider a maximal left-submodule M, of M. From the
exact sequence in R-mod :
o—*1L—«-M-i-M/M.,—>o
we obtain an exact sequence in R [X] -mod :
o->-M1 [X]—>M[X] —*M/M1 [X] >o .
If N is an R [X] -submodule of M[X] then NnfL [X] is generated by less than n elements,
whereas N/NnM [X] is a submodule of M/M., [X] and may therefore be generated by a
single element. Consequently N may be generated by less than n+1 elements. □
Let D be a graded division ring, then we denote by d(D) the minimal positive
i such that D. /o or d(D) =0 if D. =0 for each i/o. Obviously D has trivial grading
if and only if d = o.
7.8. Proposition. Let R be a graded ring, SeR-gr a simple object. Then :
1. S is a semisimple R -module.
— o

65
2. Put D = ENDD(S), then the number of isotopic components of the R -module ï> is
K O
less than or equal to d.
Proof : 1. follows from Lemma 7.2.
2. Let aSD,, a^o. The mapping a :S-*-S is an R-automorphism of degree d, so for
any n£Z, the mapping S -'■S, sending x£S to a(x) is an R -isomorphism, whence 2
follows.
7.9. Example. Let K be a trivially graded field and let V = ® V be a graded
n>o n
K-vector space, where Vn = K(n) ®... ®K(n), n+1 terms. Let R be ENDKV with KCRq,
Clearly V is a simple object in R-gr. and as V is a graded R-module which is left
limited it results that V is a simple R-module. If n/n' then dinuV/dirruV, and
thus V #V , in R -mod. Consequently as an R -module V has infinitely many
isotopic components.
II.8. THE JACOBSON RADICAL OF GRADED RINGS.
Let R be a graded ring, Me R-gr. A graded submodule N of M is said to be
(graded) maximal in M if M/N is a graded simple R-module. Clearly N is maximal in
M if and only if N is proper and for all xeh(M), N + Rx=M. Of course, if N is
maximal in M, N need not be maximal in M.
8.1. Lemma. Let S be simple in R-gr. There exists a left graded ideal I of R which
is maximal and an integer n<=Z, such that Ss(R/I)(n).
Proof : Let x/o, xeh(S) with deg x=m. Then the morphism a :R-»-S(m) defined by
a(r) = rx is of degree o. Since S(m) is simple, a is surjective i.e. S(m) =R/I with
I maximal in R and thus S = (R/I) (-m). □
8.2. Definition. If Me R-gr then the graded Jacobson radical of M, denoted by Jg(M)
is the intersection of all maximal graded submodules of M, and we define Jë(M) =M if
M has no proper maximal graded submodules.
Jg enjoys properties which closely ressemble the corresponding ungraded

66
properties, let us just mention :
8.3. Lemma. Let R be a graded ring, Me R-gr. , then :
1. If M is of finite type then Jg(M) /M or M = o.
2. Jg(M) = n{Ker f, feHomR_ (M,S), S simple object in R-gr}
= n{Ker f, f HOMj^fM.S), S simple object in R-gr}.
3. If feHOMj^CM.N) then f(Jg(M)) cjg(N).
4. Jg(RR) = n(AnnRS, S any simple object in R-gr}.
5. Jg(RR) is an ideal of R.
6. Jg(DR) is the greatest graded proper ideal satisfying : if aeh(R) is such that
the image of a in R/Jg(RR) is invertible then a is invertible in R.
7. Jg(RR) =Jg0y.
In view of 5. and 7. in Lemma 8.3 we will denote by Jg(R) = Jg(RR) = jg(.\) the
graded Jacobson radical of the ring R.
8.4. Lemma. (Nakayama's lemma).
Let lie R-gr be of finite type, then Jg(R)M/M.
8.5 Corollaries. 1. Let J(R) denote the Jacobson radical of R, then J(R) cjg(R).
2. Let R be positively graded, then Jg(R) =J(R ) ®R+-
Proof : 1. Directly from Theorem 7.5, 5..
2. The ideal R = ® R. has the property that RM/.M for every finitely generated
i>o x
graded left R-module M. So if SeR-gr is simple, R+S = o and R+cjg(R) follows. More-
we have seen that S=S„ for some n eZ and S is a simple R -module. Hence
n o n r o
°
J(R )Sn =o and thus J(R ) +R+cjg(R). On the other hand Jg(R) =JQ®J>0 i.e.
o
Jg(R) = J ®R+. Consider a simple R -module J_ and consider SeR-gr with S =T, S^ =o
if i/o. Since S is simple in R-gr : J„T=J S=o, hence J cj(R ). n
8.6. Theorem (Hopkins' theorem).
Let R be a graded ring which is left gr.Artinian, then R is a left Noetherian graded

67
ring.
Proof : Formally similar to the ungraded case one shows first that R/Jg(R) is a semi-
simple object of R-gr and that there exists a positive n such that (Jg(R))n = o. Then
for any i, 1<i<n-1, (Jg(R) ) V (Jg (R) )1+1 is annihilated by Jg(R) hence an R/Jg(R)-
module. It follows that (Jg(R))1/(Jg(R))1+ is semisimple in R-gr and because it is
also left Artinian it has to be of finite type, hence left Noetherian. It is then
p
also obvious that R is a Noetherian object in R-gr. □
8.7. Remark. If R is left gr.Artinian then R is not necessarily left Artinian. For
example k[X,X ] with non trivial grading according to the degree in X.
II.9. SEMISIMPLE GRADED RINGS. Goldie's Theorem for Graded Rings.
A graded ring R is said to be semisimple if R-gr is a semisimple category i.e
any object of R-gr is semisimple. It is clear that R is semisimple graded if and
only if : (*) R = L1 ®- © L , 1^ , 1 <i<n, being minimal graded left ideals of R. The
graded ring R is said to be simple if it has decomposition (*) but with HCMnfL.,L.)^o
for any i,j =1,.»,n. The latter condition means that for any couple (i,j) there
exists an integer n^. such that L. = L.(n..) in R-gr. The graded ring R is said to be
uniformly simple if R admits the decomposition R = L, S
ideals but with L. =L. for any i,j.
5L in minimal graded left
9.1. Graded Matrix Rings.
If R is a graded ring, ne IN , then M (R) denotes the ring of n by n matrices
with entries from R. Fix an n-tuple of integers (d,,... ,d ), which will be written as
d. To any xez, associate the following abelian group of matrices.
Mn(R)x(ïï) =
\
Rx+d2"dl
Rx+d1-d2 \
>+d -d,
n 1
Vd -d .
n n-1
RX+d1-d
1 n

68
It is obvious that M (R) = ® M (R). (d). In this way we obtain a graded ring
n xez n
M (R) (ïï) ; note that M (R) (o) will be denoted by M (R) as there will usually be no
confusion whether the ring is considered as a graded object or not. With these
notations we have :
9.1.1. Lemma. 1. For anymez, M (R) (d) = M^(R) (d+in).
2. M (R) (d) sM (R) (a (d) ) where a is a permutation of d.
3. If M is a free graded R-module with finite basis e-,,... ,e then ENDD(M) =M (R) (d)
in k n
where d. = deg e., i = 1,.»,n.
4 Let D be a nontrivially graded division ring. Then for any system of integers
d^,.-,d' there corresponds an n-tupel d^.-.d such that o<d1 <d7<.„ <d <d(D) and
M (D)(d') ^M (D) (d), where d(D) is positive and minimal such that D,™ /o.
9.1.2. Theorem (Wedderburn's Theorem).
Let R be a graded ring. The following statements are equivalent :
1. R is graded simple (resp. uniformly simple).
2. There exists a graded division ring D and dez such that RssM (D) (d) (resp.
R=Mn(D)).
Proof. As in the ungraded case. □
In connection with this the problem arises to determine the number of
isomorphism classes of simple graded rings of seize n over a given graded division ring D.
The following proposition will solve this problem. Let us denote d(D) =L and let
X » be the set {ïïez11, o=Sd, <d9<... <d <£}. On X „ we introduce the following
n j-C i ù n nj-t-
t
equivalence relation : we put d~d if and only if there exist t,q1,... ,q^ <=Z and
osSn such that d^t =^+d'an) » i = 1»-»n. Let C (D) be the quotient set X J~
and it is clear that this is a finite set (as I is finite), so let c (D) be the
cardinality of C (D).
9.1.3. Proposition. Let D be a graded division ring with I = d(D) >o and let n be
fixed natural number. The number of isomorphism classes of graded simple rings of

69
size n over D is exactly in c (D).
Proof : From Lemma 9.1.1. it follows that cT-cF" yields M (D)(3) as M (D) (cT). Conversel>
suppose that \(ÏÏ) (3) =M (D) (d7^). Consider the graded D-modules : V = D(-d ) ®...
.- ©D(-d ), W = K(-dp $ ... «K(-d'). Then END (V) = ENDD(W) . Since D is principal we
may use a graded version of Theorem 1.6 p.37 in [ 19] , to obtain the existence of :
an isomorphism of degree o, a : D-»-D, an integer p and the a-isomorphism of degree
o : f :V(-p) -*-W. Therefore we get an a-isomorphism
Df-d^p) ®... ®D(-dn-p) =D(-d]) ®... ®D(-d').
Choose now d"ezn such that H+p=Iq + cF and with d'.'<£ for all i = 1,.-,n. Then for
any integer m there is a D-isomorphism of degree o; D=D(-&i), hence, due to the
existence of f, an a-isomorphism of degree o :
D(-dV) ®.~ ®D(-d") =D(-d') «... ®D(-d').
Since D1 = D ■
:D„ =o there exists o<^S such that d'.'=d' .. 1 <i<n. □
-t-1 n i a(i)
9.1.4. Example. Let D be a graded division ring. If d(D) =1 then c (D) =1. If
d(D) =2 then cJD) = [§]+!.
9.1.5. Example. Let D be a trivially graded division ring and take n>2. Let
m>o be an integer and consider the n-tupel (o,o,._,o,m) =ïiî. Let R. = o for all X /o,
m and put :
D
D
D
0 0
DO
DO
D 0
0 D
0 ..
D D
0
.. 0
D 0
Notice that there exists an infinity of classes of simple graded rings of size n
over D.

70
9.2. Rings of Fractions and Goldie's Theorems.
Recall that if R is any ring, S a multiplicative!/ closed set in R, then the
left ring of fractions S R with respect to S exists if and only if R and S satisfy:
i) If s es, r£R are such that rs = o then there exists an s' es such that s'r = o.
ii) For r£R, s<=R there exists r'£R, s'ES such that s'r = r's.
These conditions are known as Ore's conditions on the left; usually, e.g. if R is
left Noetherian one only has to check ii).
9.2.1. Lemma. Let R be a graded ring, S a multiplicatively closed subset of R
consisting of homogeneous elements. Then S satisfies Ore's conditions on the left if
and only if :
i) If rs=o with r£h(R), s^S then there is an s'£S such that s'r=o.
ii) For any reh(R), s^S there exist r'eh(R), s'eS such that s'r = r's.
Proof : Obviously the Ore conditions for S imply i) and ii). Conversely, consider
s £S, rER, write r = r. + ._ + r with deg r-, < ._ <deg r , r. /o i = 1,._ ,n.
If n = 1 then the proof is finished. Now we proceed by induction on n, so there
exist r1eR, s1eS such that : s1 (r, + ._ +r J =r1s. Pick s2eS, r2eR such that
I n-1
s r =r s. Consider u^S, v^h(R) with us =vs =t and put ur +vr =w, then tr-ws.
For the second statement suppose as = o with a = a, + ... + a , s es. Thus (a., + ... +a ,)s=c
and a s =o. The induction hypothesis yields the existence of t, es such that
t,(a., + ...+a i) =o. However, since t,a s=o also, there exists t?£S such that
tot.,a_=o, hence, putting s' -t,t2 we get s'a = o. □
Let R be a graded ring and S a multiplicatively closed subset of h(R)
satisfying Ore's conditions. Define a gradation on S R by putting :
(S_1R)X ={a/s,aeh(R), s€S; deg a - deg s=X} .
It is easy to see that S R thus becomes a graded ring. In a similar way, for any
M<=R-gr, the graded module of fractions S M is obtained by equipping s M with the
grading : (S M)x ={m/s, meh(M),seS, deg m - deg s=X}. Results about these
constructions follow also from the general theory of graded localizations in II.

71
A graded ring which is of finite Goldie dimension in R-gr and which satisfies
the ascending chain condition on graded left annihilators is called a graded Goldie
Ring. If R is trivially graded then we get the usual notion of Goldie ring, cf. p1 ] .
Let us mention Goldie's Theorem : a ring R has a semisimple (simple) Artinian
classical ring of fractions if and only if R is a semiprime (prime) Goldie ring.
Exactly as in the ungraded case one can prove that, if R has a simple (semisimple)
Artinian graded ring of fractions then R is a prime (semiprime) Goldie ring. However
the converse is not true, as the following shows :
9.2.2. Example. Let k be a commutative field, R a graded ring such that R =k [X],
R" =k[Y] and XY=YX=o, i.e. R =kXn, R =k. R = kYn for n>o.
' n ' o ' -n
It may be easily verified that R satisfies :
i) Every xeh(R) with deg x/o is non-regular.
ii) The ideal (X) + (Y) is essential in R and it does not contain any regular
homogeneous element.
The ring R is a semiprime graded Goldie ring but it has no semisimple graded ring of
fractions. Still we may prove the following :
9.2.3. Proposition. Let R be a semiprime graded Goldie ring which satisfies one of
the following properties :
a. R has a central regular homogeneous element s with deg s>o.
b. R is positively graded and minimal prime ideals of R do not contain R+.
c. R is positively graded and R has a regular homogeneous element of positive degree.
d. Homogeneous elements of R of positive degree are nilpotent.
Then R admits a semisimple Artinian ring of fractions.
Proof : Let L be a nonzero graded left ideal of R. Under the hypothesis a,b,c, it is
easy to establish that L contains a homogeneous element of positive degree which is
not nilpotent. Under the hypothesis d, it follows that L contains a non-nilpotent
element of degree o.
The essential part of the proof is to show that a left graded essential ideal contains

72
a regular homogeneous element. Under the assumptions a,b,c, we find that there is
an a- eh(L) with deg a. >o and aV fo v = 1,.- ,.- . For some n e IN : Ann (a-) =
AnnCa1?4"1) = .-. Put b. =&}, then deg b >o and AnnfbJ = AnnCb^). If Annfb^ fo then
there is an homogeneous b-eLnAnnfb..) with deg b2>o and Ann(b,) =Ann(b2) ,v = 1,2,... .
Proceeding in this way we obtain a direct sum Kb. ®Rb2®._ of nonzero graded left
ideals and since the Goldie dimension of R is finite, there exists an re IN such
that :
Annfb..) n ... nAnnfb ) =o .
d/d d/d
Let d- = deg b- >o, 1 <i«r and d = d-._d . Then c=b- + ... + b is a homogeneous
l 6 i ' 1rr rr
element of L with deg c = d. However Ann c = n Ann(b- i)= n Ann(b-) =o; so Re is
i-1 i=1
an essential graded left ideal, c is a regular element and deg c>o. Under the
assumption d one may easily prove that L contains a regular homogeneous element of
degree o. From here on the proof will be similar to the proof of Goldie's theorem's
in the ungraded case. □
9.2.4. Corollary. Let R be a left and right Noetherian graded ring which is either
positively graded or commutative. Let P be any prime ideal of R then either :
ht(P) =ht((P) ) or ht(P) =ht((P) ) +1 (where ht(-) denotes the height of the ideal
mentioned).
Proof : Suppose that Q is a prime ideal of R such that (P) cqcp. Up to passing to
R/(P) we may assume that (P) =o and hence that R is a prime ring.
Let S be the set of regular homogeneous elements of R. Clearly, R being Noetherian
prime and graded, R is a graded Goldie ring hence, the foregoing proposition implies
that S R exists and that it is a simple Artinian graded ring. From our results on
the Krull dimension we retain . K.dim S R<1. On the other hand S" Q and
-1 -1 -1 -1
S P are nonzero prime ideals of S R, therefore S Q = S P. This means, becauxe
of the right Noetherian property that sPCQ for some ses, s ^o hence P = Q or s e Q;
would imply s e (P) =o the latter is impossible hence P=Q is the only remaining
possibility. NowifP=(P) thenht(P) ) =ht(P) ) so suppose that P f (P) and assume

73
that ht(P) =n<°°. For n = 1, (P) is a minimal prime ideal, i.e. ht(P) = o. We pro-
ceed by induction on n. Let P cp c ._ cp =P be an ascending chain of length n of
prime ideals contained in P. By the induction hypothesis we have ht(P _..) =n-1 if
p 1 = (p„ 1) but in this case (P) =P , and also ht(P) =ht((P) ) +1. On the other
n-1 n-rg v Jg n-1 ^ J vv JgJ
hand if Pn_1 Î (Pjj.^g then ht(Pn_.,) =ht((Pn_1)g)+1. Now (P)g + (Pn_.,)g since
otherwise P . would be properly contained in P and P . 3 (P) , so ht((P) ) >ht((P *) ).
n-1 f f j n-1 wg' g n-rgJ
This ht((P) ) >n-1 and ht(P) <ht((P) ) + 1. D
9.2.5. Corollary. Let R be a Noetherian commutative graded ring. If P is a graded
prime ideal with ht(P) =n, then there exists a chain of graded prime ideals :
P ÇP &••• cp = p-
o^1^ n
Proof : The statement is true for n = 1. If ht(P) =n>1 then there is a chain of
distinct prime ideals : 0 ÇO, ç .- en =P.
If 0 _1 is graded then application of the induction hypothesis finishes the proof.
Suppose that 0 _1 is not graded and then the foregoing corollary yields that
ht((0 _1) ) =n-2. The induction hypothesis asserts that we may find a chain of
distinct graded prime ideals P cpc.„cp = m ) . Consider the graded ring
R/(0 _..) =R and denote by PjCLi the image of P, 0 -, in R. Choose a nonzero
homogeneous element a of P, hence lï£ÏÏ _... Let V . be a minimal prime ideal containing
Rlî. Then F . is graded and V _1 ^V because otherwise ht(P) <1 follows from the
principal ideal theorem. Putting V _. =P _1/((L_1) we find a chain of distinct
prime ideals ; P cp,c,.cp ,cp =P. n
^ o 1 n-1 n
9.2.6. Corollary. Let R be a Noetherian commutative graded ring and let I be an
ideal of R, then :
ht(I)<ht(f) , ht(I) <ht(IJ.
Proof : Since ht(I) =ht(rad(I)) and (rad(I)) Crad(I ) (see Lemma 1.1.) we may reduce
the proof to the case where I is a semi prime ideal, I=P. n... np , p. prime ideals of
R. Then P...P ci yields P~.P~...P~c i~ (Lemma 1.1.). Let Q be a prime ideal

74
containing I~ such that ht(I~) = ht(Q). For some i, P^CQ, but IcP. hence we may
assume that I =P is a prime ideal in proving the statement. If P is graded then
P=P~. If P is not graded then ht(P) =ht((P) ) +1. Clearly, (P) cp~ and (P) ff.
It results from this that ht((P) ) <ht(P~) and therefore ht (P) < ht (P~). In a similar
way, ht(I)<ht(IJ may be established. □
11.10. PRIMARY DECOMPOSITION
Let R be a graded ring, P a graded ideal of R. It is obvious that P is a
prime ideal of R if and only if for a,beh(R)-P, aRb£P. We write spec R, spec R for
the set of prime ideals of R, resp. graded prime ideals of R. In view of Lemma 1.5.,
the minimal prime ideals of R are graded and the radical of a graded ideal is also a
graded ideal. If M is a nonzero R-module the a prime ideal P of R is said to be
associated to M if there exists a nonzero submodule M' of M such that P =Ann M' =
Ann M" for every nonzero submodule M" of M1.
Let Ass M be the set of prime ideals of R which are associated to M. In case R is a
commutative ring then Pespec R is associated to M if and only if there is an x^M,
x^o, such that P=Ann x. A well known result in ring theory states that, if R is a
left Noetherian ring and M^o eR-mod, then Ass M^o.
10.1. Theorem. Let R be a graded ring, M nonzero in R-gr.
Suppose that peAss M, then :
1. P is graded.
2. There is an xeh(M), x/o, such that P =Ann Rx.
3. If R is a fully left bounded left Noetherian ring (cf. [ 36] for definition and
properties of these rings) then there exists xeh(M), x^o, such that P=Ann Rx =
Ann M' for any nonzero submodule M' cRx.
Proof : 1. By hypothesis P =Ann M' for some nonzero R-submodule M' of M. Take X ep,
x eM' and write X = 2' A-, x= 2' x.. Let m, resp. n, be the largest integer such
i ez j ez J
that X ^o, resp. x fo. Pick aeh(R). Since Xa^P we have ax = o, hence X ax =o,
implying X Rx =o.

75
Let beh(R). From Xbx = o we deduce that, i bx , +X ,bx =o. Replacing b by
1 m n-1 m-1 n ±- & /
bX a in the above equality yields X bX ax , =o, hence (X R) x , =o. The number of
m n ' ' mm n-1' v m n-1
nonzero components of x being finite, it follows that (X R)P x- =o for all x- fo
P P
appearing in x, for some p£Z. Consequently (X R) Rx = o, thus (X R) cp and X Rep.
Thus X-X ep and repetition of this procedure yields that X. ep for all X. appearing
in the decomposition of X i.e. P is graded.
2. Take x^o^M', Then P=Ann Rx. Set x=x, + ... + x, where x. 4o for all j,
— 1 k j •"
deg x1 <... <deg Xj..
Let J. be Ann Rx-, 1 <i<k. It is easy to verify that P=J-i n ••• nJ,., thus P=J. for
11 I K 1
some i, therefore P=Ann Rx-.
3. According to 2 we can find xeh(M), x^o, such that P=Ann Rx. Put N = Rx. By our
assumption N is a faithful R/P-module, where R/P is a fully left bounded left
Noetherian ring. By [19] , Lemma 2.1 and Proposition 1.4., it follows that N is a
nonsingular R/P-module. However N is graded hence there exists yeh(N), y^o, such
that Ann„ ,„ y is not an essential left ideal of R/P, meaning that there may be found
a graded left ideal isomorphic to a graded submodule A of N. Let z£h(A), z^o.
Clearly, P=Ann„Rz =Ann„M' for any nonzero submodule M' of Rx. □
Let MeR-mod; let M[X] be the R[X] -module of polynomials with coefficients
in M. We have M [X] =R[X] ®M. We know that : M[X] . = {mX1,meM}, defines a grading
R x
on M [X] .
10.2. Theorem. Let R be a graded ring, MeR-mod. Then :
AssR[x]M[X] ={P[X] , PEAss M} .
Proof : If PeAss M then P =AnnJl' for some M' cm. Then it is clear that P [X] =
AnnRfXlM' ^ " Let m'xV be a nonzero element of hpl1 [X] ). Since P =AnnRRm' it
follows that P[X] =AnnRrX]R[X] .m'Xv. Hence P [X] eAssRK1M[X] ■ Conversely, let
QeAsSp py-,M[X] . Theorem 10.1.,2., implies that there exists a nonzero homogeneous
element mXv with Q=AnnRK,R[X] .mXv. Put P =AnnRRm. It is obvious that Q = P[X] ,
P is prime and PeAssnM. □

76
Let NCM be R-modules. We say that N is a primary submodule of M if Ass M/N
is a singleton, and N is said to be a P-primary submodule of M if Ass M/N ■ {P}. If R
is Noetherian commutative and M is of finite type then N is a primary submodule of M
if and only if N is a classical primary submodule, (cf.[ 3],ch.4).
10.3. Lemma. Let R be a left Noetherian ring, MeR-mod and Pespec R. The following
statements are equivalent :
1. 0 is a P-primary submodule of M.
2. The set (0 :P)={xeM, Px=o} is an essential submodule of M and P contains any
ideal which annihilates a nonzero submodule of M.
Proof : 1 =*2. If X is a nonzero submodule of M then Ass X = {P} implies that
Xn (0:P)M^o. If I is an ideal which annihilates a nonzero submodule M' cm. Since,
Ass M' = {P}, there is an M"CM' with M" fo and P =Ann M". However LM' =o yields
IM" =o, hence icp.
2*1. Since R is left Noetherian, Ass M^Q. Let QeAss M. From our assumptions it
follows that Qcp. Consider M' fo, M1 cm, with Q=Ann M\ The fact that M'n(0:P)M^o
entails : P cAnnpl1 n (0:P),,) =Q. Therefore P = Q. □
10.4. Remark. If R is a graded ring, MeR-gr then in Lemma 10.3 we may replace
"ideal" in statement 2 by graded ideal.
10.5. Lemma. Let R be a graded ring and MeR-gr. Let NCX be submodules of M. If
X/N is essential in M/(N) then X~/(N) (resp. X_/(N) J is essential in M/(N) . In
g b b b
particular, if X is a left essential submodule of M then X and X_ are left esstenial infr
Proof : Take xeh(M)-(N) and let x be its image in M/(N) . Since x£N, its image x
is M/N is nonzero. By the assumption we find XeR such that Axe X/N and Xx^o.
Writing X =X. + ... +X with deg X. < .„ <deg X we find that Xx=X.x+ ... +^x and we
may assume that ^-x^N because otherwise we may replace X by m =^~^n while Xx=Xx.
Since X x^N we have X x^o. On the other hand, Xx eX/ii yields XxeX, hence ^x^X .
But X x£(N) , therefore X xeX~/(N) and X x^o. Lemma 1.3.3.13. finishes the
n r *■ Jg' n g n

77
proof. The statement about X^/(N) follows in a similar way. □
10.6. Theorem. Let R be a left Noetherian graded ring, MeR-gr and N a P-primary
submodule of M. Then (N) is (P) -primary submodule of M.
Proof : Set X/N= (0:P)w,„; clearly this is left essential in M/N. By the foregoing
lemma we deduce that X~/(N) is essential in M/(N) . (P) X~c(N) follows from PXcn,
hence X /(N) c (0: (P) ) .,, . and therefore the latter left module is essential in
M/(N) . Consider an ideal I which is graded and such that it annihilates a nonzero
Y/(N) , i.e. IYC(N) . Since I is graded : IY~ c (n) and since Y f (N) we have
(N) fX i.e. Y /(N) fo. Therefore, we may assume that I annihilates the graded
nonzero submodule Y/(N) of M/(N) • As before we may deduce that IYc (N) i.e. IYCN.
b b b
The fact that Y is graded yields YjZlN and we arrive at the conclusion that
I(Y+N/N) =o with Y+N/N^o. By hypothesis I cp and I c (P) follows. Then apply
Lemma 10.3. □
10.7. Corollary. Let R be a left Noetherian ring and M a nonzero left R-module.
The o is a P-primary submodule of M if and only if o is P [X] -primary in M [X] .
Proof : Follows from 10.6 and 10.2. □
A finite family (N.) .of primary submodules of M is a primary decomposition
for the submodule N of M if N = n N-. The decomposition is said to be a reduced
iel x
decomposition if the following requirements are fulfilled :
i) n N. £N. for all iel.
ii) If Ass M/N- =P. then the ideals P. are mutually disjoint.
For more detail about primary decomposition cf. [ 3 ] , chapter 4. It is a well known
result that, if R is left Noetherian and if MeR-mod is of finite type, then any sub-
module of M admits a reduced primary decomposition.
10.8. Corollary. Let R be a left Noetherian graded ring, M a graded R-module of
finite type, N a graded submodule of M such that N= n N- is a primary decomposition
iel x

78
of N in M. Then we also have the following :
1. N = n (N-) is a primary decomposition of N in M.
iel x g
2. If the primary decomposition N= n N. is reduced then the decomposition
iel x
N= n (N-) is reduced and for any iel, Ass M/N- =Ass M/(N-)g-
iel x &
Proof : 1. Follows from Theorem 10.6.
2. By Theorem 10.6, Ass M/I^ =Ass M/(N-) for all iel.
Suppose that n (N.) c (N.) , then N = n (N-) . There exists an injective homomor-
j ti -1 g 1 g jH -1 s
phism, if :M/N-> ® M/(N-) » thus we may conclude that :
j# J g
Ass M/NcAss $ M/(N.) = u Ass M/(N.) = u Ass M/N- ,
jYi J g j^i J g jYi J
contradiction.
Hence the decomposition N= n (K-) is reduced. □
iel 1 S
By a classical P-primary submodule of M eR-mod we mean a submodule N of M
which is P-primary and has the property that PpIcn for some natural number n Theorei
10.6 implies that if MeR-gr, and N is a classical P-primary submodule of M, then
(N) is a classical (P) -primary submodule of M. Corollary 10.8 still holds if we
replace "primary" by "classical primary". That, for a left Noetherian ring R and an
R-module M of finite type it is not true in general that 0 has a classical primary
decomposition in M, is just old hat.
Finally let us point out that all results in this section remain valid if R
is a graded ring of type A where A is any commutative torsion free group.
11.11 EXTERNAL HCMDGENIZATION.
Let R be a graded ring. The ring of polynomials R[T] may be turned into a
graded ring by putting : deg T = 1, R[T] - { 2 r- T-1 ,r- eR.}. In the same way we
i+j=n
build the module of polynomials M[T] starting from an MeR-gr. It is easily checked
that M[T] is an R[T]-module and that M[T] =R[T] ®M, where the tensor product is
R

79
graded in the usual way. If we decompose xeM into homogeneous components,
x = x_ + ... +x + ... +x , then we may associate to it a homogeneous element x of M[T]
which is given by :
x*=x T^ + x, r^^xf^^x .
-m 1-m o n
We say that x is the homogenized of x. If a is a homogeneous element of M [T] , say
u=u , Tk+J + ._ +u TJ + ._ +u., with u- eM, -k<i<j, then u. =u , + .- +u + ... +u- eM
-k o 2 1 * -k o j
is said to be the dehomogenized of u. We have the following :
i) If f £R[T] , u£M[T] are both homogeneous then (fu)A = f^uA .
ii) If u,veh(M[T]) have the same degree then (u+v)A =uA+vA.
iii) If xeM, (x*)A =x.
iv) If u£M[T] is homogeneous, then (uA) T =u, where k = deg u-degfu^) .
If p eR, inEM then d(p) or d(m) will denote the highest degree appearing in
a homogeneous decomposition of p or m.
* * k *
v) p m =T (pm) , where k = d(p) + d(m)-d(pm).
vi) Let x,y£M and suppose d(x)>d(y), then : T (x+y)* = x*+Ty , where £ = d(x)-d(y)
and k = max{d(x) ,d(y) }-d(x+y).
If N is an R-submodule of M then by N we mean the submodule of M[T]
generated by the n , n£N. Of course N is a graded submodule of M [T] , it is called
the homogenized of N. Any n£N is of the form T n , n£N, v>o.
To a graded R[T] -submodule L of M[T] we may associate LA = {ujt,u£h(L)} and from the
properties mentioned above it follows that LA is an R-submodule of M. The
correspondence N-»-N satisfies :
1) (N*)A=N.
2) If L£N then L* £N*.
3) If N is a graded submodule of M then N =N[T] .
4) If I is a left ideal of R then (IN)* = I*N*.
5) If x£M then (N : x)* =N* : x*

80
6) ( n N.)* = n N*.
iel x iel
The correspondence L"*LA satisfies :
1) (LJ*=>L.
2) If LCL' then L^cL^.
3) ( n L.)A= n (L )A.
EJ x* iej x*
4) If L is a graded left ideal of R[T] then (JL)A = J^.
5) If u£M[T] is homogeneous then [L:u]A=LA :uA.
6) ( 2 L)A= £ (L ) .
iej x iej x
From this it is clear that we may define a functor,
E : R[T]-gr ->R-mod ,
such that E(M) =R[T]/(T-1) ® M = M/(T-1)M.
R[T]
11.1. Lemma. With notations as above we have :
1. E is an exact functor.
2. If N is an R-module (of finite type, of finite presentation then there exists an
MeR[T]-gr (of finite type, of finite presentation) such that E(M) =N.
3. If R has negative grading and if P is a projective R-module, then there is a
projective object Q in R[T] -gr such that E(Q) =P.
Proof : 1. Right exactness of E is obvious. Take MeR[T] -gr and let M be a graded
submodule of M. For left exactness of E it will be sufficient to have that
M' n (T-1)M= (T-1)M'. Pick x£M' n (T-1)M, then x = (T-l)mSM' with meM.
m=m + .„ +m + ... +m with m.,-t<i<s, being the component of degree i of m. From
(T-l)meM' we get :
-m . +Tm . -m _. +Tm ^. - .- -m +Tm £M' .
-t -t -t+1 -t+1 s s
Because M' is graded, m eM1; then m - eM' and so on till we find that meM' i.e.
xe(T-1)M\

81
2. Write N = F/L where F = rA ' is a free R-module equipped with the grading induced by
the grading of R. The R[T]-module F [T] is free graded. If L is the homogenized of
L in F[T] then we put M = F [T] /L*. One easily verifies that M/(T-1)M = N.
3. Let P=F/L be projective i.e. F=L®K with P = K. Since R is negatively graded
F. = o for all i >o. Take the canonical basis (f.). . of F, i.e. f. = (o,._ ,1 ,o,.- ,o),
then (f.). , also is an R[T] -basis for F [T] . Since F =L©K we may write u. = x. +y-
with x. eL, y. eK, However deg u. =o and F. =o for i>o yields that either d(x.+y.)=
d(x.)=o or d(x.+y.) =d(y.)=o. in both cases (x.+y.) =x- +y. and hence u. =x. +y..
^ iJ v i 'i wi v i 'i/ l 'i ii'i
Thus L* + K* =F* =F[T] . Since L* nK* =o it follows that F [T] = L* © K* and therefore,
if QSF[T]/L* then Q is projective in R[T]-gr while E(Q) =F/(L*)A =F/L = K = P. □
Note. If 3. holds for positively graded rings then it would be possible to derive
from it an easy proof for Serre's conjecture and it would also yield an affirmative
answer to the Bass-Quillen conjecture. Again from this one can deduce that the
positively graded analoge of 3, must be false for non-commutative R. For commutative
R the question remains open; an affirmative answer would entail deep results in K-
theory.
11.2. Theorem. Let R be a graded ring, then :
gr.gl.dim R<gl.dim R<1+gr.gl.dim R.
If R has a limited grading then :
gr.gl.dim R = gl.dim R .
Proof : 1. The inequality gr.gl.dim R<gl.dim R is just Corollary 7.7. Now if
M£R[T]-gr then we have the following exact sequence in R[T]-mod :
o - M [T] -i-* M [T] -Î-* M — o
where i and j are homogeneous, i is multiplication on the right by T-1. Consequently,
gr.gl.dim R<1+gr.gl.dim R. If NeR-mod then let MeR[T]-gr be such that N^E(M)-
Exactness of E entails that gr.dirru N<gr.p.dim„m M<gr.gl.dim R[T] hence

82
gl.dim R< gr.gl.dim R.
2. This part follows directly from Corollary 7.7. □
11.3. Theorem. Let R be a left Noetherian positively graded ring and assume that
gl.dim R<». Let P e R-mod be projective and of finite type, then there exist projec
tive R -modules of finite type Q and Q' such that :
P e(R » Q) ^R 8 Q' .
R R
o t>
Proof : Choose MeR[T]-gr such that E(M) = P, because of 11.1 M may be chosen to be
of finite type. Since gl.dim R<°° we obtain the following exacts sequence in
R[T]-gr.
(*) 0-^-*...-^-^-
►M-
where 0 ,Q-._Q are projective and of finite type in R[T]-gr. The grading of R[T]
is positive so Nakayama's lemma may be applied to deduce that any projective object
of finite type in R[T] -gr is of the form R [T] <& S where S is a projective R -module
R °
of finite type. Substituting Q. =R[T] ® S. in (*) and then applying the functor E
to the sequence, right exactness of E yields an exact sequence in R-mod :
dn dn-1 d2 d1 do
o —ECcy—-ECQ^.,) ►.. — E(Q.,) — E(Qo) — P-0
where E(Q.) =Q./(T-1)Q. =R ®S .
%
By the projective of P we retain from all this that : P©Ker d =R ® S ;
o p o'
Ker d ® Ker d1 =R ® S1 ;.»;R ® S . =Ker d . ®R ® S .
o 1 d 1 d n-1 n-1 d n
"o "o Ro
So we find that there exist projective R -modules Q and Q' such that P ®(R ® Q] -
Ro
R »o<. a
Ro
11.4. Theorem. Let R be a left Noetherian ring with gl.dim.RO. Let P be a
projective left P. [T.,... ,T ] -module of finite type. There exist projective R-
modules Q and Q', both of finite type, such that : QIT.,... ,T ] ®P = Q' [T-.-.T ] ■

83
Proof : Apply Lemma 11.3 to R[T ... JJ . □
Note that in K-theoretic wording the above theorem states that KQ(R) =K (RIT.,...,! ]).
11.12 GRADED RINGS AND MODULES OF QUOTIENTS
12.1. Torsion Theories over Graded Rings.
Let C be a Grothendieck category (for general torsion theory C need only be
abelian,). A torsion theory in C is a couple (T,F) of nonvoid classes of
objects of C such that :
TT1 : TnF = {o}
TT2 : T is closed under homomorphic images in C
TT3 : F is closed under taking subobjects.
TT4 : For any MeC there is a subobject t(M) of M such that t(M) eT, M/t(M) e F-
A torsion theory (T,F) is said to be hereditary if T is closed under taking subobjects
as well. We will only consider hereditary torsion theories here. A kernel functor
on C is a left exact subfunctor of the identity in C. A kernel functor k is said to
be idempotent if and only if k (M/k (M)) =o for all MeC. In this terminology,
elementary results in torsion theory yield that there is a one-to-one correspondence
between idempotent kernel functors k in C and torsion theories in C. These also
correspond in a one-to-one way to Gabriel-filters (or-topologies) in a chosen generatoi
G for C; such a Gabriel-filter in G will then consist of the subobjects L of G such
that G/Lef, in casu : k (G/L) = G/L because the torsion class of a torsion theory
associated to an idempotent kernel functor k is given as the class of MeC such that
k (M) =M, while Me F if and only if k (M) =o. For more detail about torsion theory cf
[ 9 ] , [10 ] , [32 ] •
Notation. In the sequel (T,F) will always be a herditary torsion theory in C and k
will be the corresponding idempotent kernel functor on C.
Evidently we wish to apply these techniques to R-gr = C. Since R-gr actually

84
is a Grothendieck category this presents no real problems, however R is not a
generator for R-gr so the beautiful theory as developped in case G =R-mod does not
generalize completely. Moreover we will encounter (solvable) problems in relating
torsion theories in R-gr and torsion theories in R-mod.
12.1.1. Lemma. Let R be a graded ring, (T,F) a torsion theory in R-gr. The
following statements are equivalent.
1. If Met then M(n) eT for any nez.
2. If NEF then N(m) eF for any mez.
Proof : Easy. □
12.1.2. Example. Let R be a positively graded ring. Consider for T the class of
Me R-gr such that M. =o if i<o and take F to be the class of Me R-gr sucli that
M. =o if i>o. Clearly (T,F) is a hereditary torsion theory which does not have the
properties mentioned in Lemma 12.1.1. On the other hand it is not difficult to
generate torsion theories which do have that property by adding to T all suspensions
of objects from T.
A torsion theory satisfying the equivalent conditions of Lemma 12.1.1. is
said to be a rigid torsion theory. A kernel functor is rigid of the corresponding
torsion theory is. For any torsion theory (T,F) in R-gr we have that Homn (M,N)=o
K-gr
for every Mer, Ne F. However it should be noted that HOM^(M,N) =o for all Met,
NeF, is equivalent to (T,F) being rigid.
12.1.3. Lemma. A kernel functor k in R-gr is rigid if and only if k (M(n)) = « (M) (n)
for all nez .
Proof : Suppose that k is rigid, then k (M/(n) cK (M(n)). But (M/K (M)) M =M(n)A 00 (n)
is «-torsion free i.e. in F, therfore k (M(n))/K (M) (n) =o.
Conversely, if Met then k (Mj =M i.e. k (M(n)) =M(n) follows and thus M(n) eT for all
nez. □

85
A non-empty set of graded left ideals £ of R a graded filter if the following
requirements are fulfilled :
F : If Le£ and L. is a graded left ideal of R such that LCL. then L1 e£.
F2 : If L-.I^ejC then L n^ej:
F, : If Lei then [L : x] e£ for all xeh(R).
F4 : If L. e£ and [L : x] ex for all xehfLJ then Lex.
As a consequence of the fact that R is not the generator of R-gr, we cannot
expect that there is going to be a one-to-one correspondence between graded filters
in R and hereditary torsion theories in R-gr. However we do have the following
result :
12.1.4. Lemma. There is a bijective correspondence between, on-one hand the
hereditary rigid torsion theories in R-gr, and on the other hand, rigid idempotent kernel
functors on R-gr. The latter correspond further in a bijective way to the set of
graded filters in R.
Proof *. This is mere modification of the proof in the ungraded case. □
The sequel of this section is concerned with the characterization of those
kernel functors on R-gr that may be obtained from kernel functors on R-mod. An
idempotent kernel functor k_ on R-mod is said to be graded if the filter £ (k_) of k_
possesses a cofinal set of graded left ideals.
12.1.5. Lemma. Let R be a graded ring and let k_ be a graded kernel functor on R-mod.
If M is a graded left R-module then k_(M) and M/<f_PQ are graded left R-modules, the
canonical R-morphism M-*M/k_(M) may be considered as a graded morphism of degree o.
Furthermore : k_ induces a kernel functor k on R-gr which is rigid and idempotent, the
torsion class T[k) consists of graded R-modules N such that k_0£) =N, whereas F(k)
consists of Ne R-gr, k (N) =o.
Proof : If we establish that k_(M) is graded then it will follow immediatly that

86
M/k_(M) is graded and that M->M/k_(M) is a graded morphism of degree o. Since k_ is a
graded kernel functor on R-mod, x£k_(M) means that Lx = o for some graded left ideal
L ££(/£). Write x = 2x. , L = H . Then L x = o for each nez hence L x. =o for each
— £ l ' n n ' ni
n,ieZ. Thus x. £k_(M) for all iez i.e. k_(M) is a graded submodule of M, which we
will denote also by k_(M) . Furthermore, if TO) and FO) are as stated then one
easily verifies TT. till TT so (T,F) is a hereditary torsion theory in R-gr which
is easily seen to be rigid. The kernel functor k corresponding to this torsion
theory is a rigid kernel functor on R-gr such that « (M) = k_(M) holds for every Me R-gr. C
As in any Grothendieck category we define the torsion theory regenerated in
R-gr by an object Me R-gr as follows : the torsion class TO,,) consists of the
NeR-gr. such that HomR (N,Eg(M)) = o, where Eg(M) is the injective hull of M in
R-gr. If «M is a rigid kernel functor then Ne TO,,) yields HCMR(N,Eg(M)) =o or
HomR_gr(N> ®Eg(M)(n)) =o. Thus the rigid torsion theories amongst the k Me R-gr,
* ë n ;1
may be thought of as being the torsion theories cogenerated by a suspension-invariant
M. Let K„(n) denote the torsion theory cogenerated in R-gr by M(n). The inf k of
torsion theories O .,iel} in R-gr is defined by taking for TO) the intersection of
the TO -), we denote this by k =A« . . It is then clear that Ak (n) is a rigid
i i n M
torsion theory which is cogenerated by $ Eg0I)(n) =Eg(©M(n)). In particular, if we
n n
define K-, <K 2 on the lattice °f hereditary torsion theories in R-gr by TO J cr(K2)
then a rigid torsion theory k <k for some Me R-gr if and only if k <A «M(n). The
kernel functor A «)((ii) is called the rigid kernel functor associated or cogenerated
n
by Me R-gr., we will denote it by k~F..
12.1.6. Lemma. Let Me R-gr and let k denote the kernel functor in R-mod cogenerated
by MeR-mod then : £ 0M) = (JCR, for all xeM, x fo, and aeR, (J : a) £AnnRx}
■CO^) ={LeL (R), for all xeh(M), x^o, and aeh(R), (L : a) /AnnRx}. Moreover, we
have that £0M) nLg(R) =^0y).
Proof : By definition of k,, we have :
' M
r
£0o) ={LeLCT(R),HOMI?(R/L, Egpf|) =o} .
W l gVJ' T(

87
Let Lei(jJ) and suppose there exist x^o in h(M) and aeh(R) such that (L:a) cAnnRx.
Since (L:a) e£(Ko) it then follows that Ann„xe£(k^). But this entails that there
would exist a nonzero f eHOM(R/Ann„x, E^(M)) what contradicts R/Ann„x eT(k,,).
Conversely, supposing that HONL(R/L,Eg(M)) ^o for some LeL (R) with the property that
for all x^o in h(M) and aeh(R), (L:a) £Ann„x, we get that there exists a graded
morphism f : R/L-*-E^(M), f ^o. Now since M is gr. essential in E^(M) there exists
an a£R/L with representative aeh(R) such that x = f(a) fo is an element of h(M) •
Clearly (L:a) cAnrux, contradiction. That ^OO is as stated is well known.
Further, the inclusion -COO nL (R) c£(k*) is obvious. For the converse inclusion
consider L££(k5!), x^o in M and a£R. Decompose x as x. + ._ +x , a as a- + ._ +a
and assume that these decompositions are so that we have ascending degrees. Since
x fo there can be found a X eh(R) such that X.a. eL and X.x ^o. Similarly, we
find X eh(R) such that X J a,eL and X X.x fo. Using a recurrence argument we
obtain an element X eh(R) such that X ...X_X.a eL, X ...X_X.x ^o. Putting
n l ' n 2 1 n ' n 2 1 m b
X =X ,.U. eh(R), we have that Xx^o while XaeL, hence (L:a) s£Annnx, i.e.
n z i k
12.1.7. Theorem. Let R be a graded ring and let (T,F) be a rigid torsion theory
in R-gr. Let k be the kernel functor on R-gr corresponding to (T,F) and let £{k)
be its graded filter. Denote by (T.jF.), respectively (T7,F7), the hereditary
torsion theory in R-mod generated by T, respectively co-generated by F. Let k-,
£., respectively k Z^ be the corresponding kernel functors and filters. We have
the following properties.
1.a. X. ={L left ideal of R, (L) ££(«)>
b. If Me R-gr then ^ CM) = k(M)
c. £. is the smallest filter in R containing £(k) and£. nL (R) =£(«).
2.a. ^2 = {L left ideal of R, Horn (R/L, E(M)) =o for all Mef}
b. If Me R-gr then k ,(M) =k (M)
c. X2nLg(R) = ■£(>).

88
Proof : 1.a. and then easily I.e., is proved in a straightforward way. As for
1.b., it is clear that k (M) c<t 1 (M) because for a rigid kernel functor « an element
x is in k (M) if and only if its annihilator in R is in £ (k ). Conversely, if
xEk..(M) then Anruxex. i.e. (Ann„x) e£(>). Writing x =x. + ... +x with deg x. < ...
.-<deg x we see that, for homogeneous ae (Annpx) we have ax. =o, i = 1 n.
Hence x- e«(M) and x£k (M) follows. Note that we have herewith also shown that «
is induced on R-gr by k. in the way described in Lemma 12.1.5. Therefore, all rigid
kernel functors on R-gr may be considered as being induced on R-gr by some kernel
functor on R-mod.
2.a. Is obvious since (T?,F7) is co-generated by F in R-mod.
2.b. Since M/k (M) eF, M/k (M) e F-, follows, but as «9pl) maps to a submodule of
MA (M) which also has to be in T?, this means that « 2(H) c« (M). If FeF(k) then
from k - (F) =o it follows that, «1 <«c, hence «, <inf{Kp,FeF} =K?. Therefore, for
every Me R-gr, « 1 QJ[) =k pt) c«2pf), thus « (M) =k2CQ.
2.c. Follows from 2.b. and the fact that the rigid kernel functor k is uniquely
determined by its graded filter.
By Lemma 12.1.6 we also have that k = inf {«M, M£ F} while k 2 = ±nf{K MeF}.
12.1.8. Remarks■ 1. It is clear that «. is a graded kernel functor on R-mod
whereas k _ is not a graded kernel functor.
2. The rigid torsion theories in R-gr form a set which may be partially ordered by
putting k <k ' if and only if £ (k ) c£ [k '). This ordering coincides with the
ordering induced by the ordering on the kernel functors on R-gr, however for non-
rigid functors this ordering cannot be checked on the graded filters alone.
3. The operation A restricted to rigid torsion theories coincides with the operation
A induced in the set of rigid kernel functors by the A in R-mod. Here again we have
that k =Ak . for rigid k and « - is equivalent to £[k) =c£[k .) and again this
i i 1
criterion does not work for non-rigid kernel functors.
4. More about co-generated torsion theories will be in Section 12.3.

89
12.2. Examples of Torsion Theories in R-gr.
The singular submodule Z(M) of MeR-gr will be {xeM, Ann_x is an essential
left ideal} i.e. the singular submodule of M in R-mod.
12.2.1. Lemma. If MeR-gr then Z(M) is a graded submodule.
Proof : Pick anx^o in Z(M) and write x =x. + ... +x with deg x. <... <deg x ,
assuming that x- ^o for all j =1,.„,n. Put L = Annnx, then L cAnnnx . Since L is
j k k n
an essential left ideal, L is essential and thus Annnx„ is an essential left ideal;
k n
thus x eZ(M). The lemma now follows by recurrence on n. □
It may easily be verified that Z defines a rigid preradical on R-gr. The
associated radical Z satisfies Z„(M)/Z(M) =Z(M/Z(M))- To a radical there may be
associated an idempotent kernel functor (cf. [36]) in the usual way. The kernel
functor K-, corresponding to the radical Z„ is rigid ; K-, may be called Goldie's
graded kernel functor.
The dependence of Z(-) on the module category under consideration is especially easy
to deal with in the case of polynomial ring extension, as follows:
12.2.2. Proposition. Let R be an arbitrary ring, Me R-mod. Then ZR __ (M[X]) =
ZR(M) [X] .
Proof : Take m XneZR(M) PQ with m^o. Since mezR(M) it follows that AnnRm is left
essential in R, and this implies that (Ann^jn) [X] is left essential in R[X] . Now
AnnR mXn = (Ann^) [X] , thus mXn e ZR (M [X] ), hence ZR(M) [X] c ZR ™ (M [X] ). For the
converse inclusion we only have to establish, on view of Lemma 12.2.1, that any
homogeneous element of Z ^ (M [X] ) is in Z (M) [X] . Let mXn be in ZR ^ [M[X] ), then
AnnR ra mX = (Annjjn) [X] and consequently (AnnRm) [X] is left essential in R [X] ,
therefore AnnRm is left essential in R i.e. mezR(M) and mXnezR(M) [X] follows. □
The rigid torsion theory (T,F) co-generated in R-gr by R, i.e.
T = (MeR-gr, HOMR(M, Eg(R)) =o} where Eg(R) is the injective hull of S in R-gr, is

90
called Lambek's graded torsion theory. It's filter £ is given by:
£ ={LeL (R), (L : a) 2Annnb for all a,beh(R), b^o}.
g K
Lambek's torsion theory in R-mod is co-generated by (T,F). Theorem 12.1.7 implies that
L e£ yields that L_ is dense in R.
Let S be a multiplicatively closed set in R. To S we associate a kernel
functor «_„ on R-mod, which is given by its filter
■£(ls) ={L left ideal of R, (L:r) nS^?S for all r£R.}.
12.2.3. Proposition. Let S be a multiplicatively closed system consisting of
homogeneous elements and not containing zero then k_ _ is a graded kernel functor in
R-mod i.e. there corresponds a rigid kernel functor k„ in R-gr to S, the filter of
which is given by.
£(kJ ={L£L (R),(L:r) nS ^ for all reh(R)}.
Proof : Let I^OO i.e. (L:r) nS^ for all reR. If r is in h(R) this yields
that sreL for some ses i.e. sre ÇL) and (00 : r) OS f<j>. If r is not homogeneous,
g b
write r = r. +._+r with deg r, <... <deg r . Take s„ e (L:r ) nS, s .efLrsr ,)ns,
1 n ° 1 ^ n n v n n-1 n n-1
... ,s1 e (L:Sn.. .s r.) nS, then, putting s = s .s,...s , we have sr e (L) with s eS i.e.
| " II I I i* IT £j
(L) e£(« c) or k _ is graded. The other statements follow directly from Lemma 12.1.5
g — ^ —^
and Theorem 12.1.7. □
12.2.4. Remark. If S satisfies the left Ore conditions in R then :
Kc(K) ={meM, sm = o for some seS} for any Me R-gr.
Proof : Immediatly from 12.2.3. and 12.1.6. (1.b). □
Another example of a torsion theory in R-gr may be constructed as follows.
Let (T,F) be the torsion theory in R-gr generated by the class of simple objects in
R-gr. It is not hard to see that this is a rigid torsion theory and that an object

91
M of T is a semi-Artinian object of R-gr. i.e. the Gabriel dimension of M in R-mod
is at most 1.
Recall that a torsion theory in R-mod is said to be symmetric if its
associated filter allows a cofinal system of ideals of R. Similarly we define that
a rigid kernel functor k (or rigid torsion theory) in R-gr is symmetric if £ (k ) has
a cofinal set consisting of graded ideals.
12.2.5. Lemma. If k is a rigid kernel functor on R-gr such that k_ is symmetric on
R-mod then k is symmetric.
Proof : If Le£(K) then L_e£ (V) and L_ contains an ideal I of R such that ieXQO.
Since k_ is a graded kernel functor on R-mod, (I) ex (V). Hence : (I) ££(«), (I) CL
S So
and (I) is a graded ideal of R. □
Now let R be a left Noetherian graded ring, I a graded ideal of R, P a prime
ideal of R, and consider :
£(I) ={L£L (R), L3In for some neM}
£(R-P) ={L£L (R), LDRsR for some seR-P}.
Then «■ associated to the filter £ (I), and "d_d> associated to the filter £ (R-P),
are rigid idempotent kernel functors on R-gr; they are induced in R-gr by the functors
<_, , «_„ p on R-mod, which are given by their filters
£(I) ={L left ideal of R, L3In for some ne IN}
X(R-P) ={L left ideal of R, LdRsR for some seR-P},
which have been constructed and used in [30 ] ,[ 31] .
12.3. Injective Objects and Torsion Theories
If Me R-gr, let E[M) be the injective hull of M in R-mod.

92
Let MeR-gr be such that the singular radical Z(M) =o. Put : E' ={x£E(M), there
is a graded essential left ideal I, such that LxCN for all neZ. It is easy
enougli to verify that E' = O E' is a graded R-module containing M as a graded
mez m
suhmodule.
12.3.1.. Lemma. Let MeR-gr be such that Z(M) =o, then E_' is the maximal R-sub-
module of E(M) with the property that E' is graded and it contains M as a graded
submodule.
Proof. Suppose that we have MCN in R-gr such that Mc?çce(M). If xeNm then
(M : x) is a graded left ideal of R which is also essential. Since, (M:x) .xc
MnVm=*W xeE; follows. -
12.3.2. Proposition. If MeR-gr is such that Z(M) =o then Eg(M} =E' in R-gr.
Proof. In any Grothendieck category the injective hull may be characterized as
being a maximal essential extension. By Lemma 1.3.3.13., if Eg(M) is a graded
essential extension of M then it is an essential extension of M i.e. up to
isomorphism we may assume : Eg(M) CE(M). The foregoing lemma now yields E°(M) £E'
i.e. E' =Eg(M). ■
Note that, even in case Z(M) fo, h(Eg(M)) consists of x£E(M) for which there
is a graded essential left ideal L such that LxCH for all nez while Lx^o.
n n+m
12.3.3. Remark. A graded module which is injective in R-mod is injective in
R-gr. A kernel functor k in R-gr is cogenerated by M if and only if it is cogenerated
by Eg(M) if and only if k_ is cogenerated by E(M) in R-mod.
A graded R-module M is said to be a graded supporting module for the kernel
functor k on R-gr if k (M) = o and k (JI/N) =M/N for every graded submodule N fo. A kernel
functor k on R-gr is said to be graded prime if it is cogenerated by a graded
supporting module. A kernel functor k on R-gr is said to be projective if R e£(K). The
projective graded prime kernel functors on R-gr form a set which we denote R-Pr.

93
From now on we will write k_ for the kernel functor on R-mod associated to k
on R-gr in the way described in Theorem 12.1.7 (i.e. the k there constructed is
denoted by k) . It is clear that if MeR-gr is such that k_m is a prime kernel
functor on R-mod with supporting module M, then «M is a graded prime kernel functor
on R-gr with graded supporting module M. Note that a graded kernel functor k_ on
R-mod such that the induced k on R-gr is graded prime, need not be prime. For example
take R to be k [X] , where k is a field, X a variable given positive degree, and
consider M = ReR-gr. The kernel functors «■ and k on R-gr, resp. R-mod, may both be
A — A
given by their filter which is generated by the set {(X),..., (Xn),... ,k [X] }. It is
obvious that k [X] is a graded supporting module for k but k [X] is not a supporting
x
module for k v. Warning : if M is graded then k . on R-mod need not be a graded
— A M
kernel functor. However if Me R-gr is such that k is a graded kernel functor then
it follows from Lemma 12.1.6. that «,, = «
M —M
12.3.4. Example. Let R be a left Noetherian graded ring and P a graded prime ideal
of R. It is well-known that the Lambek-Michler torsion theory with kernel functor
Kp on R-mod is co-generated by E (R/P), hence also by E*=(R/P). However k is not
necessarily a graded kernel functor; for example, take R to be a commutative integral
domain, pick seR-p and let s be non-homogeneous (this is possible if R is non-triviall>
graded), then Rs cannot contain an element of h(R-P) and Rs e£(x ).
Clearly an Me R-gr of the form © E(-n) for some EeR-gr can never be a sup-
n
porting module for a kernel functor k on R-gr, therefore graded prime k will in most
cases not be rigid. Therefore if we say that k. on R-gr is a rigid prime kernel
functor we mean to say that k =A k ' (n) where «' is a graded prime kernel functor,
—' n
i.e. k =k for some M which is a graded supporting module for k .
12.3.5. Lemma. If Me R-gr then «■ is the largest rigid kernel functor on R-gr such
that K]Jj(M) =o.
Proof : If k is a rigid kernel functor on R-gr such that k >«m, k (M) =o, then the

94
fact that the ordering of rigid functors may be checked by looking at their filters
MJ
yields that there exists an L£i(n)-i(0 i.e. R/L is a graded module which is not
«..-torsion, therefore by Lemma 12.1.6, there exists a nonzero graded morphism of
degree p say, f : R/L-*■ Eg (M) • On the other hand R/L is «-torsion and k. is rigid,
therefore f(R/L) is «-torsion. But « (M) =o yields K(Eg(M)) =o hence f(R/L) =o, contra
diction. Conversely if k is any rigid kernel functor on R-gr such that k(!I) =o then
for every I €£(*), HCMnfR/I.M) =o follows from the rigidity of k. By Lemma 12.1.6,
I e£(K ). So we have £(k) c£(k ) meaning tliat « <k5! because both are rigid. □
12.3.6. Theorem. If Me R-gr has the property that k is a prime kernel functor on
R-mod with supporting module M, then «M is a rigid prime kernel functor with
supporting module M and k . is a graded prime kernel functor, £ (k,,) =£(k ) nL (R).
Proof : Imnediately from Lemma 12.1.6. and the above. □
12.3.7. Corollary. If P is a graded prime ideal of R then we may correspond to P a
rigid kernel functor on R-gr which may be induced on R-gr by means of a prime kernel
functor k on R-mod. Let us denote that rigid prime kernel functor associated to P
by k„ (normally we would have written «p but there will be no confusion.
If R is a left Noetherian and positively graded ring then «p is projective if and
only if PeProj R = {graded prime ideals of R not containing R+}.
Proof : Consider the injective hull ED(R/P) of R/P in R-mod. It is well known that
ER(R/P) is a supporting module for the torsion theory k co-generated by it. Hence
also E°(R/P) is a supporting module for *p and k is co-generated by Eg(R/P). By
the foregoing theorem we may conclude that the rigid kernel functor «■ which
corresponds to the filter £0p) nL (R) is as stated in the corollary. If PeProj R then
R+ +P/P = (R/P) + . If there exists a left ideal of R/P which does not intersect (R/P) +
properly then there is an x^o in R/P such that (R/P)+x = o, contradiction. Thus
(R/P)+ is essential in R/P and as such it contains a regular element i.e. R+nG(P) fi
Since R+ is an ideal of R,(R+:r) nG(P) f$ for every r£R i.e. R+eL (R) n£ (k ).

95
Conversely, suppose that R ££(k ). Then it follows from Lemma 12.1.6. that for all
y€h(Eg(R/P)) , R+£Anny, evidently R+ £P follows. □
We do not expound further the analogies between the correspondences,
Spec R^-Proj R and R-Sp->-R-Pr, (where R-Sp denotes the set of prime kernel functors
on R-mod).
Let us point out that combination of our results on Ass and on the injectives in R-gr
allows to characterize graded fully bounded left Noetherian rings as follows.
12.3.8. Theorem. Let R be a positively graded left Noetherian ring. By Eg(R) we
mean the equivalence classes of graded indecomposable injective objects for the
relation : isomorphic up to suspension. The following statements are equivalent to
one another :
1. The natural correspondence : Spec R-*-Eg(R) is a one-to-one correspondence.
2. Graded indecomposable injectives in R-gr are graded isotypic (i.e. E is gr.
isotypic if E= ©E. with the E. being the suspensions of some E ).
3. For every PeSpec R and for every left graded essential L in R/P we have that L
contains an ideal I of R/P.
4. For every rigid kernel functor k on R-gr, k is symmetric.
Moreover if R is graded fully left bounded (i.e. one of the above conditions holds)
then equivalently : R is gr. left Artinian or graded prime ideals of R are gr. maximal.
12.4. Graded Rings and Modules of Quotients.
The general theory of localization in a Grothendieck category may be applied
so as to obtain the construction of objects of quotients with respect to some kernel
functor or torsion theory. In the case of R-gr the interesting point of this is that
for rigid torsion theories, the objects of quotients constructed, relate in a very
nice way to the object of quotients constructed in R-mod. The exposition of this
link between localization in R-mod and localization in R-gr is subject of this section.
Throughout k will be an idempotent kernel functor on R-gr, (T,F) will be the

96
associated torsion theory in R-gr and £(k) will be the filter of k. Let M £ R-gr.
Define Q^(M) = lim HOM„(L,M/k (M)) ; this graded R-module is said to be the object of
quotients of M in R-gr with respect to «■
One easily checks that Q°(R) is, in a natural way, endowed with a graded ring
structure, while oJ(M) is a oJ(R)-module i.e. in Q^(R)-gr. It is clear that Q^ is a left
exact functor in R-gr which commutes with arbitrary suspensions.
For further use we state :
12.4.1. Lemma. Let R be a positively graded left Noetherian ring then R is an R -
ring of finite type.
Proof : Let x.,.-,x be homogeneous elements generating R+ as a left ideal and let S
be the ring generated by R and {x-,... ,x }. By construction R es. We proceed by
induction on j such that R.cs. Suppose that R.cs for all j<ie]N. An x£R., i>o,
can be written as i r x with r £R. , , d = deg(x ). Since d >o for all s, r es
...os s i-d s 6l sJ s 's
t=1 s
for all s and x€S follows. Thus R = S. □
Let R be a positively graded ring throughout this section. Let «_ be a graded
kernel functor on R-mod and let M e R-gr. For each m£Z put :
N ={xeO CM), there is a graded left ideal L€£(k) such that L xc (MAO*)) 4.™
for all neZ} .
(Note that MA_(M) is graded because of 12.1.5).
The set N is well defined if m does not depend on the choice of L ££(«_). However if
L-, e£ 0O is graded and such that for some x/oeO (M) we have :
(LJ xc(M/k_(M)) + for all n^Z, then LnL-ejC^) is graded and for some neZ,
(in J) x^o since (inj)x = o contradicts o^xeQ (M). So we obtain :
o t (lnJ)nxc (M//i(M))n + mn (M/^(M))n+v, entailing m = v.
Obviously ® N is a graded R-module which we will call the graded module of quo-
m€Z
tients of M at «_ in R-mod, and denote it by gO (M), with these notations we have :

97
12.4.2. Proposition, g 0 (R) is a graded ring containing RA_(R) as a graded subring.
The graded ring structure of g 0 (R) is the unique ring structure compatible with
its graded R-module structure. For every MeR-gr, g 0 (M) is a graded g 0 (R)-module.
Proof : It is immediate that g 0 (R) is a graded left R-module which is contained in
Q^ (R) and which contains R/k_(R) as a graded R-submodule. Pick x,y£g 0 (R) and
consider xys(\ (R) (the latter is a ring by general localization theory in R-mod).
Without loss of generality one may take xe (gO (R)) , ye (g 0 (R)) for some n,meZ.
Let I,jay be graded left ideals of R such that :
Ikxc (RA_(R))m+k for all kez,
J£yC(R/^(R))n+£ for all £eZ.
If (J:x) ni =o then k_ is trivial and there is nothing to prove. If L = (J:x) ni ^o
then we find heZ such that there is a nonzero c£L, . lien ce cxyei ycR,
h ' n+nr n+m+n
Consequently Lhxy cRh+m+n for all heZ, thus xy e (g (^ (R)) + . It follows that
g 0 (R) is a graded ring and that this ring structure is determined by the graded
R-module structure. Uniqueness of this ring structure as such follows from the
fact that g 0 (R) is a subring of 0 (R), the ring structure of the latter being
uniquely determined by its R-module structure. Similar argumentation shows that
g (^ (M) is a graded g Q^ (R)-module. □
12.4.3. Theorem. Let k_ be a graded kernel functor on R-mod, MeR-gr. Then we have:
(g % CM) )_ = lim , H0M(L,MA (M) )m .
i- m riTynLg(R) -- m
Proof : It is well-known that 0 00 = JiBL_»Hom(L,MA (M)). By construction,
- L£iy
x£(g IM). if and only if x is represented by a graded morphism of degree m :
mx : L^MA_pf) for some L €£(«_) nLg(R). Since £ (k_) nLg(R) is cofinal in £ QO the
theorem follows easily. □
12.4.4. Corollary. Let k be a rigid kernel functor on R-gr. Then k_ is the graded

98
kernel functor on R-mod the torsion theory of which is generated by the torsion
theory of k . By Theorem 1 2.1.7. k_(M) =k (M) and£Q<_)nL (R) =£("). By definition of
Q^(M) and by Theorem 12.4.3. it follows that Q^(M) = ë% CM) • Thus the functors Q^
and gO are naturally equivalent functors on R-gr. Because of this we will use the
notation Q^(M) for the object of quotients in R-gr associated to the graded kernel
functor «_ on R-mod.
12.4.5. Proposition. Q° is a covariant left exact endofunctor in R-gr :
Moreover Q° is functorial with respect to graded morphisms of arbitrary degree.
Proof : We only have to prove the last statement. Let N,MeR-gr, f : N->M a graded
morphism of degree m. Then (^ (f) : (^ (N) ->(^ (M) is R-linear. If x£ (Q^(N))k then
there is an Is£[k) such that I xc(NApi)) +, for all ne IN. Hence for all ne IN
we have I h(x) ch((NA_(N)) +,) and the latter is exactly f((N/k O0)n+k) hence
contained in (MA (N)) +1c+_ where the graded morphism of degree m : N/k (N) -*-M/k (M),
deriving from f, has again been denoted by f. Therefore h(x) e (Q°(M))i,+rn and this
states exactly that h|Q°(N) is a graded morphism of degree m. □
Recall the following definitions. A kernel functor k_ on R-mod is said to be
a finite type if for every Le£(V) there exists an L' e£(V) , L' cl and L' is finitely
generated. We say that k has property T if one of the following equivalent conditions
is fulfilled :
i) 0 is exact and commutes with direct sums,
ii) For all Me R-mod C^ (M) =QK (R) ®M.
iii) For all Le£(V) : Q^ (R) =QK(R) j (L) where j is the canonical morphism R^RA_(R).
iv) The quotient category with respect to k_ coincides with 0 (R) -mod.
v) Every (^ (R)-module is k_- tors ion free as an R-module.
From iii) one easily derives that a kernel functor having property T has finite type.
12.4.6. Lemma. Let k be the rigid kernel functor on R-gr induced by the graded
kernel functor k_ on R-mod. If k_ has finite type then k has finite type.

99
Proof : Easy. □
MA.7. Proposition. Let « be the rigid kernel functor on R-gr induced by the
graded kernel-functor k_ on R-mod and suppose that k_ has finite type, then Q°(M) =
(^ Qf) for all Me R-gr.
Proof : Since q5(M) = lim HOM^CL.M/k(M))> it follows from Lemma 1.3.3.2; combined
^ LeJCOO *
with the fact that the finitely generated graded left ideals of £ (k ) form a cofinal
system, that Qg(M) =■ lim , Hom„(L,M/K (M)) (note k_(M) =k (M)). □
LéjCÇO
12.4.8. Corollary. If R is a left Noetherian positively graded ring and k_ a graded
kernel functor on R-mod inducing k on R-gr then 0 (M) is graded for every Me R-gr,
and (^(M) =Qf CK).
If k_ is a projective kernel functor on R-mod then let S be the set of xeQ (R)
such that there exists a graded left ideal L££(«) such that Lxcj (R) and L xcj (R)
for all n>n for some fixed n e ]N , and consider S=® Sm.
0 ° m
12.4.9. Lemma. If k_ is projective then S is a graded ring.
Proof : We only have to check that m in the definition of S does not depend on the
choice of Le£(V); the rest of the proof is then formally the same as the proof of
Proposition 12.4.2. So let Je £ («J be such that Jxcj (r) and J xc(j (R))__, for
all n>n' , some fixed n' e J\I.
o ' o
For all n>max{n ,n'} = N we have then that :
oo o
H=JnL££(«) is such that Hxc(i (R)) _,_ n (j (R)) ,. Thus if m/m' then H x = o
n jk n+m jk n+m n
for all n>N , consequently R>N Hx = o follows then from the fact that the gradation
of R is positive. However (R+) °CR>N and (R+) °e£(>) since the fact that k is
1 o
idempotent implies that £(k) is closed under taking products, thus R>„ e£(«) and
thus R>N He£«) or xe/^ <^(R)=o. □
o
under taking products, thus R>,
He£(K.) or xSk_ q,(R)=o. □
12.4.10. Proposition. Let k_ be a projective kernel functor then the ring S is
nothing but 0s (R).

100
Proof : The definitions imply that of(R) = gO (R) cs. Take x£S to be nonzero i.e.
let L££W be such that Lxcj (r) and L xc(j (R)) for all n>n , for some fixed
v J jk n J« m+n o
new. Now, if L,x=o for all k<n then L xc(j (R)) _,_ holds for all ne IN and
o ' k o n kjk K "n+m
thus xeQ°(R) follows. Suppose that L, x f o for some k<n and fix this k. Pick a
nonzero yeL.x and let zej fy). Assume first that i (R )L,x = o for all n>n -k.
' k jk " jk K nJ k o
Then R z maps to zero in j (R) for all n>n -k, consequently R> _,. z maps to zero
n°Ik
in j (R) i.e. R^ , za(R), However R. . e£(«) since (RJ ° cr thus
jk >n -k —v J >n -k — + >n -k
oo o
ze«_(R) and y = o, contradiction. Therefore we may assume that j (R )L,x^o for some
n>n -k. We obtain : o^j (R ILxcL ^xcfj (R)) since k+n>n . From this it
o ' jkk nJ k n+k VJ« v "n+k+m o
is evident that L,xcj (r) hence in all cases : L xc(j (R)) for all ne IN,
k jk l Jm+k' n kjk k "m+n '
i.e. xe(Qg(R))m.
12.4.11. Remark. In a similar way it follows that the projective module of quotients
of MeR-gr at a projective kernel functor «_ is nothing but Q°(M).
12.5. More Properties of Graded Localization.
First let us expound some results about graded morphisms and their
localizations, which for graded morphisms of degree o are trivial or covered by general
localization theory in R-gr. R is positively graded throughout and k is rigid unless
otherwise stated.
12.5.1. Lemma. Let i tea kernel functor on R-gr, let MeR-gr, Lei(«). A graded
morphism of arbitrary degree : f : L-*-M extends in a unique way to a graded morphism
of the same degree h : R->Q°(M).
Proof : By faithful «_-injectivity of Q (M) we know that f extends to an R-linear map
¥ : R->0 (M). It is clear that in proving the lemma we may suppose that M is «-torsion
free. Now ¥(L ) =L ¥ (1) and ¥ (L ) = f (L ) cm _,_ so for all ne IN : L ¥(1) cm _. i.e.
1 nJ n *■ ' l nJ l nJ n+m n n+m
f(1) GQ^(M)m where m is the degree of f. Now since R is graded ¥ maps R to Q^(M) and
it is obviously the unique extension of f.

101
12.5.2. Proposition. Let M,N,SeR-gr be such that NCM and « (M/N) = M/N for some
kernel functor k on R-gr. Any graded morphism f of degree d, f :N->-Q°(S) extends in
a unique way to a graded morphism h of degree d, h : M->Q°(S).
Proof. Let xeM and let I«(«) be such that Ixcn. The definition of Q^(M) entails
that Q^ (0s CM) ) = Q^ (M) and tlierefore Qg (Q8 00 ) = 0s (M) . Let x be homogeneous and apply
the foregoing lemma to m , i.e. right multiplication by x, we get the following
commutative diagram of graded morphisms
-<Ç00.
The graded morphism h composes with Q^(f) : Q°(N) ->-Q°(S) and yields a graded morphism
>>/ :R->-Q°(S). Define h(x) ='/' (1). It is straightforward to verify that <l> does not
depend on the choice of 1££(k) and that the map h is graded morphism of degree equal
to deg f which satisfies all requirements. □
Recall that for Me R-mod, k_ a kernel functor on R-mod, we can characterize
0 [M) by the exact sequence in R-mod :
77 (M)
o ^M/^(M) —^ CM) -*k CE (M)/M) —*o
where E(M) is an injactive hull of M in R-mod.
If Me R-gr then we have the following exact diagram in R-mod :
o —M -+ E CM) ZM, E CM) /M -► o
Il t t
o —»-M -+ EgÇM) >EgÇM)/M ~* o
and the bottom row may be considered as an exact row in R-gr.

102
12.5.3. Proposition. Let Me R-gr and let i be a kernel functor in R-gr. Then
Q^(M) is characterized by the following exact sequence in R-gr : (let M' be M/k(M)):
g
0_>M. —>Q°(M) -=->-K (Eg(M')/M') —>o
where -irg is the restriction of the canonical map Eg(M') ->-Eg(M')/M'.
Proof : The fact that Qg(M) =g Q (M) guarantees- that it is a graded submodule of
Eg(M'), such that Q^OQ/M' is K-torsion. Suppose xeEg(M') and IxCM' , i.e. ycM1
for all neZ, for some ie£(>). Put x=2' x„ and i€l , then 2' ix»=2' m- for some
£ ^ n £ £ . j
m. eM'. Since M' is a graded submodule of Eg(M') it follows that ix„ =m +» for all
£,n and iel . Consequently I x„cm' , thus x„eQg(M) and xeQ^Ol) follows. □
12.5.4. Corollary. Let k be a kernel functor on R-gr, let k_ be the cogenerated
kernel functor on R-mod, then, for every Me R-gr, QJ^(M) is the largest graded module
containing M/K (M) as a submodule, which is contained in 0 (M).
Q°(R)-modules may be considered as being R-modules by restriction of scalars
with respect to the canonical graded ring morphism R-^-Q^fR).
12.5.5. Lemma. Let MeQg(R)-mod, NeR-gr. If a map f iM^Q^N) is R-linear then it
is also Q^CR) -linear.
Proof : Pick meM, X eQ^R). For some L££(k) we have LX cr/k (r) and thus Lf(Xx) =
f(LXx ) =LX.f(x). Hence f(Xx) -Xf(x) eK ((^(N)) =o. D
12.5.6. Theorem. Let i be a rigid kernel functor on R-gr and let «_ be the kernel
functor on R-mod associated to it, then the following statements are equivalent :
1- Qg(R)JK(L) =Qg(R) for all Le£(>) ,where j is the canonical morphism R^R/k(R).
2. If MeQ8(R)-mod. then/^(M) = o.
3. If MeQg(R)-gr. thenK(M)=o.
4- Off—) =Qjf(R) ® " (natural equivalence in R-gr).
5. k_ has property T.

103
Proof : 1 =*2. If xe«_(M) then Lx=o for some L ££(*), hence Q^(R) (Lx) =o. By-
restriction of scalars this means that Q°(R)j (L)x=o i.e. x = o.
2 =* 3. Obvious.
3=*4. Let M£R-gr. The canonical R-linear map M->-0 (M) factorizes as
R
Since 0s (M) is a Qg(R)-module we nay restrict this sequence to a sequence of graded
morphisms :
M-2-»Qg(R) ®M -Qg(M) ,
R
(Note : a(m) =1 ®m, B(2q.®m.) =2q.m. and similar for ag and gg).
By 3., k (0s(R) ®M) =o hence Ker ag =k (M) that Ker agc« (M) is clear since Ker acK (M))
* R
thus Ker a=K_(M). Then g and a fortiori gg has to be a monomorphism. Because of the
foregoing lemma gg is (^(R)-linear, hence Imgg is a graded Qg(R) -submodule of Q^(M).
Note that gg is a graded morphism of degree o. Since Imgg contains M/k(M) it follows
that Qg(M)/Imgg is «-torsion and a graded Qg(R)-module, thus by 3. again, ImBg-Qj^(M)
or gg is an isomorphism.
4 => 1. We have Q^(R) -Q^(L) = gg(Qg(R) gL) = Qg(R) JK (L) for every L££(k).
1 =>5. Since Qg(R) is a subring of 0 (R) and since £ (k) has a cofinal system of graded
left ideals it follows from 1. that Q (R)j (L) =Q (R) for every Le£0<); i.e. K has
K K K ~
property T.
5*1. If «_ has property T then it has finite type, thus Proposition 12.4.7. yields
that Q^OO =0^ (M) for all Me R-gr. In particular for L«(«) we have that 0^(R) =
qf (L) =0^ (L) =0^ (R)JK (L) =Qg(R)jK (L). □
12.5.7. Definition. A rigid kernel functor k in R-gr is said to have property T if
QÇ is exact and commutes with direct sums.
12.5.8. Theorem. Let i be a rigid kernel functor on R-gr and let «_ be the associated
kernel functor on R-mod. Equivalently :

104
1. k_ has property T in R-mod.
2. k has property T in R-gr.
Proof : 1 =*2. If k_ has property T then it has finite type i.e. Q° and 0 are
equivalent functors in R-gr. Since 0 is exact and commutes with direct sums in R-mod it
will certainly by true that Q° is exact and commutes with direct sums in R-gr.
2*1. Take L£i(n), let j be the canonical morphism R-»R/« (R) and consider
Q?(R)j (L) • This is a graded Q^(R) -module thus it may be obtained as the quotiert of
a free graded Q^(R)-module which is then isomorphic to F = Q^(R)(nJ ®.- ®Q^(R) (n ) ® .-
for n.,,.,11 ,.-£2. Since Q^ commutes with suspensions and direct sums we see that
F=Q^(R(nJ ®.~ ®R(n ) ®.~), hence Q^(F) =F. Furthermore the epimorphism
it : F->-Q°(R)j (L) yields by exactness of oJ> an epimorphism :
which is the unique extension of it. Thus, since F=oJ>(F), it' = tt and Q^(Q^(R)j (L)) =
Qf(R)JK(L) follows. On the other hand Q^(L)/jK (L) is «-torsion hence Q^(L) =Q^(R)
yields that Of (R)/Q^(R)jK (L) is «-torsion too i.e. 0^ (Of (R) j K (L) ) =0f(R). Combining
these results yields that for every L££(0, we have that Q^(R)j (L) = QJ>(R). Theorem
12.5.6, 1 <*-5. then conludes the proof. □
12.5.9. Lemma. Let i be a rigid kernel functor on R-gr and let M,M' eQ^(R)-mod.
Suppose that f :M-*M' is an R-homomorphism such that Q^(R)f(M) is K_-torsion free as
an R-module, then f is Q^(R)-linear.
Proof : Pick x€M,xeQ^(R) and let L£j!(«) be such that LX c j (R). Then
f(LXx) =Lf(Xx) = LX.f(x), hence f(Xx) -Xf(x) e« (Q^(R)f(M)) =o. D
12.5.10. Lemma. Let MeQ°(R)-mod be K_-torsion free when considered as an R-module,
then the following conditions are equivalent :
1. M is injective in R-mod.
2. M is injective in Q^(R)-mod.

105
Proof : 1 =*2. If I: is the injective hull of M in Q^(R)-mod then
o^M^E_^E/M -> o
splits in R-mod, i.e. there is an R-linear f : E-*M such that f|M = 1 . Since E_ is
K_-torsion free we may apply Lemma 12.5.9. and conclude that f is QS(R)-linear,
meaning that the sequence also splits in Q°(R)-mod. So E=M follows.
2*1. Let E_ be the injective hull of M in R-mod. Since k_(M) = o we have k_(E) =o
hence 0 ÇE) = E_ and E_ is a 0 (R)-module thus, restricting scalars, also a Q°(R)-module.
By the assumption 2, the sequence
o->M->E_->E_/M-»-o
splits in Q°(R)-mod. But then M is a direct summand of E_ in R-mod too, thus E_=M. □
12.5.11. Corollary. Let i be a rigid kernel functor on R-gr, Me R-gr, and suppose
that « (M) =o and that M is also a Q5(R)-module. Then : E^(M) =Eg (M) in R-gr..
R qjf(R)
The foregoing lemmas may be applied in the study of compatibility of kernel
functors as in [ 33] , cf.loc.cit. for general theory concerning reflectors and
Giraud subcategories of Grothendieck categories. Again the fact that Q°(R) is not
necessarily a generator of the quotient category G(k) associated to k in R-gr, cannot
prevent us from describing the induced rigid kernel functors in G(k) by their filters
inQf(R).
All kernel functors « on R-gr considered will be rigid, k_ will be the associated
kernel functor on R-mod. Let k and a be kernel functors on R-gr, then «1 induces a
kernel functor v in Q°(R)-mod by taking the «.-torsion Q^(R)-modules for the torsion
class of v . It is easily checked that v. is a graded kernel functor inducing a rigid
kernel functor v on Q°(R)-gr. We say that k_. is Q^-compatible if « - Q?=Q? Ki in R"gr»
or equivalently k. Q°=Q^ k . in R-gr; in this case we also say that k is QS-compatible.
It is not hard to verify that, if « -, is Q^-compatible, then £(e..) has a cofinal subset
consisting of the Q^(L), LSi^), so if k has property T (then «_ has property T!) it

106
follows that £ 0 1 ) = (Q^(R) jk (L), Lsi^)}.
12.5.12. Example. If k . > a then k is (^-compatible.
12.5.13. Proposition. Let «,, « be rigid kernel functors on R-gr such that « -, is
Q^-compatible, then we have that Q8 Q^Ql) =Q^ Q^(M) for all Me R-gr.
Proof : By definition : v qS(M) = Qg01(M)). Put : e(-) = Eg (-),
] ] Qf(R)
N-^OO/^ 9f(M) -qfOQ/^O^CM)). (*)
By left exactness of Q^, N^Q^(M/«.(H)) is monomorphic, hence N is «-torsion free and
in Q^(R)-mod. Corollary 12.5.11 entails that e (N) = Eg(N). Furtliermore we have the
following isomorphisms in R-gr :
" ! (e OT/N) =k 1 (e (N)/N) =« 1 (Eg(N)/N) .
Applying Proposition 12.5.3. we obtain the following commutative diagram in R-gr :
o—-N-»-^ (N)-*!',(e(N)/N) —»o
= l h I*
o—N—Q^(N) —>Kl(Eg(N)/N)^o
where t is the restriction of the graded isomorphism e(N) -*-E°(N), and where rows are
K
exact. It follows from this that : Qg (N) =Q^ (N) in R-gr. However, from (*) it is
1 1
then clear that on one hand Qg (N) = të (cfCM)) while on the other hand 0g (N) =q£ (ft) ■■
11^ 1 T
Of CQ°CM) ). Lemma 12.'5.9. yields that these isomorphisms are graded isomorphisms in
Cf(R)-gr! □
Since « has property T if and only if k_ has property T the whole theory of
compatibility as expounded in [33] may now easily be adapted to the graded case, in
particular compatibility of symmetric graded kernel functors, as in [33] , may be
developped in a similar way (with an eye to projective geometry over a noncommutative
ring).

107
11.13. GRADED PRIME IDEALS AND THE ORE CONDITION.
In this section R will always be a positively graded and left Noetherian ring.
To a graded prime ideal P of R we associate a graded prime kernel functor «p as in
12.3.7 and a rigid graded symmetric kernel functor KR_p as in 12.2. For a left
Noetherian ring R, to every kernel functor «_ on R-mod there corresponds a symmetric
o
kernel functor k_ on R-mod which is the largest symmetric kernel functor smaller than k .
13.1. Proposition. Let k_ be a graded kernel functor on R-mod inducing the rigid
o o
kernel functor « on R-gr, then k_ induces a graded symmetric « on R-gr which is the
largest (in the ordering of rigid kernel functors on R-gr) rigid symmetric kernel
functor smaller than «.
Proof : Follows in a straightforward way from the fact that an ideal of R which
contains a graded left ideal of R also contains a graded ideal of R. □
For completeness sake we include some results on kernel functors associated
to m-systems (semi-multiplicative systems) i.e. a set D such that for each pair
x,y£D there exists a z£R such that xzyeD. To an m-system D in a left Noetherian
ring we may associate a symmetric kernel functor k_ given by £ («_) = {L left ideal of R,
L3RdR, for some deD.}.
13.2. Theorem. Let R be a left Noetherian graded ring and let D be an m-system in
R such that R-D is additionally closed then h(D) is an m-system and the symmetric
kernel functor «_ associated to h(D) is symmetric graded.
Proof : Pick xeh(D), yeh(D). Since D is an m-system there is a z^R such that
xzy^D. Write z = 2 z.. Ifz-^o then xz.y^R-D for all z. ^o appearing in the
expression for z, then xzyeR-D, contradiction. Thus xz.yeh(D) for some i and h(D)
is an m-system in R. Obviously, the kernel functor associated to h(D) is symmetric
graded on R-gr. □
If k_ is a kernel functor on R-mod then (k_) will be the rigid kernel functor

108
on R-gr given by the filter :£((V) ) =£Q0 nL (R).
g b
13.3. Lemma. Let P be a graded prime ideal of R and consider the following kernel
functors on R-mod :
Kp associated to the multiplicatively closed set G(P).
k, , . associated to the multiplicatively closed set h(G(P)).
K„_p associated to the m-system R-P in R.
Kj |-d_d-| associated to the m-system h (R-P) in R.
Then the following relations hold :
2' [KRIP)g>[^i(R-P))g=Kh(R-P)
Proof : Trivial. □
13.4. Theorem. Let R be a positively graded left Noetherian graded ring and let
PeProj R then k_=k, . . and "R_p = Ki,fR-PV tw^ere Kp an^ kr_p are defined as in
12.3.7 and 12.2. resp).
Proof : Since k_ is induced on R-gr by «p (cf. 12.3.7.) we have that £((k ) ) =-C(«p)
i.e. «=(/<) >«,,.. Now let Le£(« ), let it : R->R/P be the canonical epimorphism.
From L££(k ) it follows that ir(L) is a graded left ideal of R/P containing a left
regular element, hence, that ir(L) is a graded left essential left ideal of the prime
left Noetherian and positively graded ring R/P, with (R/P)+ fo.
First we prove that if J is a graded left ideal of R/P such that h(J) consists of
nilpotent elements then J=o. Note that if Ra has this property for some aeh(R)
then aR has this property too, so it will be sufficient to prove that no right ideal
J of R/P has that property! So suppose that J is a right ideal of R/P such that h(J)
consists of nilpotent elements. If a^o in h(J) then Ra^o ,
i.e. for all a^o in h(J), Ann^(a) ^R/P.
The Noetherian hypothesis allows to choose a b^h(J), b^o, such that AnnRb is

109
maximal amongst left annihilators of elements of h (J) which are nonzero. Take r, eR,
p £ t t-1
then Ann^b =AnnRbr, but br, eh(J) hence (br,) =o for some t such that (br,) fo.
t t t-1
Thus AnnRb =AnrL (br, ) , but thenbr,b=o. Hence bRb=o, contradiction.
So we have shown that if J is a nonzero left ideal of R/P then I contains a non-nil-
potent homogeneous element. Since (R/P) fo and tt(L) is essential in R/P, it follows
that there exists a non-nilpotent homogeneous element a. in it (L) n (R/P) . Now repeat
the argumentation of the proof of Proposition 9.2.3. to conclude that ir(L) contains
a homogeneous regular element of positive degree q, say c. There is thus an element
of degree q in R, d say, such that d+p£L for some peP and ir(d+p) =c. Since L is
graded d+p eL, and still it (d+p ) =c (since p eP). Moreover d+p is homogeneous of
degree q and d+p €G(P) if and only if deG(P); hence d+p eh(G(P)) because deG(P)
follows from the fact that ir(d) is left- (hence right-) regular in R/P. Therefore
o o
«p=«wp-, now follows from statement 1 in Lemma 13.3. Ey Proposition 13.1, K-^(p-\ =Kp
o o o
is induced in R-gr by Kp="R.p hence k^ = CKR.p)g>Kh(;R.p) whereas Kh(R-p)>Kh(P) is
obvious from h(G(P)) ch(R-P). The fact that (k p) is exactly kr_p as defined in
12.2 finishes the proof of the second statement. □
13.5. Remark■ A special case of this Theorem has been proved in [ 30 ] for prime
ideals P of R which have the property that, for all x,y£R-P, xRynyRxn (R-P) f$.
Exactly the extra information drawn from the graded version of Goldie's theorems
made the complete solution to the problem given here possible.
13.6. Proposition. Let R be a left Noetherian positively graded ring and PeProj R
a graded prime ideal such that the left Ore conditions with respect to G(P) hold,
then the left Ore conditions also hold with respect to h(G(P)).
Proof : Given seh(G(P)), reR, consider (Rs : r). Write r = r. + ... +r with deg r. <
n
< ._ <deg r and put L - n (Rs : r.) c (Rs : r). Clearly L is a graded left ideal which
n i=1 x
is in £00 because of the left Ore condition with respect to G(P). Now Lej:(« )J =
JC(«p) and it follows from Theorem 13.5 that Le£(«, „,), consequently there exists
s^hfGfP)), r1 eR such that s.r -r..s. □

110
13.7. Remark. The analogues of Theorem 13.4 and Proposition 13.6 hold in the non
Noetherian case for those PeProj R such that R/P is a graded Goldie ring. (R is of
course still positively graded!).
11.14. THE PRESHEAVES ON Proj R.
In this section R is again always a positively graded and left Noetherian ring.
First we deduce a representation theorem for finitely generated modules over Proj R.
14.1. Theorem. Let MeR-gr be a finitely generated object then QR_p(X) =0 f°r aH
P e Proj R if and only if Q (M) = o if and only if there is an n e IN such that M = o
for all m>n.
Proof : By Lemma 12.4.1, R is generated as an R -ring by the finite number of chosen
homogeneous generators for R+ i.e. R=R <r1,.-,r > with deg r >.~>deg r- >o. If
PeProj R then QR_p(M) =o means that I M = o for some graded ideal Ip£P- Put
I = S I then rad I =P. n ._ np with P. 3R+ for all i = 1,.~r. Consequently
PeProj R F I r i
vr
R' cI for some Ne IN. Hence M = «n (M) or Q„ (M) =o. Conversely M = k (M) yields
K+ K+ K+
that QR_p(M) =0 for a11 P^Proj R since R+iz!P.
Now let M be generated by the homogeneous elements m.,._ ,m with deg m. <deg m. < ... <
vr
deg m . If M =o for all m>n then R M= o for N>n deg m., hence QR (M) =o. Con-
vr
versely, suppose that R1, M= o for some NSH. If N >N.deg r then R=R <r-,... ,r >
/Jr + om olm
entails that R^ cr^( thus for all N >N.deg rm : R„ M- o . Take N, >deg m +N.deg rm
o l o
and N1 >N deg r and consider x^NL i.e. x = x1 m1 + ... +x m with deg x. >N.deg r =N
for all j. But then x.m, eRj. M = o for all j,k i.e. x = o, thus M = o for all m>N.,. □
•^ o
The following result is fundamental in the study of the projective spectrum :
14.2. Theorem■ Fix n e IN . For each n>n let there be given an additive subgroup
p of R . The following statements are equivalent :
1. There is a unique PeProj R such that PnR =p for all n>n .
J n rn o
2. a. For all n,k and for all m>n we have R-P-Ri. cp t.

111
b. Let re IL, t£R and suppose that n,m>n are such that rR,tcp for all k,
then rep or t £tl.
n m
c. p 4R for some n>n„.
*n r n o
Proof : 1=>2. It is trivial that a. and b. are fulfilled if 1. holds. Now suppose
that p =IL for all n>n . Then for all k^o and all aeR, we have that :
a R£ a...aR£ aGRk(r+1) +2 £. ■
1 r • i
for all (£-,._,£) e ]Nr. Choose r such that k(r+1) >n , then the foregoing comes
down to : a R. a... a R. a cp for all (£.,.„,£ ) e ]Nr , hence aep or aeP
contrail *-T 1 r k
dieting R+£P.
2=>1. From c it follows that we may fix n>n and an element a£R -p . For each me IN
' o n rn
put
Pm={xGRm' for large t> for a11 [£1 ' " 'V G ^ ' a R£ a - a R£ xCpm+tn+S£.
Obviously, if m>n then P' =p because of b. For anymeiN, P' is additive. Put
' ' o m Mil ' m
P=2 P'. Clearly P is an ideal of R because of a. and P ^R since a^P . Suppose
m ' n n n rr
seR^, weR are such that sR,wep'x+ for all ke]N. Since R is left Noetherian
RsRw is generated by a finite number of homogeneous elements which may obviously be
chosen to be of the form sr,w. Thus we may choose t fo large enough such that : for
all (£.,._ ,1 ) e ]N , for all chosen generators sr, w we have :
aR„ a ._ aR sr^cp^^
it j_
For any t' larger than t (fixed as above) we obtain :
((RaR)t\sRkwcP7+x+M+k ,
for all k,T. Let us write A - RaR, then for all k.-yeiN :
(A1 ) s (A1'), w cp .

112
t' t'
Now, (A ) fo implies that 7>nt' because deg a = n>o. Further : A fo, because
otherwise a£p would result from b. Again from b. it follows now that either,
t ' t '
(A ) s is in p +. or (A ) w is in p. where both t+m and k+M may be taken to
be larger than t'n>n . The construction of P yields that either s or w must be in
P and therefore we have established : PeProj R. Assume that QeProj R is different
from P and suppose Q satisfies the conditions of 2.
Pick b£Q-P and suppose that b is homogeneous. If deg b>o then for all m>n and
all (£.,.- ,1 ) s ]Nm we have :
1 m
b R „ b ... b R„ bcQnRCp^cp
1 m-1
where X =md + 2£. . So (Rb)mcp, but this contradicts b£P. □
i x
14.3. Remark■ The above theorem is true if R is commutative, even in the absence of
the left Noetherian condition. We will prove a similar theorem for special but non
left Noetherian rings further but it is not known to us whether the above theorem
holds for arbitrary positively graded rings in general.
14.4. Proposition. Let I be a graded ideal of R and let C be the set {J, J an ideal
of R maximal such that J££(I)}. Put gC = {(J) , JeC}. Then gC consists of graded
prime ideals and£(I) = n £ (K ). Moreover if I CR. then the elements of gC are
P e gC R"F
in Proj R.
Proof : It is straightforward to check that C consists of prime ideals (see also [32] ),
thus gC consists of graded prime ideals. Obviously an ideal He£(I) if and only if
H£P for all PeC. Since the filters we have to compare are both graded filters it
follows that a filter basis for £ (I) is obtained by taking the graded ideals not
contained in any of the (P) egC. This proves that £(.!)= n £ (k ). If Icr then
g PegC R"P +
for any PegC we have that P f I and a fortiori P7iR+, whence PeProj R. □
14.5. Proposition. Let I be a graded ideal of R, then : R+nrad(I+) =R+nrad(I).
Proof : If P is a graded prime ideal containing I+ then either PDI or P3R+. Note

113
that this property does hold if R is not necessarily left Noetherian. □
14.6. Corollary. Put V+(I) = {P eProj R , PDI}. Then :
V+(D -V+(I+) -V+(rad(I)) =V+((rad(I))+) -V+(rad(I+)) .
Let k . be the symmetric graded kernel functor associated to £ (I), Q° the
graded localization functor on R-gr corresponding to k,, j, : R-*-Q?(R) the corresponding
canonical graded ring homomorphism of degree o. If Icr then «. is a projective
kernel functor on R-gr, because of Lemma 12.4.9 then Q§(M) for Me R-gr may also be
given by (Qf(M))m = (xe Qt(M) , there is aJei(I) such that Jx c j (r) andJnxC
«I»»n+mfcralln>V-
Note that if ICR+ then je£(l) if and only if J e£(I). An easy strengthening of
Proposition 14.5 yields that : R nrad(I) =R+nrad( © I ) for each graded ideal
n>n n
I of R and each natural number n .
o
Endow Proj R with tlie topology induced by the Zariski topology of Spec R as
follows. Put, for any ideal I of R : V+(I) ={PeProj R, PDI}, X+(I) =X-V+(I) ivhere
X = Proj R. In these definitions we may replace I by the smallest graded ideal of R
containing I, so from now on, when we write V+(I) or X (I), I is understood to be
graded. By Corollary 14.6. we may assume further that ICR+ and V+(I) =V+( © I )
n>n
o
for each n e IN. Again from Corollary 14.6 it follows that V (I) remains unaltered
under taking radicals and positively graded parts of I in any order. The following
relations are easily checked :
V+(I+J)=V+(I)nv+(J)
V+(U) -V+(inj) =v+(I) uv+(J) .
This shows that the sets X (I), I varying in the set of graded ideals (or those
contained in R+) of R, exhausts the open sets of the topology induced in X.
To an open set X+(I) the projective kernel functor RT is associated i.e.

114
£(I+) =£(ki ) ={L€L (£), LeL (£), L contains an ideal J in L (R) such that
rad(J)3I+}.
14.7. Lemma. For any open set X+(I) of Proj R we have :
£(I+)=n{£(KR_p),pex+(I)}.
Proof : If P* £X+(I) then P' ££(« ) hence P' cp for some PegC(I+). Thus £("R_p) <=
£(k ,). Applying Proposition 14.4 yields :
£(«T )= ^ £^R-p)= ^ £(KR-P')- D
+ PegC(I+) K P* sx+(I) K r
It is easily verified that in case R is left Noetherian, X+(I) is quasi-compact
in X. The following proposition may be verified by mimicing, step by step, the proof
of the corresponding properties for Spec R mentioned in [ 32] , taking care to use the
up till now established graded theory where necessary :
14.8. Proposition. Assigning Q° (R) to X+(I) for each graded ideal I of R; defines
a presheaf of graded rings q£(R) : Open (X)°PP->-R-gr, over the Zariski topology on
Proj R = X. If R is a left Noetherian graded ring then every open set is quasi-compact
and the presheaf q£ (R) is separated.
If ÇçJJL) is separated then it may be embedded in a sheaf of rings which is
obtained by applying the well known sheafification functor L (sheaf reflector) which
associates to a separated presheaf P a sheaf LP which on UeOpen(X) is given by :
L (U) = t lim lim, p(V) ,
UeCov(U) VeU
where Cov(U) is the poset of coverings of U by open sets of X. Since lim and lim
are inner in R-gr and also in the category of graded rings (with graded morphisms of
degree o) it follows immediately that LQ^(R) is a sheaf of graded rings. For Me R-gr
we obtain a presheaf Q^(M) and a sheaf LQ^(M) of graded R-modules. The following

115
theorem is again an easily verified graded version of a similar theorem in [32] .
14.9. Theorem. Let R be a positively graded left Noetherian prime ring then
Lc£(R) =(£(R) i.e. q£(R) is a sheaf.
14.10. Remark. The graded ring of "functions" for q£(R) is defined to be :
lim Q°(R). This ring may be obtained by localizing R at « =V{kt ,1 graded ideal
x+(i) x K
of R}, where the supremum (V) is defined as being the kernel functor induced on R-gr
by V_kt on R-mod.
14.11. Theorem. Let P eProj R then we have :
1. KR_p=v^KI . l+ such that Pex+(I)}
2. The stalk of the sheaf LQ^(R) at P is exactly QR_p(R).
Proof : 1. PGX+CI) if and only if I+£P , thus kj<kr_p for every graded ideal I such
that PeX+(I). Conversely, if J£-C(«„p) then J contains a nonzero graded ideal I
such that l+sz:p , henco PeX+(I) and also je£(K ).
2. The stalk of LQ^(R) at PeProj R is defined to be lim , q|(R) =s. since R is
Pex+(I)
left Noetherian "n_p(R) is finitely generated hence J KR_p(R) =o for some graded ideal
J£P, consequently Qj(R) ^Q|_p(R) is a monomorphism. This remark entails that in the
directed system considered, the maps f'Q^fR) ->Q|_p ,PeX (I) , eventually turn out to
be monomorphisms for small I. Therefore we obtain a monomorphism of degree o,
f : S^-q|_p(R) . An element xe (Q|_p(R)) may be represented as a graded morphism of
degree m , m : I->-R for some l€£()i ). Now in also represents an element y, of
q|(R). By construction of the f, it is clear that fT(yT) =x, therefore the image of
yT in S, y say, is such that f(y) =x and this states exactly that f is an isomorphism
of degree o (note that both the gradation of S and Qn_p(£) extend the gradation of R,
so the bijective morphism of degree o is an isomorphism.
If X is any topological space, fl a (pre-) sheaf of graded rings defined over
X. Define a presheaf flQ by «0(U) =«(U) for each UeOpen(X), then.

116
14.12. Lemma, fi is a sub (pre-) sheaf of fi.
Proof : Let Py denote the restriction morphism fi(U) ->fi(V) with respect to open sets
Vcu in X. Since p.. is graded of degree o, the restriction of p.. to R(U) maps R(U)
into fi(V) . As a subpresheaf of fi, fi is separated. Further if U={U.}. is a covering
of UeOpen(X) and if f. efiflj.) are elements of degree o such that :
pUi (f.)=PUJ (f,), for all i,j;
U. nU. U. nu. J
1 J i J
then there is an fefifU) such that p.. (f) =f. for all i. Again, the fact that pv has
i
degree o implies fefipj) , consequently « is a sheaf.
The sheaf of rings (LQ^(R)) defined over Proj R is called the structure
sheaf of Proj R; it will be denoted simply by Proj. Since direct limits of graded
morphisms of degree o "respect" taking homogeneous parts of degree o we find that :
Proj(X+(I)) = (Q|(R))0 for every graded ideal I of R. Projp = (Q|_p(R))0 for every
PeProj R.
In the commutative case graded prime ideals can be related to common prime
ideals in some of the rings appearing in the structure sheaf of Proj. This relation
is fully expressed by saying that Proj is a scheme i.e. Proj R has a covering by open
sets X+(I), I in some set of graded ideals 6, such that : Proj| X+(I) =Spec(Q|(R)) ;
indeed it suffices to take for 6 the set of graded ideals generated by a single
homogeneous element of R. The general non-commutative case looks to be difficult and it
is still non-solved up to now. However in the sequel we show that this property
does hold for the class of Zariski central rings, introduced in [33 ] . We need some
ring theoretic introduction about these rings.
II. 15. GRADED ZARISKI CENTRAL RINGS.
15.1. Basic Facts About Zariski Central Rings.
Zariski central rings, studied in [ 33] , are a special case of the

117
birational algebras considered in [31 ] . In this section we repeat some definitions
and fundamental properties, referring to [31] and [33] for more details.
In this section R is a ring with unit and C denotes the centre of R. The ring R is
said to be Zariski central if the Zariski topology of Spec R = X has a basis of open
sets X(S) ={PeX, P 75S} where S varies through the subsets of C; equivalently : R is
Zariski central if rad(I) =rad(R(lnC)) for every ideal of R. Examples of Zariski
central rings are, semisimple Artinian rings, Azumaya algebras, rings R=A[X,y>,5] where
A is a simple ring and where the centre of R is not a field : in particular A [X,y>] is
Zariski central.
15.1.1. Lemma. If R is a Zariski central ring then the filters :
£(I) ={L left ideal of R, L d J ideal of R, rad(J) 3 1}
£(R-P) ={L left ideal of R, LdJ ideal of R, J£P},
are idempotent filters for every ideal I of R and every prime ideal P of R.
Because of the above lemma we are able to localize at «,, %_p> the kernel
functors on R-mod associated to £(I), £ (R-P) resp., even if R is not necessarily left
Noetherian. If « is a symmetric kernel functor on R-mod then let £ (k ) be the filter
generated by the ideals I of C such that RI ££(«)• The following statements are
equivalent, for some MeR-mod :
1. M is k-torsion,
Q
2. Jl is k -torsion.
The class of Zariski central rings is closed under taking epimorphic images.
15.1.2. Theorem. Let 1 be a symmetric kernel functor on R-mod, R a Zariski central
ring, and suppose that k has property T then k has property T. Moreover :
1. 0 (R) is a central extension of R and 0 (R) and Q (R) are isomorphic rings.
K
2. For every Me R-mod the C-modules 0 (M) and Q (Jf) are isomorphic.
3. If I is an ideal of R then Q (R) j (I) is an ideal of 0 (R), where j : R^O (R) is
K K ri K ti
the canonical morphism.

118
4. There is a one-to-one correspondence between prime ideals of 0 (R) and prime
ideals of R which are not in £ (k) .
15.1.3. Lemma. If D is a graded division ring then D is a Zariski central ring.
Proof. D=D [X,X"1,y>] for some skewfield D and an automorphism f of D , cf.
Theorem 6.3, unless D=D is a skewfield and in the latter case the statement is
o
trivial. Now S = D [X,y>] is Zariski central, cf. [31] and localizing S at the Ore
2
set {1,X,X ,...} yields that D is Zariski central too. □
15.2. GZ-and ZG-Rings.
A positively graded ring R is said to be a GZ-ring if it is a Zariski centra]
ring. A ring R, positively graded, is said to be a ZG ring if for every graded ideal
I of R we have that rad(I) = rad(R(I nC)). A GZ-ring is a ZG-ring; we will deal with
GZ-rings here, however, most results may also be obtained for the larger class of
rings.
15.2.1. Theorem. Let R be a positively graded Zariski central ring and let P be a
graded prime ideal of R. Then :
1 • kd_p coincides on R-gr with «„_p where p=PnC i.e. «npPQ =Kp_p(pt'Q f°r every
Me R-gr.
2. K„_p has property T and Qf_p(M) = oJ_p(M) for all Me R-gr.
3. If J is a graded ideal of R then Qn_p(R) j (J) is a graded ideal of Q§_P(R) (here
j : R^Qn_p(R) is the canonical graded morphism of degree o).
4. There is a one-to-one correspondence between graded prime ideals P of R such that
Pe£(h „_p) » and proper graded prime ideals of Qn_p(R)-
Consequently, if I is a graded ideal of R (hence rad(I) is a graded ideal of R too),
then :
Q|_P(R)JK(rad(I)) =rad(Q|_p(R)JK(I)) .
Proof : 1. Since Kn_p is induced on R-gr by f_p_ p and since ^_D_p and ^r_p coincide on
R-mod. (Theorem 15.1,2.)

119
it follows that KR_P and «r_p coincide on R-gr. Consequently '•'n_p may be regarded as
being the kernel functor on R-gr associated to the central (Ore) set h(C-p)■
2. From 1 it follows that, ie£(« ) if and only if there is an xeinh(C-P), but
then x~1 £Q|_p(R) follows from x~1 £QR_p(R) and (Rx)x~1 CR with Rxe£(KR_p). Thus
Q|_p(R) JR-pW =Q|_p(R). where JR_p : R^Q|_p(R) is the canonical epimorphism of
degree o. For Me R-gr, QR_p(M) -Qf_p(M) follows from Zariski centrality of R. Since
both %_p and ^r_p are graded and of finite type it follows that (cf. Proposition
12.4.7) Q|_p(M) ^QR_p(M) =0^,01) sQ8_p(M),
3. Let J be any graded ideal of R and take xe Q|_p(R) j p(J)Q|_p(R). Write x = 2' qi Ji q]
with qi ,q^eh(Q|_p(R)), Jieh(J). We may fix an element ceh(C-P) such that
cqjejR_p(R) for every i. Since c is central in Q|_p(R), we get : cx€Q|_p(R) jR_p(J)
or x€Q|_p(R)jR_p(J) by left multiplication with c~1€Q|_p(R).
4. For all graded prime ideals Q of R we have that Q = Qn_p(R)JR_p(Q) is an ideal of
q|_p(R), which is non-proper if and only if Q££(R-P). It is easily checked that Qe
is a graded ideal and that QSnjR_p(R) =jR_p(Q) if Q££(R-P). If I is an ideal of
oJ_p(R) which is not in Qe then InjR_p(R) is an ideal of jr;_p(R) which is not in
JR_P(Q) because c|_p(R) (I n jR_p(R)) = I by property T for *R_p.
Therefore, if I and J are ideals in oJ_p(R) such that IJCQ then
(InJR_P(R))(JnJR_p(R)) cQSnJR_P(R) =JR_p(Q) and this entails that mjR_p(R) or
J nj (R) is contained in jn_p(Q) since the latter is a prime ideal. Thus, I or J
is contained in Q , Combination of the foregoing properties yields that every prime
ideal of Qn_p(R) equals Qe for some graded prime ideal Q of R, The second statement
in 4. is a trivial consequence of the first, □
15.2.2. Remarks. 1, The same holds for ZG-rings.
2. In the same way similar properties may be established for any kernel functor
associated to a central multiplicative set. In particular, such properties hold on a basis
for the Zariski topology of Spec R, indeed one may consider kr , c^o in C, which is
2
associated to {1,c,c , .„ },
3. From the general theory of graded localizations it follows that : if R is prime

120
then Q°(R) is prime for any rigid kernel functor on R-gr, if k has property T then
Q^(R) is left Noetherian if R is left Noetherian.
15.2.3. Proposition. Let R be a positively graded Zariski central ring. Then R is
a Zariski central algebra.
Proof : Let I be an arbitrary ideal of R . Clearly C is in the centre of R „■ If P„
o / o ' o oo
is a prime ideal of R such that P0:)R0(I nCQ) then P + R+ is a prime ideal of R such
that :
P +R3rad(R (I nc ) + R ) = rad(R((R (I ncl+Rjnn)
O + 00 O + ^^00 O +
Now, (R (I nc )+R+) nC=>RI RnC is easily checked by compairing points of arbitrary
degree; therefore
PQ+R+Drad(R(RI RnC)) =rad(RI R) ,
hence P dI . □
o o
The first main advantage of Zariski central rings is that, even in the non
left Noetherian case, Proj is having the nice properties one can hope for. First let
us establish the following extended version of Theorem 14.2,
15.2.4. Theorem. Let R be a positively graded Zariski central ring. Let nQe U be
fixed, and suppose we are given additive subgroups p of R for all n>n . The
following statements are equivalent :
1. There is a unique graded prime ideal P of R, P7SR , such that PnR =p for all
n>n .
o
2.a. For all n,k£]N and for all m>n we have Rp R, Cp .
b. If r£R , teR with n,m>n are such that rR,tCp for all keM then rep
n' m ' o K ^n+m+k n
or tepm.
c. Pn^Rn for some n>n .
Proof : Repeat the proof of Theorem 14.2. for 1 =»2. For 2 =» 1, repeat the construction
of the graded ideal P, P^R^ as in Theorem 14.2, If we prove that P is a prime

121
ideal then the uniqueness argument used in Theorem 14.2 stays valid and thus the proof
will be finished. So, suppose that s£R , weR are such that sR,wcp' for all
keiN. Put T=RsRnC, then RTRwcp. Suppose that w£P yields TeP then it also yields
rad(RT) cp i.e. RsRcp or se P. This means that in the proof of Theorem 14.2 we may
assume that s eC but then RsRw is finitely generated (by sw) and the same argumentation
carries over. □
The equivalent of Proposition 14.4 does hold for Zariski central but not
necessarily left Noetherian rings, the proof goes by reduction to the commutative case.
11.16. SCHEME STRUCTURE OF Proj OVER A GRADED ZARISKI CENTRAL RING
16.1. Lemma. Let R be a positively graded Zariski central ring with centre C and let
k be a rigid kernel functor on C-gr, To k we may associate a rigid kernel functor
k on R-gr, which is given by £(«), the filter generated by the ideals RI, I££(k ).
Then Q° is a central graded localization and k has property T; moreover the centre of
Q°(R) is the graded ring of quotients of C at k .
Proof : The first assertions are just modifications of the properties mentioned in
Section II. 15.2, The statement about the centre is a straightforward consequence of
the fact that j (Z), where Z is the centre of j (R) and where j : R^-oJ (R) is the
canonical ring homomorphism of degree o, is a graded C-module such that j (Z)/C is
k -torsion ; hence the localizations in R-gr of j (Z) and of C at «■„, coincide. □
16.2. Lemma. Let R be a positively graded GZ-ring, then Proj R has a basis for the
Zariski topology which consists of open sets X+(Rc) where c runs through the set of
homogeneous elements of C+. The graded kernel functor k„ on R-gr, associated to
X+(Rc), has property T. It is clear that «„ is obtained from k„ in the way described
in Lemma 16.1. Let ojj be the localization functor corresponding to k , let j : R-"Q^(R)
be the canonical ring morphism. Then graded ideals I of R extend to graded ideals
of Q^(R). Furthermore Q^(R) is a Zariski central graded ring with centre cj> (C), Q^c
being the localization functor on R-gr corresponding to «r ,

122
As a consequence of Proposition 15.2.3. we have
16.3. Lemma. Let R be a positively graded GZ-ring. If p is a prime ideal of C ,
C being the centre of R, such that C nR p =p then rad(R p) is the unique prime ideal
of R lying over p.
16.4. Lemma. Let R be a positively graded Zariski central ring with centre C. Let
k be a rigid kernel functor on C-gr associated to a multiplicative set in C and let
Q° be the localization functor on C-gr corresponding to k. Denote S = Q^(R), D=Q°(C).
fel
Suppose that q is a graded prime ideal of C and put qv ' = ©a , for some fixed eeM.
Then we have rad(Sq) = radfSq1-6-1).
fel
Proof : The inclusion rad(Sq) Urad(Sq)v ' is clear. Conversely, since q is central, it
e fe~l fe~l
follows that for any xeo we have that x ea eqv ', whence x€rad(Cqv J), thus
qCradO^r6-1) follows. For any ideal I of R we have that : Q^(R) rad(I) =rad(Q^(R)I)
(cf. Theorem 15.2.1, 4.). Applying this to radtRq^) yields SqCradCSq^) and thus
rad(Sq) =rad(Sq(e)) holds. □
The last lemma we need is the non-Noetherian version of the Stalk Theorem 14.1
16.5. Lemma. Let R be a positively graded Zariski central ring. Let Proj R be the
topological space with its structural sheaf as constructed in Section 14. Then the
stalk of Proj at PeProj R is exactly Q§_p(R).
Proof. There is a basis for the topology consisting of open sets X+(I) such that the
graded rigid kernel functor K, associated to X+(I) is central and Qj(R) =Q§nC(R)
(ring isomorphism of degree o). Consequently lim , Qf(R) equals lim Qf(R),
P€X+(I) PeX+(I)€B
where 6 is the selected basis for the topology, hence
lim Qf(R) = lim , Q? nr(R) = c| (R) (where Y = Proj C)
P€X+(I) i peY(lnC) i L ^"p
Since R is graded Zariski central Q§_ (R) =Qn_p(R) and the statement follows. □
16.6. Theorem. 1. If R is a prime positively graded Zariski central ring then Proj R

123
is a scheme. This means that there exists a basis 6 for the Zariski topology of Proj R,
consisting of open sets X+(I) where I is a graded ideal of R contained in R+, such that
each X+(I) endowed with the induced topology and sheaf is isomorphic to Spec(Qg(R))
with its usual topology and structural sheaf (cf. [32]). If X+(I) and X+(J) are in 6
then the ring (QfT(R)) is generated as a ring by the restrictions of the rings (Q°(R))
and (Qf(R))0.
2. If R is a positively graded Zariski central ring, then the presheaves of X+(I) and
Spec(Qg(R)) are isomorphic. In this case the scheme structure of Proj R follows by
sheafification methods.
Proof : For the basis 6 we will choose {X (Re), c a homogeneous element of C+}. Since
R is a Zariski central ring this 6 is a basis for the topology on Proj R as well as a
covering for it. We split the proof in the following three parts :
a. The bijection between X+(Rc) and Spec(Qg(R)) ,
b. The topological homeomorphism, and
c. The (pre-) sheaf isomorphism.
a. Bijective correspondence between X+(Rc) and Spec(Qg(R)) .
Let qQeSpec(0s(R))o. Since Qg(R) is Qg(C)-Zariski central, it follows from
Proposition 15.2.3. that (Qg(R)) is Zariski central over (Qg(C)) (which is contained
in, but not necessarily equal to the centre of (Qg(R)) . So if p = q n (Q (R)) then
p is prime and q is the unique prime ideal of (Qg(R)) lying over p i.e. :
q0 = rad((Qg(R))0pQ).
Define q' CC by, q^={deCm, dec~meqo, e = deg(c)}. It is easily verified that q'
remains unaltered if one substitutes p for q in this definition. Now let us first
Fo no
establish that ci1J1=WGCm> there exist N,Me]N with eM=mN and dV'^c^}. That q^ is
contained in the latter set is obvious. Conversely, suppose N>e, then M>m and
(d c ). (d1 c ) eq_- Since q is prime and both elements considered are central
it follows that either dVmeq or else d _ec~ +meq . In the first case we are done.
In the second case we repeat the procès and in the end we have to face the case N<e,
M<m with eM=Nm. Then : (d d ) (c c ) = (d c ).(dc ) is an element of

124
(Q(C)) q Cq .thus also d c eq . The characterization of q' just obtained makes
vxc o o o o
it clear that q' is a graded prime ideal of C.
Consider CnRq' and pick c1 = 2r. y. eCnRq', with y. €q' .r^R and, as CnRq' is
graded, we may suppose that c., r., y. are homogeneous elements. Since y. commutes witli
N "1 "i~e e
r. for all (i,j), we may choose N large enough such that c, = 2r. ... r. y. ... y. y-
j 1 i.| ijj I i i
with c . >e, i.e. each term in the sum has the form given above with respect to at least
one index i appearing in the expression for c.. Enlarging N if necessary, we may assume
that N = e.c .
Thus c1- c " with m = deg(c.) is in (Qg(R)) q cq because each term in C1., c " may be
written in the form :
-,m + deg(y.) * 1 Ve e _deg ^i
c BWV . r ...r y. ... y. . y. c
x1 XN '
which obviously is in q because y. eq.
It follows that Rq' nc =q' , hence rad(Rq') = Q is a prime ideal of R by Zariski centrality
of R. Notation : write Qex for Qg(R)Q.
By Theorem 15.2.1. : Qex = rad(Qg(R)q') and then by Lemma 16.4 : Qex = rad(Qg(R)q'^e^).
Clearly Q is a graded prime ideal of R which does not contain c because c e Q yields
ceq' and 1 eqQ. Thus Q£X+(Rc) and we have obtained a well defined map <l> : Spec(Qg(R)) -►
-+X+(Rc), given by qQ ^Q. Conversely, given Q e X+ (Re), consider (Qex) n(QS(Q) . If
y is in the latter set then cny€QnC for some n. Put q' = QnC and form q' *-eK
Then pQ = {de (Qg(C))o, cndeq^e for some n}, is a prime ideal of (Qg(C)) . So the
relation cny€QCCnc translates to yep i.e. (Qex) n(Qg(Q) =p . By Zariski cen-
trallity of (Qg(R))Q over (Qg(C))Q it follows then that q =rad((Qex)o) is a prime ideal
of (Qg(R)) • One checks that Q corresponds to q in the way described first, hence the
map <l> is clearly surjective. Injectivity of i// will follow from, q =q" =rad((Q ) )
where ^(qQ) =Q and ^(qQ) =Q. It has been established already that QeX = rad(Qg(R)q' ^) .
ex
Now take x£ (Q ) . Since in any positively graded ring T any ideal I has the property
rad(I ) 3 (rad I)Q we get that xe rad((Qg(R)q' ^)0) •

125
However we have
But
(Q^(R)q'(e))0-(Q^(R))0 q^(Qg(R))_.e q:e .
ye(Q^(R))_ie means that cXye (c£(R))0, hence (Qg(R))_ie = (Qg(R))0 c"1, whereas for
each i, zeq'. means that zc ep . Taking into account that q'cp we obtain
xerad((OJ;(R)) p ) = q^ Finally, we arrive at : p c (Qex) cq and therefore q =
rad((Q x) ) follows from the fact that q is prime and p =q n(Qg(C)) .
b. The homeomorphism X+(Rc) Spec(Qg(R)) .
Since X+(Rcd) =X+(Rc) nx+(Rd), c.dec , the open sets X+(Rcd) with d varying
through C+ forms a basis for the topology induced in X (Re) by the Zariski topology of
Proj R. Let <l> :X+(Rc) -+Spec(Q*(R))o, P^rad((Pex) ) be the bijective map constructed
in a. Let dec , then dec"me (Qg(R))Q. If Pex+(cd) then PeProj R is such that id£P.
This implies that dec"m£ rad((Pex)Q) ; indeed (dec~m)Ne (pex) would yield deNepex or
M eM
c d £P for some Me IN. Both c,d are central, thus c0P yields dep and cdep.
Conversely, if i(/(P) is in the open set {qe Spec(Qg(R)) , q does not contain d c m}, then it
is equally easy to derive from this that cd^P. So it becomes evident that ^ is a
topological homeomorphism, as claimed.
c. The sheaf isomorphism.
First consider the case where R is a prime ring. In this case the presheaf
induced on X+(Rc) is actually a sheaf. On the other hand (Q^(R)) need not be a prime
ring, however since (Q°(R)) is a Zariski central ring it is easy to prove that the
usual presheaf on Spec(Që(R)) is a separated presheaf. To prove the sheaf isomorphism
in this case it will therefore be sufficient to find a basis for the topology such that
over open sets in this basis, the ring of sections of the sheaf on X+(Rc) is isomorphic
to the ring sitting over the image of that open set in the presheaf on Spec(Q^(R)) .
Let c,dec+ and put deg(c) =e, deg(d) -m, c' =cmd~ee (Q^(c))Q and d' =dVme (o|(C))o.
To the open set X+(Rcd) in X+(Rc) there corresponds the open set Y(d') = {qe Spec(Qg(R))0,

126
d' £q} in Y = Spec(Qg(R)) . In the (pre) sheaf structure of X (Re) the ring of sections
over X+(Rcd) is (Qg,(R)) , while the presheaf on Y associates Q,, ((0s(R))Q)) to Y(d'),
where Q,, is the localization functor in (Qg(R)) -mod corresponding to the central
multiplicatively closed set generated by d'. It will now be sufficient to prove the
following :
Qd,((Qf(R))0)-(Q^d(R))0-Qc,((Q|(R))0).
Since c and d commute, the second isomorphism will follow from the first by a symmetric
argument.
Define a ring morphism <p : Qd, ((Qg(R))0) - (Q|d(R))0 hr
t [(d')"N (xc"f)] = (x ce+m df)(cd)"(f+eN\ where x£ (Qg(R))fe.
The inverse homomorphism \j> : (Qg,(R)) -+Q,, ((Q^(R)) ) will then be given by
■Hy- (cd)"f) = (d')~M. (y df(e"1))c"f(m+1\ where ye (Qgd(R))f ,e+m, . All implicit
assumptions may easily be verified (this part of the proof is identical to the proof in the
commutative case).
That (Qg,(R)) is generated by the images of (Qg(R))0 and (Q§(R)) is again mere
verification. In case R is not prime, one may repeat a similar proof, taking care to
replace R by j (R) or Jd(R) where necessary. Since torsion parts will behave nice we
will obtain, over a basis of the topology, isomorphisms between the rings making up
the presheaves on X+(Rc) and Spec(Qg(R)) . Sheafification then yields the desired
result, n
16.7. Remark. If in the situation of the theorem we have that R cc i.e. R =C then
o oo
it follows that R ->-(Qg(R)) is an extension of R and therefore, to the canonical ring
morphism R0 "*■ (Q? (R) ) „ there corresponds a presheaf morphism : Spec (Qg(R))->-Spec RQ.
In this case Proj R may be viewed as a Spec R -scheme.
16.8. Remark. If in the situation of the theorem C is generated as an C -ring by Ci
then we have :
1. If P€X+(Rc) then (PeX)0 is a prime ideal of (Qg(R)) i.e. the bijection <l> may
then be given by .// (P) = (Pex) .

127
2. If R is a prime ring then (Qg(R)) is prime and the presheaf Spec(Qg(R)) is a sheaf.
Proof : 2. will follow from 1. applied to P=o.
1. Pick c£C+. It has been noted that X+(Rc) depends only on rad((Rc) ), hence the
fact that C =C7 implies that we get a basis for the Zariski topology of Proj R by
taking (X+(Rc) ,ce C,}. If I and J are ideals of (Qg(R)) such that IJ c (Pex) for some
PeX+(Rc) then pick is I, j ej. For any qe (Qg(R)) we see that c"m q£ (Qg(R))
hence iqj ecm(pex)o c (Pex)m. Since Pex is a graded prime ideal of Qg(R) it follows
ex ex
that i or j is in P ; hence i or j is in (P ) .
16.9. Remark. The statement of 16.8 holds for Azumaya algebras over polynomial rings
but fails for rings of twisted polynomials unless these are already Azumaya algebras.
Indeed if R=A[X,a] , where A is a simple ring, has center k [T] then T=X implies that
R is an Azumaya algebra thus if not, then T is a polynomial in X of degree higher than 1.
III. LOCAL CONDITIONS FOR NOETHERIAN GRADED RINGS.
In this chapter, all rings considered will be commutative and Noetherian
unless explicitely mentioned otherwise.
III.1. Injective Dimension of Graded Rings.
Let us introduce the following terminology : a ring R is termed a gr.-local
ring if it has exactly one maximal graded ideal (these ideals are said to be gr.-maximal)
in the set of proper graded ideals of R.
Let P be a prime ideal of R and let S be the multiplicatively closed set
h(R) -P=h(R-P). Since R is Noetherian, our results in 11.12 and 11.13 imply that
localization in R-mod at the kernel functor associated to S is inner in R-gr i.e. for
any Me R-gr , QS(M) = Qg_p(M). If Q= (P) then S=h(R)-Q and QSÇM) = 0f_Q(M) follow.
Since k_ has property T in R-mod, k has property T in R-gr and therefore Q§_ p(R) is
gr.-local and Q|_p((P) J = Qg_p(R)(P) is gr-maximal.

127
2. If R is a prime ring then (Qg(R)) is prime and the presheaf Spec(Qg(R)) is a sheaf.
Proof : 2. will follow from 1. applied to P=o.
1. Pick c£C+. It has been noted that X+(Rc) depends only on rad((Rc) ), hence the
fact that C =C7 implies that we get a basis for the Zariski topology of Proj R by
taking (X+(Rc) ,ce C,}. If I and J are ideals of (Qg(R)) such that IJ c (Pex) for some
PeX+(Rc) then pick iei, j ej. For any qe (Qg(R)) we see that c"m qe (Qg(R))
hence iqj ecm(pex)o c (Pex)m. Since Pex is a graded prime ideal of 0s(R) it follows
ex ex
that i or j is in P ; hence i or j is in (P ) .
16.9. Remark. The statement of 16.8 holds for Azumaya algebras over polynomial rings
but fails for rings of twisted polynomials unless these are already Azumaya algebras.
Indeed if R=A[X,a] , where A is a simple ring, has center k [T] then T = X implies that
R is an Azumaya algebra thus if not, then T is a polynomial in X of degree higher than 1.
III. LOCAL CONDITIONS FOR NOETHERIAN GRADED RINGS.
In this chapter, all rings considered will be commutative and Noetherian
unless explicitely mentioned otherwise.
III.1. Injective Dimension of Graded Rings.
Let us introduce the following terminology : a ring R is termed a gr.-local
ring if it has exactly one maximal graded ideal (these ideals are said to be gr.-maximal)
in the set of proper graded ideals of R.
Let P be a prime ideal of R and let S be the multiplicatively closed set
h(R) -P=h(R-P). Since R is Noetherian, our results in 11.12 and 11.13 imply that
localization in R-mod at the kernel functor associated to S is inner in R-gr i.e. for
any Me R-gr , QS(M) =oJ_p(M). If Q= (P) then S = h(R)-Q and QSÇM) =q|_q0Q follow.
Since k_ has property T in R-mod, k has property T in R-gr and therefore Q§_p(R) is
gr.-local and Qg_p((P) J =Qg_p(R)(P) is gr-maximal.

128
Let kg(P) denote the graded division ring Qg_p(R)/q|_p(P)•
1.1. Lemma. Let P be a prime ideal of R, then :
1. For any Me R-gr, QR_pM =0 if and only if QR_m-) (M) = o if and only if Qg(M) =o.
2. Put supp M= {Pespec R, Qn (M) =o}, then Pesupp M if and only if (P) e supp M.
K-r ^- g
3. If P is a graded prime ideal then : Eg (kg(P)) = Eg(R/P).
4. Let P be a graded prime ideal and let A. = {xeEg(R/P), P1x=o} then the following
properties hold :
a. A- is a graded R-submodule of Eg(R/P) and Eg(R/P) = U A..
1 a i>1
b. A.+1/A. is a graded kg(P)-module.
c. A1 =kg(P).
5. Any injective object in R-gr is a direct sum of some Eg(R/P)(n), for some graded
prime ideals P of R and some nez.
6. Denote by gr.inj .dinuM the injective dimension of M in the category R-gr. If
SCh(R) is any multiplicatively closed set, Me R-gr, then : gr.inj.dim _. S M <
S 'R
gr.inj .dimJ-1.
Proof : 1., 2., 3. are easy consequences of the foregoing chapters. 4., 5., 6., follows
as in the ungraded case, cf. [22 ] . n
Let He R-gr. and consider minimal injective resolutions in R-gr and R-mod
resp. :
o ->■ M ->■ Eg ->■ Eg ~* ...
o 1
o ->■ M ->■ E ->■ E1 -> ... .
o 1
For a prime ideal P of R, let Mg(P,M), resp. m (P,M), be the number of copies of
Eg(R/P) in Eg, resp. of ER(R/P) in E .
1.2. Proposition. Let Me R-gr, P a graded prime ideal of R. For notational convenience
we write Qg for the localization functor on R-gr associated to h(R)-P, then :
1. The group 0s Ext^(R/P,M) is a free graded kg(P)-module of rank Mg(P,M).
R n

129
2. m|(P,I1) =Mi(P,M]
3. inj.dim Qg(M) = gr.inj .dim qS(M) •
Qg(R) Qg(R)
Proof : We have
Ext^R/P.M) -EXT^(R/P,M) =HOM^(R/P,q|) = HomR(R/P,Q?) ,
hence :
Qg Extp (R/P,M) = QgHonu (R/P,Qg) = Horn (kg (P),Q?)
R * x Qg(R) x
and the latter is a free k^P)-module.
If P., is a graded prime ideal of R such that P. jtP then 0s Hom(R/P, Eg(PJ) =o, hence
Qg Honu(R/P,Qg) is a free kg(P)-module of rank equal to Mg(P,M). Since ExtR(R/P,M) is
a free k(P)-module of rank m-(P,M) whilst Q Ext(R/P,M) is the localization of
0s Ext(R/P,M) at P (Q the localization in R-mod associated to P), 2. folloivs immediatly
and 3 follows from 2. □
1.3. Proposition. Let P be a prime ideal of R such that P f (P) , then for all ne IN
and MeR-gr we obtain : m ((P) ,M) = Mn+1(P,M).
Proof : Because of Corollary 2.4 in[ 1 ] we have : m ((P) .M) =m (Qg((P) ),Qg(M))
n g n &
while Mn+, (P,M) =M +1 (Qg(P) ,Qg(M)). So we may replace R and M by Qg(R) and QgCM) resp.,
i.e. we may suppose that R is gr-local and P the gr.-maximal ideal of R. In that case
P is a maximal ideal of R. Since R/(P) is a graded division ring, there exists aeR
such that P = (P) + Ra. From the exact sequence :
(*) o^R/(P)g -R/(P)g— R/P — o ,
where m denotes multiplication by a, we obtain :
a
m .
.-^Ext (R/(P) ,M)—^Extn(R/(P) ,M)^Extn '(R/P.M)-
.- -Extn+1 (R/(P)g)M)-^Extn+1 (R/(P)g,M) - ...
Note that Extn(R/(P) ,M) =EXTn(R/(P) ,M) is a graded R/(P) -module, hence Extn(R/(P) ,M)

130
is a free graded R/(P) module. Consequently,
m :Extn(R/(P) M)-Extn(R/(P) M) ,
a g g
is monomorphic. The long exact sequence obtained above yields a short exact sequence :
o^Extn(R/(P) ,M)—^-►Extn(R/(P) ,M) ^Extn+1(R/P,M) ^o .
In this sequence V = Ext (R/(P)_,M) is free graded of rank m ((P) ,H), whereas V =
Ext (R/P,M) is a vector space of dimension ix (P,M) over the field R/P. In R/(P) -
mod we have the exact sequence :
ma
(**) o—»V —=>V —V —>o .
%((P)g,M) *«n((P)e.M)
Since Va(R/(P) ) g we deduce from (*) and (**) that V * (R/P) n g . On
g %+1(P'^
the other hand V = (R/P) , therefore Mn((P) ,M) =M +1(P,M). □
1.4. Corollary. In the situation of 1.3, n ((P) ,M) =o if and only if m +1(P,M) =o.
1.5. Corollary. Let P be a graded prime ideal of R. The minimal injective resolution
of Eg(R/P) in R-mod is the following :
o-+Eg(R/P)—>-E(R/P) —* ® E(R/P')-*-o
P'
where the direct sum is over all prime ideals P' fV of R such that (P') =P.
Proof : That resolution has the form :
o -» Eg (R/P) —■* E (R/P) —* Q1 —♦ o
M1(P',Eg(R/P))
where 0, = 2 E(R/P') . Applying Proposition 1.3. and Corollary 1.4.
P' eSpecR
we get Q, = ® E(R/P') where (P') =P. D
1 pi S
1.6. Remark. If M is a maximal ideal of R which is graded then Eg(R/M) =E„(R/M), e.g.
if R = k PC. ..Xn] and M = (X., .-X ).

131
1.7. Corollary. Let R be a commutative Noetherian ring with left limited grading. If
M£R-gr is a finitely generated graded module then : gr.inj .dim_M = inj .dim„M.
Proof : Put n = gr. inj . dinuM < °°. If inj.dinuM^n then inj . dirruM = n+1. Now m , (P,M) fo
for some prime ideal P of R, so by the foregoing results P is maximal. Clearly P is
not graded and by Proposition 1.3. it results that m ((P)„,M) fo hence M^((P) ,M) fo.
We claim that (P) is graded maximal, indeed if not, then (P) cp P graded yields
that m^+1(P.,,M) to, contradiction. Since R has left limited gradation it follows that
(P) is a maximal ideal, i.e. P = (P) , contradiction. Consequently, inj .dim„M = n.
o S
In going from R to R[X] we need the following lemma where R has trivial
gradation and R [X] is graded as usual :
1.8. Lemma. If P is a graded prime ideal of R[X] then either P=pR[X] or P =pR[X] + (X),
where p = P n R.
Proof : Since P=>pR[X] and R[X]/pR[X] = (R/P) [X] we may assume p=o, i.e. R is a domain.
Let K be the field of fractions of R. Since PnR = o, PK[X] is a graded prime ideal of
K[X] i.e. PK[X] =o or PK [X] = (X). □
1.9. Proposition. Let R be a Noetherian commutative ring and take P eSpec R. Put
E = E„(R/P). Consider the ring R[X] graded corresponding to the degree in X. Denote
M = E[X] = E®R[X] .
R
The minimal injective resolution of M in R [X]-gr is :
o ->M -+e|(R[X]/PR[X]) -+Eg(R[X]/PR[X] + (X))->o
and the minimal injective resolution of M in R[X]-mod is :
°^-^ ER[X] (RM/PRM) -® er[X] (R[X]/P') ^o
the direct sum being taken over the prime ideals P' of R[X] such that P' nR = P.
Proof : Consider the minimal injective resolution of H in R[X]-gr :

132
o-*M-*IB—>-I°-»I°
o 1 2
Since Ass M={PR[X]}, Ig =Eg(R[X]/PR[X] ). Taking into account, Corollary 2.6. [ 1 ] , we
obtain :
Mi+1(PR[X] +(X),M)=Mi(PR[X] + 00/00, E[X]/XE(X3) =Mi(P,E) .
Since Mi(P,E) =1 if i =o and Mi(P,E) =o if i>o, we have m.,(PR[X] + (X) ,M) =1 and
M-(PR[X] + (X),M) =o for all i>1. Let P' be a prime ideal of R[X] such that
M-(P' ,M) >o. Then necessarily P' e Supp M and therefore P' nRDP. If P' nR^P, put
P' nR = Q. By Lemma 1.8. it follows that P' =QR[X] or P' =QR[X] + (X).
In case P' =QR[X] we have :
Z'iCP'.M) =Mi(QR[X],E[X])=dimk(p) QR_p (Ext* ^ (R [X] /QR [X] , EM)) .
However,
ExtR[x] (R[X]/QR[X] ,E[X]) = Ext*(R/Q,E)®R[X]
R
and as Q^P we obtain Ext^(R/Q,E) =o for all i>o. Thus m.(P',M) =o for all i>o.
R 1
In case P' =QR[X] + (X), we have an exact sequence
m
o^R[X]/QR[X] ^RPq/QRtX] -+R[X]/P' —*o
yielding a long exact sequence :
in . -
.- Ext1(R[X]/QR[X] ,M)-^-»Ext1(R[X)/QR[X] ,M) ^Ext1 ' (R[X]/P',M)-
^Extl+1(R[X]/QR[X] ,M)—^Extl+1(RM/GR[X] ,M) -*
(all Ext are Ext„ m ) •
From the above we deduce : Ext (R[X]/P',M) =o for all i>1. Finally this amounts to
if = Eg(R[X]/PR[X] +(X)) and Ig = o for all k>1.
For the second statement of the proposition, let the minimal injective resolution for
M in R [X] -mod be given by :
— o 1 2

133
It is clear that I = En m (R [X] /PR [X] ). From Proposition 1.2 and 1.3 it follows that
O K1AJ
I- contains ® E (R[X]/P'), where P' varies in the set of prime ideals of R[X] such
I p, k LaJ
that (P') =PR[X] or P' =PR[X] + (X). Therefore (P') =PR[X] if and only if P' nR = P.
Indeed if P' nR = p and P' f (P') then ht(P') =1 +ht(P') ), thus if (P') ?^PR[X] then
ht(P') >1 +ht(PR[X]) =1 +ht(P). Thus ht(P') >2+ht(P). On the other hand, since R is
Noetherian and P' nR = P, ht(P') <1 +ht(P), cf. [17 ] , contradiction. The converse
implication being obvious this proves our statement.
Let P, be a prime ideal of R[X] which is not graded and such that Pi nR^P. Then
(P^ 3PR[X] and (Pp i*PR[X] + (X) because (P.,) =PR[X] + (X) would imply P., nR =
(P^ nR = p. From Proposition 1.3. we retain that M-((P,) ,M) = Mi+1 (P., ,M)) • However
we have already established m • ((Pi) ,M) =o for all i>o, hence a' • (P, ,M) =o for all i>1.
This yields thus that I, = ® En rvl (R[X]/P' ) where P' is as stated. If I, were nonzero,
I p, K|AJ ^
then there exists a prime ideal V? of R[X] such that ^(PtjK) ^°- If p2 ^s graded then
from Proposition 1.2, we deduce m2(P2,M) fo and thus l|^o, contradiction.
If P2 is not graded, then we deduce from Proposition 1.3. that m2(P?,M) = m..((P2) ,M) to.
In this case it is necessary that (P?) =PR[X] + (X), hence P nR=(p ) nR = p. since
P2?^(P2) , ht(P2) =1 +ht((P2) )=2+ht(PR[X] = 2+ht(P), but fromP2nR = P, ht(P2) <1+ht(P)
follows (cf. [17] as before), contradiction.
Therefore m2(P2,M) =o and I?=o follows.
1.10. Corollary. Let R be a commutative Noetherian ring, E injective in R-mod, then :
gr. inj . dimR [x, E [X] + 1 = inj . dimR m E [X] .
III.2. Regular, Gorenstein and Cohen-Macaulay Rings.
Let R be a commutative Noetherian ring and M a nonzero finitely generated
R-module. We put : V(M) =supp(M) =V(AnnJI) ={PeSpec R, AnnRMcp}. Recall that the
Krull dimension of M, denoted by K.dinuM is defined to be the supremum of the lengths
of chains of prime ideals of V(M) if this supremum exists, and °° if not. (the reader

134
may verify that K.dirruM coincides with the Krull dimension of M defined in Section 1.5.).
We have hy = K.din^ (R) (QR_p (M) ).
If I is an ideal of R such that IM^M then the least r for whicli ExtR(R/I,M) fo will be
called the grade of I on M, denoted by : grade (I,M). It is easily checked that
grade(I,M) is exactly the common length of all maximal sequences contained in I.
If R is a local ring with maximal ideal n, then M is said to be a Cohen-Macaula
module or a CM.-module if grade (fi,M) =K.dimRM. A not necessarily local ring is said
to be a Cohen-Macaulay ring or CM.-ring if Qn_p(R) is a Cohen-Macaulay Q p(R)-module
and in that case M is a CM.-module if Q 0Q is a C.M.-Q D(R)-module. It is easy to
verify that M is a CM.-module if and only if for each maximal ideal fieV(M), p.(fi,M)=o
whenever i<ht,(fi), cf. [ 37] . We say that M is a Gorenstein-module if for every maximal
ideal fleV(M), m^.M) =o if and only if i ^htM(fi). In [ 37] it has been established
that M is a Gorenstein-module if and only if QR_p(M) is a Gorenstein-QR_p(R)-module for
all peVQf).
A Gorenstein ring R is a ring which is a Gorenstein-module when considered as a module
over itself. For a detailed study of these classes of rings we refer to [37] . First
we give local-global theorems in the graded case.
2.1. Theorem. Let R be a commutative Noetherian graded ring, M^R-gr a finitely
generated object. The following statements are equivalent :
1. M is a Cohen-Macaulay module.
2. For any graded prime ideal P of V(M), Qn_p(M) is a Cohen-Macaulay Q„_p(R)-module.
Proof : 1 =»2. is trivial. To prove the converse let us assume first that QR_p(M) is
a CM. QR_p(R)-module for every graded PeV(M). Let P'eV(M). If (P') =P' then we
are done. So suppose that P' f (P') . In this case we have that ht. P' =ht, (P') +1.
Choose Khtj^CP'). Then i-1 <htM((P') ), hence v-^. ((P') ,M)=o. By Proposition 1.3
it follows then that m-(P',M) =0 and therefore M is a CM.-module. □
2.2. Theorem. Let R be a graded commutative Noetherian ring, MeR-gr a finitely generatec

135
object. The following statements are equivalent :
1. M is a Gorenstein-module.
2. For each graded PeV(M), QR_p(M) is a Gorenstein QR_p(R)-module.
Proof : 1 =»2. Easy. For the converse, suppose that Qn_p(M) is a Gorenstein QR_p(R)-
module for each graded P£V(M). Let P' be arbitrary in V(M). If P' = (P')g then the
assertion is clear. Supposing that P' f (P') and i^ht^P'), we obtain i-1 ^ht.((P') )..
Since QD fDn (M) is a Gorenstein Qn ,DM (R)-module it follows that :
R-(P Jg R-(P Jg
"i-lCCP^g^^nCOR^p.) ((P')g), QR_(pI) 00)-o .
g b
From Proposition 1.3. we deduce that m - (P' ,M) =o. In a similar way it may be deduced
from m • (P',M) =o that i^htu-CP'); hence M is a Gorenstein module. □
A local ring with maximal ideal n is said to be regular if gl.dim R<» or
2
equivalently, if K.dim R = dirru ,_(fi/fi ). If R is not a local ring, then R is termed
to be a regular ring if for every maximal ideal n, Qn_o(R) i-s regular.
2.3. Theorem. Let R be a graded commutative Noetherian ring, then the following
statements are equivalent :
1. R is a regular ring.
2. Qn_p(R) is regular for every graded prime ideal P of R.
Proof : The implication 1. =>2. is obvious.
2=>1. If P' is an arbitrary prime ideal of R then the ring QR_m,-. (R) is regular i.e.
gl.dim Qnfpii (R) <■»• Since QR_p,(R) is the localization of s=Qd_cpm (r) at tne
prime ideal generated by P' in it, it follows that we may assume that we have chosen R
to be gr-local with gr.maximal ideal (P') . By hypothesis gl.dim S=n<°°. Consider a
finitely generated R-module M. By Proposition 1.2. we have that inj.dirruCXwp,-, (M) <n
and hence gr.inj dimRM<n. Therefore gr.gl.dim R<°° and Corollary 1.7. yields gl.dim R<»,
hence gl.dim QR_pt (R) <"•

136
Our following result shows that regular and positively graded rings are close
to being polynomial rings over regular rings.
2.4. Proposition. Let R be a positively graded regular ring with a unique graded
maximal ideal tt, then R is a regular local ring and R=R K.,.-,X,] , the X., i = 1._k,
being indeterminates over R , which are taken to be homogeneous elements of positive
degree.
Proof : We proceed by induction on n = K.dim R. If n = o then R=R is a field. If n>o,
7
choose a nonzero homogeneous element x of positive degree in fi-fi (if there is no such
x then R = R follows). Put R. =R/(x), n =fi/(x). Now R is a graded regular ring and
K.dim R1 =n-1.
The induction hypothesis yields : (R.,) is a regular local ring and R. s (R. ) [Y.,... ,Y, ]
with deg Y. >o, i=1.„k. Since (x) or =o it follows that (RJ =RQ. So, if K.dim RQ=c
then d + k=n-1. Choose representatives U-,.-,U, in R for the images of Y.,.-,Y, in R.,
and put U,+, =x. One easily checks that R=R [U.,... ,U,+J . Therefore we have an epi-
morphic graded ring homomorphism : ^ : R [X.,... ,X, .] -*-R ILL ,... ,U, .] =R. However, the
fact that K.dim R = k+d+1 = K.dim R [X.,... ,X, J yields that *p is an isomorphism. □
2.5. Remarks. 1. If R is a positively graded regular ring then R is regular. Indeed,
if ai is a maximal ideal in R then fi = cj®R+ is a maximal graded ideal. Since the
localization Q n(R~) is regular and also positively graded (note that it coincides with the
graded ring of quotients Q° o(R))> it follows from the foregoing that (Qd_^(r))0 i-s
regular. However (QR_o(R)) =Qd (R ) • It follows that R is regular too.
o
2. See also J. Matijevic [ 24] . If R= ® R. is Gorenstein, hence Cohen-Macaulay, then
i >o 2
it is not necessarily true that R is Gorenstein or CM.. Indeed, put T=k[X,Y]/(X ,XY),
where R is any field. Put Q1 = (Y), Q? = (X), then the image of 0, nQ2 in T is zero. Let
W,V be indeterminates of degree 1 and put S =T[W] , Q^Q^, Qe=Q2S+WS and I =Q®nQ|.
Put U=S/I, R=U[V]/(xV+yV,V2) where x,y denote the images of X,Y resp. This
construction yields a graded ring R with K.dim R= 1 and R =T. However T is not a CM.-ring.

137
If fi denotes the unique maximal graded ideal of R then QR_n(R) ^s Gorenstein.
III.3. Graded Rings and M-sequences
R is a commutative graded ring throughout this section. Such a ring R is
said to be completely projective or is said to have property op. if for any graded
ideal I and for any finite set of graded prime ideals p.,...,p with h(I) Cp. u ... Upn
with h(I) Cp u ._ Up it follows that I is contained in at least one p.. It is not
hard to verify that, if R is completely projective then so is S R for every multipli-
catively closed subset S of h(R). It is also straightforward to check that epimorphic
images of a completely projective ring have property cp. too.
3.1. Lemma. Let R be a commutative graded ring such that all graded prime ideals of R
are in Proj R, then R is completely projective.
Proof : Assume I£p-, 1 <i<n and assume that the p. are not comparable one to another.
If n = 1 then h(I) jzîp. follows. We proceed by induction on n. The induction hypothesis
yields h(InP.)£ u p. ,1<i<n. Hence there is an a. eh(lnP.) such that a. £p. for
j^i J 1 -1
each j ^i. Let x. = n a.. Clearly x. e I and x. £p. for j ?i but x. £p.. Since no
J i ... j ' i i j i i
Jfi J J
graded prime ideal of R contains R+ is easy to see that we may choose deg x. >o for
each i, 1 <i<n. Put d. =deg x. and d = d1._d , y. =x.' i. Then deg y. =d and it is
' l ° l 1n'7ii ° J i
obvious that y=y1 + .- +y is a homogeneous element such that yeI and y^P, u ... uPn- a
3.2. Examples. 1. If R is a positively graded ring such that R is a field then R is
completely projective.
2. Let R be a gr.local ring with gr.maximal ideal fi such that R/fi is a graded division
ring with non-trivial grading. Then R has property cp. Indeed, if P is a graded prime
ideal then Pcfi, Since R/fi is non-trivially graded, there is a homogeneous t of
positive degree, t^fi, hence t^P.
3. Let 0 be a discrete valuation ring with maximal ideal w generated by t say. Put
R = 0[x] sû[X]/(tX) i.e. in R we have tx=o. Then (t,x) is a graded ideal and every

138
homogeneous element of (t,x) is a zero-divisor; however t+x is not a zero-divisor. If
Pi,.-,P are the prime ideals associated to R, then these are graded and clearly h(I) c
P. u ._ up but I£P. for each i, 1 <i<n. So 0 [x] does not have property c.p.
Let R be a commutative Noetherian ring and M a finitely generated R-module.
A sequence a.,.-,a in R is said to be an M-sequence if fa, ,.-,a )M^M and for each
i = 1,.~,n, a. is not an annihilator in the module M/(a., ,.„ ,a. JM. If I is an ideal
of R such that IM^M, then we shall denote by grade (I,M) the common length of all
maximal M-sequences contained in I. As it was already mentioned in Section III.2, it is
well known that grade(I,M) is the smallest number r such that Extn(R/I,M) ?o.
3.3. Corollary. Let R be a commutative Noetherian ring which is graded and completely
projective. Let MeR-gr be a finitely generated object and I a graded ideal of R such
that IM^M. Putting grade(I,M) =m, then there exists in I an M-sequence of length m,
consisting of homogeneous elements. Moreover any M-sequence formed by homogeneous
elements of I has the same length i.e. grade(I,M).
Proof : For m = o the assertion is clear. Let m>1 and let P^.-.P be graded prime
ideals associated to M. Since grade (I,M) >o, I;z!Pu._up and thus because of Lemma 3.
h(I) jz:P1 u ._ uPg. Hence there exists f eh(I) such that f-, ^P1 u ._ up , therefore £.
is a non-annihilator of M. Let M, be the graded R-module M/f.M, then grade(I,M) = m-1 and
we apply the induction hypothesis. The second statement may also be proved using a
similar induction argument. □
3.4. Corollary. Let R be a commutative Noetherian graded ring having property c.p.
Let P be a prime ideal of R having height n. Then there exist homogeneous elements
a, ,.~,a in R such that P is minimal over Ra, + ._ +Ra .
Proof : For n = o there is nothing left to prove, so we may assume n>o and proceed by
induction on n. Let (Q-), —,Q, } be the set of minimal prime ideals of R and as R is
graded, each one of the Q^, i = 1,... ,k, is graded. Since ht(P)>1, P is not contained
i-
m any Q. and therefore, Lemma 3.1. entails h(P) jZQ u ._ uq . Pick a1 eh(P)- u Q..
1 k I i=1 i

139
In R/Ra1 , P/Ra. is a graded prime ideal with ht(P/Ra.,) <n-1. The induction hypothesis
yields that P/Ra. is minimal over (a,,,.-,a ), where the a., i = 2,._,n, are homogeneous
elements of R/Ra,. Choosing representatives a. for a. in R it is obvious that P is
minimal over (a.,... ,a ). □
1 n
Consider a gr.local ring R with maximal graded ideal tt and suppose that R is
completely projective. Let MeR-gr have finite type. The ring R=R/AnnRM is a gr.-
local ring with maximal graded ideal n = fi/AnnDM and R has property c.p. Write n for
the Krull dimension of M, then n=K.dim R = ht(H). By Corollary 3.4., there exist
homogeneous elements L,.,ï eîî such that H is minimal over Ra7+ -. + Râ~" in R. Our
hypothesis yield that M/a.M + ._ + a M =M/â,M + ... +â M. Furthermore R/ÏÏâ.. +.-+R ïï is a
gr.Artinian ring, hence M/ïï.M + ... + a M is a gr. Artinian (R/Ra. +._+Ra )-module. On
the other hand we have that R/Ra. + .- +Ra =R/Ann3l + Ra1 + .- + Ra , thus it follows
1 n R 1 n'
that M/a.M+ ... + a M is a gr.Artinian (R/Ann„M + Ra.. + ... +Ra ) -module; consequently,
the latter module actually is gr. Artinian in R-gr.
A system of homogeneous parameters for the module M with K.dim M=n is a set of elements
a.,.-,a eh(îî) such that M/a..M + ... + a M is Artinian in R-gr (hence of finite length in
R-gr!).
3.5. Remarks. 1. As in the ungraded case one may establish that every M-system may be
included in a system of homogeneous parameters for M.
2. If R is positively graded and such that R is a field then there exists a system
of homogeneous parameters for R+. However if R is not a field but a local ring with
maximal ideal cj then the graded ideal n = cj+R+ may not have a system of homogeneous
parameters, as the following example shows.
3.6. Example. Let A be a discrete valuation ring with maximal ideal w and uniform
parameter t. Let R=A[X]/(tX); then R is Noetherian and positively graded. Moreover
R is a gr. local ring with Krull dimension 1. One easily verifies that there cannot
exist a homogeneous a- e (co + R+)/(tX) such that {a-} is a system of homogeneous
parameters .

140
In the light of the foregoing observations we now suppose that the graded
Noetherian ring R has the form R ®R. ®._, where R is a field. Then R is gr.local
& o 1 ' o &
with gr.maximal ideal R+.
3.7. Proposition. Let MeR-gr have finite type and let {a...... ,a }eh(R+). A necessary
and sufficient condition for a..,.-,a to be a system of homogeneous parameters for M is
that M has finite type as an object of R [a..,...,a ]-gr , while n is minimal as such.
Proof : If a.,.»,a is a system of parameters for M then M=M/a.,M + .- + a M is Artinian
in R-gr. Since M has finite type its grading is left limited. On the other hand, since
M is Artinian in R-gr and since R is positively graded, the grading of M is also right
limited. Consequently M has to be a finite dimensional R -vector space. Nakayama's
lemma then yields that M is finitely generated as a R [a..,...,a ]-module. Conversely,
if we start from the assumption that M is a finitely generated R [a..,...,a ]-module,
then M/a.M + ._ +a M is a finite dimensional R -vector space, thus Artinian in R-gr.
Next let us check minimality of n as such. Indeed, if {b,,._,b } is a set of homo-
I m
geneous elements such that M is a finitely generated R [b.,._ ,b ] -module then M/b..M + ...
._ +b M is Artinian in R-gr, i.e. of finite length too. But then K.dim M<m excludes
the possibility m<n. □
3.8. Proposition. Let MeR-gr have finite type. If a-,.-,a is a system of homogeneous
parameters for M then the elements a- a are algebraically independent over R .
Proof : Letting a^,.-,a be such a system of parameters, consider the ring S=R [X.,... ,X
with gradation defined by putting deg X. = deg a., i = 1,.-,n. Specializing X. -*-a.
defines a surjective ring homomorphism of degree o, <p :R [X.,... ,X ] ->R [a-,.-,a ] . We
have that K.dim R [a.,...,a ] = K.dim M = n. However if Ker <p ?o, then the Krull dimension
should have dropped, so y is monic and an isomorphism. □
3.9. Proposition. Let MeR-gr be finitely generated. The following statements are
equivalent :
1. M is a Cohen-Macaulay module.

141
2. If a-,.-,a is a system of homogeneous parameters for M then M is a free (graded
free) R [a,,... ,a ] -module.
3. Every system of homogeneous parameters for M is an M-sequence.
Proof : 1 =*2. Put S = R [a.,.^,aA . By Proposition 3.8., S is a regular ring with
K.dim S=n. Since M is a Cohen-Macaulay module, grade(R ,M) = K.dim M=n, and from
Corollary 3.3. there exists an M-sequence b.,b ,._,b withbeh(R) i = 1,.~,n. For
i = 1,._,n, put M,.. = M/b,M + ._ +b.M and put M, . =M. Considering the fact that S is
mi+1
a regular ring, we may derive from the exactness of o-*-M.- >M. ->M.+1, > o, (where
m. + 1 denotes multiplication by b. + 1) that : (*) p.dirruM. +., = 1+p.dirndl.*
On the other hand, M = M/b.M + .. + b M is an S-module of finite length. Therefore M
is a finite dimensional R -vectorspace. Because S is gr.local we have that p.dinicM ,=
p.dim„R =gl.dim S=n. However from (*) it follows that p.dirruM = n+p.dirruM . Now
p.dinuM =n yields p.dinuM = p.dinuM =o, consequently M is a free S-module.
2*3. As well as 3 =» 1 are rather easy. □
In the sequel we aim to give some applications of the theory just expounded
to the study of filtered rings and modules.
3.10- Proposition. Let R be a filtered ring with exhaustive filtration FR and let
MeR-filt have the exhaustive filtration HI. Suppose that all submodules of M are
closed in the topology defined by the given filtration. Let r1f.-,r be elements of
R and denote by a., i = 1,._ ,n, the image of r. in the graded ring G(R). If the
homogeneous elements a.,.-, a form a regular G (M) -sequence then r.,...,r form a regular
M-sequence. Moreover for each k, 1 <k<n we have :
GtM/r-jM + ._ + rk M) ^GfMJ/a^CM) + ... + ak G(M).
Proof : We proceed by induction on n.
If n = 1 ; let x^M xj* o. For some p£Z, xe F M and x^F .M. Since x fo (x is
' ' P P"1 P P
the image of x in G(M)> for notation etc., see Section 1.4), we have that :

142
o^a. x = (r.x) . ,--, .where p(1) is defined as follows, r, eF ,..R and r, £F ,.-, - R.
1 p 1 p+p(1) 1 p(1) 1 P(1J"1
Therefore r..x^o and r- is M-regular. On the other hand we have : G(M/r.,M) =G(M)/G(r..M)
and one sees that G(r.M) =a-G(M). Assume now that our assertion is valid for n=k.
Since r1M+ ._ + r, M is closed it follows that M/riM+ ._ + r, M is separated in the quotient
filtration. The fact that a,+. is not an annihilator of G(M)/a..G(M) + ... + akG(M),
together with Proposition 1.4.1. yields that r. . is not an annihilator of M/riM+ ...
.- + r,M. Using the induction hypothesis in the case n = k and n = 1 we obtain :
G(N0/aiG(M) +— + ak+1 G(M) » (G(M)/a.,G(M) + .- + ^GÇM))/(\+^W/^W +„ +8^00)
= G(M/r M + .- +rkM)/ak+ .G(M/rM + „ +r]V\)
= G(U/r.U+ ._ +rkM)/G(r..M+ .- +rk+. M/r.M + ._ + r M) =G(M/r1M+ - +rk+1M). □
3.7. Theorem. Let A be a Noetherian local ring with maximal ideal " and suppose that
FA is a filtration satisfying F.A = o, i>o, F A=A, F ,A=", and F , n<o defines the
' a i ' ' o ' -1 ' n
w-adic topology on A. Let R=G(A) = © F.A/F. -A, n= ® F.A/F. -A the canonical
i<o 1 i<-1 1
maximal graded ideal of R. Then :
1. If Qd_o(r) is a Cohen-Macaulay ring then A is CM.
2. If QR_o(R) is a Gorenstein ring then A is Gorenstein.
Proof : 1. If QR_n(R) is CM. then by Theorem 2.1. it results that R is CM. Now R is
completely projective, hence we may select homogeneous elements a-,.-,a eR which form
an R-sequence with n = ht(n) =K.dim R. Proposition 3.6. entails that grade(",A) =K.dimA,
i.e. A is a Cohen-Macaulay ring.
2. Since Q„_„(R) is a Gorenstein ring then R is a Gorenstein ring. The argumentation
of the proof in 1. allows to reduce the proof of 2. to the case where A is an Artinian
local ring. We need :
Sublemma. If A is an Artinian local ring then equivalently :
1. A is a Gorenstein ring.
2. (o :w) ={a£A, a" = o} is a minimal ideal.
3. There exists z£A, z^o such that for every x^o in A, there exists an element yeA
such that xy = z.

143
Proof of the sublemma : 1 «■ 2, cf. [ 1 ] . 3 =>2 Obvious. 2 =>3 Take z£(o:u), z ^ o
and take o^xeA. If xu = o then xe (o : cS) and by 2. there is an y£A such that xy = z.
If xcj^o then pick x. ecj such that xx. fo. Consider the ideal xx.,", then obviously
xcj3xx-cj since equality would entail <^ = x-co, a contradiction if co^o. Repeating this
construction and using the descending chain condition we find that there exist elements
x..,.~,x, such that xx-.X7.-x, ^0 and xx1x7._x,cj = o. Thus xx-.-x, e (o : <S) and minimality
of (o : w) yields that there is a y£A such that xx-.x-.-x, y = z.
Returning to the proof of the theorem, we have already established that R = G(A) is a
Gorenstein ring which is a local ring with maximal ideal cj. So we may apply the sub-
lemma, i.e. suppose that a is a homogeneous element of R satisfying the condition 3
of the sublemma. (Note that the fact that fi is a graded maximal ideal of R plus the
fact that (o : oS) is also graded allows indeed to find homogeneous elements satisfying 3.)
Pick aEA, say a£F A, such that a =a. Then for every o^teA, say r£F A, there
exists a homogeneous beR such that y b = a. Let gEA be such that g maps to b in G (A).
Then y b = a yields that rg -aEF .A. Taking y EF , Awe find that b has positive
q ' p-1 b p-1
degree i.e. b=o but this is impossible, hence F _.A = o. Consequently for arbitrary
y /oEA and a and 8 as above, it follows that Tg -a=o. This states exactly that a
satisfies condition 3 of the sublemma and therefore A is a Gorenstein ring. □

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Subject Index
Artin-Rees property 21
Cauchy sequence 4
Central system 2 2
Classical graded Krull dimension 53
Classical P-primary 76
Dimension
- classical 5 3
- Goldie 50
- left 19
- Krull 18
Element
- homogeneous 5
- homogenized 7 9
Essential objects 1 1
Filtered finitely generated 1 8
Filtered free objects 16
Filtered injective objects 26
Filtered module 1
Filtered ring 1
Filtrations 1
Filtration
- complete 2
- deduced 3
- discrete 2
- exhaustive 2
- I-adic 1
- I-fitting 46
- separated 2
- trivial 1
completion 4
exhaustion 4
kernel 83
suspensio 3
- left limited 7
- trivial 5
Graded Artinian 41
Functor
Gradation

147
Graded bi-endomorphism ring 61
Graded Jacobson radical 65
Graded fully left bounded 9 5
Graded kernel functor 85
Graded local ring 1 2 7
Graded maximal ideal 1 2 7
Graded module 5
Graded module of quotients 9 6
Graded Noetherian 41
Graded ring 5
Graded semisimple objects 6^
Graded simple objects 59
Graded supporting module 92
Graded pure sequence 1 3
Hereditary torsion theory 83
Homogeneous parameters 139
Homogenization 3 6
Kernel functor 83
Kernel functor
- cogenerated rigid 86
- compatible 105
- Goldie's graded 89
- graded 85
- graded prime 9 2
- idempotent 83
- Lambek's graded 90
- rigid prime 9 3
- symmetric 91
Module
projective 9 2
Cohen-Macaulay 1 3 4
critical 19
Gorenstein 1 3 4
graded 5
graded supporting 9 2
P-primary submodule 76
projective (graded) 9
Rees 4 6
uniform 49

148
M-sequence 136
Ore-conditions (primary decomposition 83) 70
Projective objects 9
Property T 96,103
Ring
- Cohen-Macaulay 1 3 4
- completely projective 137
- Gorenstein 1 3 4
- graded 5
- graded division 40
- graded Goldie 71
- GZ - ring 11 8
- principal graded 3 8
- Rees 4 5
- regular 1 3 5
- Zariski central (graded) 117
- ZG - ring , 118
Strict exact sequence 1 4
Strict morpliism 1 4
Structure sheaf on Proj 116
Torsion theory 83
Torsion theory
- cogenerated 86
- hereditary 83
- rigid 84
- symmetric 91
Uniform simple 6 7
Keyl algebra 5 6