Algebra through practice
A collection of problems in algebra, with solutions
Book 6
Rings, fields and modules
T. S. BLYTH ° E. F. ROBERTSON
Algebra through practice
Book 6: Rings, fields and modules
Algebra through practice
A collection of problems in algebra with solutions
Book 6
Rings, fields and modules
T. 8. BLYTH о E. F. ROBERTSON
University of St Andrews
CAMBRIDGE UNIVERSITY PRESS
Cambridge
London New York New Rochelle
Melbourne Sydney
Published by the Press Syndicate of the University of Cambridge The Pitt Building, Trumpington Street, Cambridge CB2 1RP 32 East 57th Street, New York, NY 10022, USA
10 Stamford Road, Oakleigh, Melbourne 3166, Australia
© Cambridge University Press 1985
First published 1985
Library of Congress catalogue card number: 83-24013
British Library cataloguing in publication data
Blyth. T. S.
Algebra through practice; a collection of problems in algebra with solutions.
Bk. 6; Rings, fields and modules
1. Algebra—Problems, exercises, etc.
I. Title II. Robertson, E F.
512'.0076 QA157
ISBN 0 521 27291 2
Transferred to digital printing 2001
Contents
Preface vii
Background reference material viii
1:	Ideals 1
2:	Divisibility 6
3:	Fields 11
4:	Modules 16
Solutions to Chapter 1 27
Solutions to Chapter 2 35
Solutions to Chapter 3 45
Solutions to Chapter 4 58
Test paper 1 88
Test paper 2 92
Test paper 3 96
Test paper 4 98
Preface
The aim of this series of problem-solvers is to provide a selection of worked examples in algebra designed to supplement undergraduate algebra courses. We have attempted, mainly with the average student in mind, to produce a varied selection of exercises while incorporating a few of a more challenging nature. Although complete solutions are included, it is intended that these should be consulted by readers only after they have attempted the questions. In this way, it is hoped that the student will gain confidence in his or her approach to the art of problem-solving which, after all, is what mathematics is all about.
The problems, although arranged in chapters, have not been ‘graded ’ within each chapter so that, if readers cannot do problem n this should not discourage them from attempting problem n+1. A great many of the ideas involved in these problems have been used in examination papers of one sort or another. Some test papers (without solutions) are included at the end of each book; these contain questions based on the topics covered.
TSB, EFR St Andrews
Background reference material
Courses on abstract algebra can be very different in style and content. Likewise, textbooks recommended for these courses can vary enormously, not only in notation and exposition but also in their level of sophistication. Here is a list of some major texts that are widely used and to which the reader may refer for background material. The subject matter of these texts covers all six of the present volumes, and in some cases a great deal more. For the convenience of the reader there is given overleaf an indication of which parts of which of these texts are most relevant to the appropriate sections of this volume.
[1]	I. T. Adamson, Introduction to Field Theory, Cambridge University Press. 1982.
[2]	F. Ayres, Jr, Modern Algebra, Schaum’s Outline Series, McGraw-Hill, 1965.
[3]	D. Burton, A first course in rings and ideals, Addison-Wesley, 1970.
[4| P. M. Cohn, Algebra Vol. 1, Wiley, 1982.
[5]	D. T. Finkbeiner II, Introduction to Matrices and Linear Transformations, Freeman, 1978.
[6]	R. Godement, Algebra, Kershaw, 1983.
[7]	J. A. Green. Sets and Groups. Routledge and Kegan Paul, 1965.
[8]	I. N. Herstein, Topics in Algebra, Wiley, 1977.
[9]	K. Hoffman and R. Kunze, Linear Algebra, Prentice Hall, 1971.
| IO] 8. Lang, Introduction to Linear Algebra, Addison-Wesley, 1970.
[11]	8. Lipschutz, Linear Algebra, Schaum’s Outline Series, McGraw-Hill, 1974.
[12]	I. D. Macdonald, The Theory of Groups, Oxford University Press, 1968.
[13]	8. MacLane and G. Birkhoff, Algebra, Macmillan, 1968.
[14]	N. H. McCoy, Introduction to Modern Algebra, Allyn and Bacon, 1975.
[15]	J. J. Rotman, The Theory of Groups : An Introduction, Allyn and Bacon, 1973.
[16]	I. Stewart, Galois Theory, Chapman and Hall, 1975.
[17]	I. Stewart and 1) Tall, The Foundations of Mathematics, Oxford University Press. 1977.
References useful for Book 6
1:	Ideals [3, Chapters 2—5], [4, Section 10.1], [8, Chapter 3], [13, Chapter 4].
2:	Divisibility [3, Chapter 6], [6, Chapters 9. 31, 32], [8, Chapter 3], [13. Chapter 4].
3:	Fields [1, Chapters 2, 3], [8, Chapter 5], [16, Chapters 3,
4, 7-12].
4:	Modules [3, Chapter 12], [4, Sections 10.2-10.6], [13, Chapter 6].
In [4], [6] and [13] all rings have a 1, and a subring must contain the ring identity. Ring morphisms (called ring homomorphisms in [4] and [6]) must map the identity element. In [4] ring morphisms are written as mappings on the right. The definition of an integral domain given in [6] does not require it to be commutative.
ix
1: Ideals
In this chapter we concentrate mainly on ideals of both commutative and non-commutative rings. An ideal / (left, right, or two-sided) of a ring R is maximal if the only ideal (left, right, or two-sided) of R that properly contains I is R itself. An ideal I of R is prime if I R and xy G I implies x G I or у G I. These notions are important to the development of ring theory, and in the questions that follow are linked with such ideas as nilpotency and chain conditions. A nilpotent element of a ring R is an element x of R such that xn ~ 0 for some positive integer n. A nilpotent ring (ideal) is a ring (ideal) every element of which is nilpotent. A ring R satisfies the ascending chain condition on (left, right, two-sided) ideals if every ascending chain
h C Z2 C I3 C -  • C In C /n+1 C   
of (left, right, two-sided) ideals is such that there exists a positive integer к with In — Ik for all n > k. The descending chain condition is defined similarly.
1.1	Let J? be a commutative ring with al. If I is an ideal of R prove that
(a)	I is prime if and only if R/1 is an integral domain;
(b)	I is maximal if and only if R/I is a field.
Deduce that every maximal ideal is prime. Give an example of a ring containing an ideal that is prime but not maximal.
Now let R be a non-commutative ring with a 1. Prove that if M is an ideal of R such that every non-zero element of RfM is invertible then M is a maximal ideal. Show that the converse is false by considering
Book 6	Rings, fields and modules
the ideal
.. (Тйа 2?>] .	, .1
M=([2c 2d]1 a’i,’C’‘iG2J
iu the ring Mat2x2(2).
1.2	Prove that (8) is a maximal ideal of the ring 42 but 42/(8) is not a field. Explain why this is possible.
1.3	A boolean ring is a ring with a 1 in which every element is multiplicatively idempotent (in that я2 = x). Prove that a boolean ring is
(a)	of characteristic 2;
(b)	commutative.
If / is an ideal of the boolean ring A prove that the following conditions are equivalent :
(1)	/ is prime;
(2)	A/Z=2/22;
(3)	I is maximal.
1.4	Let F be a field and let
J = {/eF[x,y] | /(о,У) = о}.
Prove that J is a principal ideal of F[X, У]. Is J prime? Is J maximal?
1.5	Let 1? be a commutative ring and let A be an ideal of R. Define
r(A) = {xeR | (3n > 1) яп G A}.
If I, J are ideals of R prove that
(a)	r(I + J) - r[r(Z) + r( J)] Э r(I) 4- r( J);
(b)	if In C J for some n G IN then r(7) C r(J).
Show that if r(T) — I then R/1 has no non-zero nilpotent elements. Is the converse true? Prove that if P is a prime ideal of R then r(P) = P.
1.6	If I, J are ideals of a commutative ring R define
I:	J — {я G R | xJ C J).
Prove that, for ideals 1, J, К of R,
(a)	if I C J then I: К C J : К and К : I D К : J;
(b)	(tfn G IN) I: J"+‘ = (7 : Jn) : J = (7 : J): Jn-
(с)	I: J = R <=> J C /;
(d)	I : J = I: (7 + J).
2
1: Ideals
1.7	Let R be a commutative ring with a 1. An ideal Q of R is said to be primary if ab E Q with a$Q implies that bn E Q for some n> 1.
Prove that the primary ideals of the ring 2 are precisely the powers of the prime ideals.
Show that (4, X) is a primary ideal of 2[X] but is not a power of any prime ideal of 2{X].
Show also that a power of a prime ideal need not be a primary ideal by considering the following example. Let R be the subring of 2[X] consisting of polynomials in which the coefficient of X is divisible by 3. Show that P = (3X, X2, X3) is a prime ideal of R but that P2 is not primary.
1.8	Let F be a field. Prove that, in P[X, Y, Z],
(Y2Z2,XYZ) = (У) n (Z) n (X, VJ2n (X,Z)2.
Is it true that (Y2Z2,XYZ) = (y)(Z)(X,y)2(X,Z)2? Justify your assertion.
1.9	A set is said to be inductively ordered if every non-empty totally ordered subset has an upper bound. Zorn’s axiom says that eveiy inductively ordered set has a maximal element.
Let A be a ring with a 1 and let 1(A) be the set of ideals of A. Given I G L(A) with I A define
F{ = {JE 1(A) | ICJcA).
Prove that F? is inductively ordered. By applying Zorn’s axiom, deduce that every ideal J of A with I A is contained in a maximal ideal of A. Hence show that if A is commutative then an element of A is a unit if and only if it does not belong to any maximal ideal of A.
1.10	If F is a field prove that the ring Matnxn(F) satisfies both chain conditions on right and left ideals.
Consider the subring
of the ring Mat2X2(IR)- Prove that every proper right ideal of R is of the form
for fixed x, у E IR. Show that R satisfies both chain conditions on right ideals but neither chain condition on left ideals.
3
Book 6	Rings, fields and modules
1.11	An element ж of a ring R is said to be right quasi-regular if x+y+xy = 0 for some у E R.
Prove that x E R is right quasi-regular if and only if the set {r+ xr | r 6 R} coincides with R.
Suppose now that R has a 1. Prove that if x belongs to every maximal right ideal of R then xr is right quasi-regular for every r E R. Show also that if M is a maximal right ideal of R and x M then — 1 = m + xr for some m E M,r E R. Deduce that the intersection of the maximal right ideals of R consists of those x E R such that, for every r E R, xr is right quasi-regular.
1.12	Let R be a ring and S a subring of R. Suppose that R* ~ S* + Rt+l for some i e IN. Prove that R3 = S3 + _R5+1 for all j > i. Deduce that, for all j, к E IN with j > i, R3 = S3 + R3+k.
Suppose now that R is nilpotent. Deduce that
(a)	if S is a subring of R with Rl — Sl + Rt+1 for some i E IN then R3 = S3 for all j > i\
(b)	if R/R? is generated by a single element then so is R-,
(c)	if M is a maximal ideal of R then J?2 С M;
(d)	if M is a maximal ideal of R then R/M has a prime number of elements.
1.13	A ring R is said to be a nilring if every element of R is nilpotent.
Prove that every subring and every quotient ring of a nilring is also a nilring. Prove also that if A is a two-sided ideal of a ring В with A and В/A both nilrings then В is a nilring. Deduce that the sum of two nil two-sided ideals of a ring is also a nil two-sided ideal.
Let Rp be the set of infinite sequences 01,02,03,... where each ej E 7Lp, and ot- = 0 for all but finitely many suffices i. Given that Rp forms a ring under component-wise addition and multiplication, find all the nilpotent elements of Rp. Prove that this set of nilpotent elements is a nil two-sided ideal that is not nilpotent.
1.14	(a) Let Аг be an ideal of a ring A2. Suppose that / is an ideal of
Ai and that J is the smallest ideal of A2 that contains I. Prove that J3 CI.
(b)	Let R be a ring. Call a subring I of R a subideal if there is a chain
I = Ao C Ai С Л2 C EAn = R
where At-_j is an ideal of A, for i = 1,..., n. Prove that if I is a subideal of R and I is the smallest ideal of R containing / then J3” С I.
(c)	Deduce from the above that if a subideal I of a ring R is nilpotent then the smallest ideal of R containing I is nilpotent.
4
1:	Ideals
1.15 An ideal Q of a ring R is said to be semiprime if Л2 C Q implies A C Q for every ideal A of R.
Prove that if Q is a semiprime ideal of R and A is an ideal of R such that An C Q for some n G IN then A C Q.
Let Q be semiprime and let X be a nilpotent ideal of R/Q. If Y is the inverse image of X under the natural morphism 1}: R —♦ R/Q, prove that Y C Q and deduce that R/Q contains no non-zero nilpotent ideals.
Show conversely that if Q is an ideal of R such that R/Q contains no non-zero nilpotent ideals then Q is semiprime.
5
2:	Divisibility
Here we are mainly concerned with the notion of divisibility in integral domains. If R is an integral domain and x E R divides у E R (i.e. there exists z E R with у = xz) then we write sc|y. If sc|y and y\x then sc, у are said to be associates. The associates of 1 are called units; they form a group under the multiplication of R. For example, the units of Z[x/n] are the elements of length ±1 where the length of a + byfn is £(a 4- byfn) — a2 — nb"2.
The usual notion of a prime integer gives rise in an arbitrary integral domain to the two distinct concepts of irreducible element and prime element. An element x is irreducible if it is neither zero nor a unit and its only divisors are its associates or units. We say that x is prime if it is neither zero nor a unit and is such that if it divides a product then it divides one of the factors. These two concepts coincide in the case of a principal ideal domain (i.e. an integral domain in which every ideal is principal).
Associated with the idea of divisibility is that of highest common factor (h.c.f.) or greatest common divisor (g.c.d.), and least cq^nmon multiple (Lc.rn.). These need not exist in general, but are important in euclidean domains (i.e. an integral domain D with a function N : P\ {0} —+ IN, called a norm function, such that if b\a then N(b) < N(a) and if a, b 0 then a ~ bq + r where r — 0 or 2V(r) < N(b)}.
2.1	Let к,у, и be elements of an integral domain R. Prove that
(a)	x\y if and only if (у) C (sc);
(b)	x, у are associates if and only if (x) — (y);
(с)	и is a unit if and only if (u) — R\
(d)	у is a proper factor of x if and only if (sc) С (у) C R\
2: Divisibility
(e)	x is irreducible if and only if (ж) is maximal among the principal ideals of R.
2.2	In the following statements R is an integral domain and R* = R \ {0}. Give a proof for each statement that is true and a counter-example for each that is false.
(а)	И | is an equivalence relation on R* then R is a field.
(b)	If x ~ у for all x, у e R* then R is a field.
(c)	If a ~ b and c ~ d then ac bd.
(d)	If a ~ b and c ~ d then (a + c) ~ (fe + d).
(e)	If every element of R* is a unit or a prime then 1? is a field.
2.3	Determine whether 5 is irreducible in each of
2, Z[X], Z[tJ, Z[y/^2].
2.4	Write 43г' — 19 as a product of irreducibles in Z[t].
2.5	Express as products of irreducibles
(a)	ll + 7x/=4 inZh/^I];
(b)	4 + 7\/2 in Z[v/2];
(c)	4— У'”з in Z|v/=3j.
2.6	(a) Does \/в- \/б = 3-2 violate unique factorisation in Z[\/6]?
(b) Show that Z[VTo] is not a unique factorisation domain.
2.7	Prove that Z|\/—6] is not a unique factorisation domain by finding two different factorisations of 10.
Find examples of each of the following in 6] and justify your assertions:
(a)	an irreducible element that is not prime;
(b)	non-zero elements a,b such that g.c.d.(a,6) does not exist;
(c)	non zero elements a, b for which g.c.d.(a, b) = 1 but no a, (3 exist in Z[\/^6] such that aa, + (3b = 1.
2.8	Prove that in Z[x/—7] the element 8 can be written both as a product of two irreducible elements and as a product of three irreducible elements. Deduce that for every positive integer к there is an element of Zf'/—7] that can be written as products of £-j-1, i+2,t+k irreducible elements for some integer t.
Give an example of an irreducible element of Z[\/~7] that is not prime. Give an example of elements a,(3 e Z[y/^7] such that g.c.d.(o,/3) = 1 but no 7,6 G Z[\/—7] exist with 70 + 6(3 = 1. Justify your assertions.
7
Book 6	Rings, fields and modules
2.9	If I is a non-zero ideal of Z[ij, use the fact that I is principal together with the euclidean algorithm to show that Z[i]/1 has only finitely many elements.
2.10	Let a e Z[i] be prime. Prove that precisely one of the following statements holds :
(a)	a ~ 1 4- i\
(b)	a e Z is prime and a 3 mod 4;
(c)	a divides a prime Z with p = 1 mod 4.
2.11	For the quadratic domain Z[>/n] where n is square-free prove that
(a)	if n < -1 then the gronp of units of is {-1,1};
(b)	if n > 1 and the group of units of Zf^/n] contains more than two elements then it is infinite;
(c)	the group of units of Z[\/2] is {±(1 + \/2)fc | к > 1}.
2.12	Let p E Z be prime and consider the snbring of Q given by
В = { — | m,nE%, n^O, p does not divide n}.
Prove that every ideal of R is principal and that R contains a unique maximal ideal.
2.13	Let D be a principal ideal domain. Prove that
(a)	if x, yi,..., yn E £\{0} are such that x and y, are relatively prime for each i then x and yv -  • yn are also relatively prime;
(b)	if pi,..., pn are irreducible elements of D, no pair of which are associates, then for all positive integers n,..., rn the elements p,*,..., pjf are pairwise relatively prime.
Hence show that if F is the field of quotients of D and x = a/b E F where b — p^1    p£* is the unique factorisation of b as a product of irre-ducibles, no two distinct p4- being associates, then there exist ay,..., an E D such that
Gi	an	л
x =	+ ... + -f~.	*
P?	Pn
2.14	If A is a commutative ring with a 1 then A is said to be
(a)	noetherian if every ascending chain of ideals of A terminates finitely;
(b)	satisfy the maximum condition if every family of ideals of A has a maximal element.
Prove that the following are equivalent :
(1)	A is noetherian;
(2)	A satisfies the maximum condition;
(3)	every ideal of A is finitely generated.
8
2: Divisibility
Deduce that if A is an integral domain then the following are equivalent :
(o) A is a principal ideal domain;
(/?) A is noetherian and the sum of any two principal ideals is a principal ideal.
2.15	If R is a euclidean domain with norm N, prove that
(a)	if a|6 and N(a) — then a and b are associates;
(b)	if a,b are non-zero elements of R neither of which divides the other then there exist Ct}f3td G R with aa + fib = d and N(d) < mmfjVfa), N(b)).
2.16	Prove that in the definition of a euclidean norm b quotients and remainders are nnique if and only if
6(g -f- b) < max(6(a), 6(6)).
Give an example in Z[i] where quotients and remainders are not nnique.
2.17	(a) Write -1 + 3i as a product of primes in Z[t].
(b)	Let R be a unique factorisation domain and let
f(X) = a0 + atX + a2X2 + - - • + anXn G
be such that g.c.d.(aOj ... > ®n) = 1- Suppose that there is a prime p G R such that p\a,t for i — 0,..., n — 1, p2 does not divide ao, and p does not divide an. Prove that f(X) is irreducible in 7?[X].
(c)	Deduce from (a) and (b) that
Х3 + 8гХ2-6Х-1+Зг
is irreducible in Z(t] [X].
2.18	Does every pair of non-zero elements of Z[-\/3] have a greatest common divisor? Find the greatest common divisor of 13 and 7 + 5\/3.
2.19	In the ring	prove that
(a)	the units are 1 and —1;
(b)	3,2 -ь >/^5,2 — \/—5 are irreducible;
(c)	9 has two different factorisations into a product of irreducibles;
(d)	the ideals (3,2 + y/~5) and (3,2 — y/^5) are prime;
(e)	although 3 is irreducible the ideal (3) can be expressed as a product of prime ideals.
9
Book 6	Rings, fields and modules
2.20	Consider the following ideals in 2 [>/10] :
Pi = (2, \Zio), f2 = (3,4 + P3 = (з,4 - v<io).
Show that each of these ideals is prime. Prove that the two prime decompositions
6 = 2 - 3 = (4 + \/10)(4 - >/10)
lead to the same decomposition of (6) as products of the prime ideals Pi,P2,P3.
10
3:	Fields
3.1
This chapter is mainly concerned with the notion of a field extension; i.e. an embedding of one field into another. If К is an extension of F then К can be considered as a vector space over F. The dimension of this vector space is called the degree of the extension and is written (K : F). When this is finite, the extension is said to be finite. An element a of К is said to be algebraic over F if p(a) = 0 for some p(X) G F[Xj. The minimum polynomial of a is the monic polynomial of least degree with this property. An extension К of F is said to be algebraic if every element of К is algebraic over F; otherwise К is said to be transcendental.
When К is an extension of F and ,..., an G К we use the (standard) notation F(ai,..., an) to denote the smallest subfield of К that contains Fu{ai,..., an}. In particular, F(a) is called a simple extension of F. A field К is called a splitting field of /(X) G F[X] if К is an extension of F of minimal degree in which f can be expressed as a product of linear factors. К is called a normal extension of F if it is the splitting field of some f(X) G F[X|.
Finally, we assnme that the reader is familiar with the statement of the fundamental theorem of Galois theory, which relates subgroups of the Galois group Gal(X, F) with subfields of К containing F.
Let F be a field of characteristic p. Prove that
(Va,beF) (a ± b)p = ap ± bp.
Hence deduce by induction that, for every positive integer n,
(Va, b e F) (a ± b)p" =	± bp”.
Book 6	Rings, fields and modules
3.2	Let К be a field and D : F[Xj —» F[X] the differentiation map described by
D(ao + o-i X + a^X2 + • • • + anXn] ~ Oj + 2a$X +    + iuinXn \
If F is of characteristic 0 prove that Df — 0 if and only if f is a constant polynomial. If F is of characteristic p prove that Df = 0 if and only if f is of the form
во + a>pXp + • • • + arpXrp.
3.3	If F is a field prove that the units of F[X] are the non-zero constant polynomials.
Let i : F —* F[X] be the canonical injection. Prove that if <p : F[X] —* F[X] is an automorphism then there is an automorphism $ : F F such that the diagram
F-----h—>F[X]
t9
F-----—~>F[X]
is commutative (in the sense that tp о i — i о 0). Deduce that there exist a, b G F with a f- 0 such that <p(X) — aX + b.
3.4	Let F be a field. Given f,g G F[X] we say that f,g are equivalent if /(a) = g(ot) for every ct G F. If F is infinite, prove that f,g are equivalent if and only if f — g. If F is finite, say F — {ai,...,an}, prove that /, g are equivalent if and only if / — g is divisible by
m(X) = (X - 01)(X - c2) - - • (X - an).
«f
In the case where F = GF(p) — 2LfpZ, show that m(X) = Xp — X.
3.5	If the field F is not of characteristic 2, prove that X2 + У2 — 1 is irreducible in F[X,yj.
3.6	By considering the ring morphism Z[X] —» 2n[Xj that is induced by the natural morphism 7L —> 7Ln, show that if / is irreducible over 7Ln then it is irreducible over Z
By taking n — 5, show that if
/(X) = X4 + 15 № + 7
then f is irreducible over Q.
12
3: Fields
3.7	Determine a basis for Q(\/2, \/2) over Q. Deduce that \/2 G Q(x/2, x/ty. Hence show that Q(\/2, v^2) is a simple extension of Q.
3.8	Let F be a field and consider
X3 z = xTieFm
Prove that Z is transcendental over F but that F(X) is a simple algebraic extension of F(Z). What is the minimum polynomial of X over F(Z)?
3.9	Show that X2 — 3 and X2 — 2X — 2 are both irreducible over Q and have the same splitting field.
3.10	Show that (X2 — 2X — 2)(X2 +1) and X5 —ЗХ3 + X2 — 3 have the same splitting field К over Q and find (K ; Q).
3.11	Let if be a field and let a, b E K. Prove that X + a + b divides X3 — 3abX + a3 + b3 in the ring K[X], and determine q(X) E К [X] such that
X3 - 3abX + a3 + bs = (X+a + b)q(X).
Find a splitting field S for Xе — 6X3 + 8 over Q. What is the degree of S over Q? Show that S contains the element \/4 + \f2. Find the minimum polynomial of this element over Q.
Is it true that Q( + x^2) is a splitting field for X5 — 6X3 + 8?
3.12	Find quadratic factors for
f(X) = X4 + 2X3 - 8X2 - 6X - 1
in Q[X] and hence show that Q(\/2,x/3) is a splitting field for f over Q. Show that V*3 Q(\/2). Deduce that (ф(\Д,\/3) : Q) — 4 and find a basis for Q(a/2, \/3) over Q.
Prove that Q(\/2, \/3) — Q(\/3 — \/2) and find the minimum polynomial of x/3 — \/2 over Q(\/3).
3.13	Find the irreducible factors in ®[X] of
f(X) = X4 - X2 ~ 2.
Show that Q(i, \/2) is a splitting field for f over Q. Show also that (Q(®, x/2) : Q) = 4 and write down a basis for Q(i, x/2) over Q. Find the
13
Book 6
Rings, fields and modules
minimum polynomial of i + y/2 over Q. Deduce that	+ x/2) : ©) = 4
and hence that ©(«, y/2) = Q(i 4- \/2).
Noting that Q(t), Q(\/2) and are all subfields of Q(i+\/2) over each of which the degree of Q(i + \/2) is 2, find the minimum polynomial of i 4- \/2 over each of
(a)	ОД;
(b)	®(V2);
(c)	®(.V2).
3.14	Given r € Q consider the polynomial f(X) = X4 + r Q[X]. Prove that the following statements are equivalent :
(1)	/ is reducible;
(2)	/ has a quadratic factor;
(3)	either r = —p2 (p € ®) or r = |g4 (g € Q).
Suppose now that / is irreducible over ®. Let f G C be a root of / and let К = Q(£). Prove that
(a)	if т/r © then the only subfield of K, other than К and Q, is ®(€2);
(b)	if \fr G © then the subfields of K, other than К and ®, are
®(«2), ®(\/?e+e3), «(-V^e + e3)-
3.15	Determine the order of the Galois group of Q(w) over Q when
(а)	w = e2w,/°;
(b)	w = </2.
3.16	Find a splitting field for the polynomial X4 — 2 over ®. Show that the Galois group of X4 — 2 is non-abelian and of order 8.
3.17	Determine a subfield of C that is a splitting field over © for the polynomial X3 — 2. Construct the Galois group of this polynomial and identify the subfields of the splitting field.
3.18	Show that if о is a root of
X3-3X+l = 0
then a2 - 2 and 2 — ex — a2 are the other roots.
Let /(X) = X3 — 3X 4-1. Show that <D(o) is a splitting field for / over ® and find (O(ot) : Q).
Show that there is a unique ©-automorphism •& of Q(a) with $(a) = c? — 2 and determine Gal(®(a), Q). Is it true that every permutation of the roots of / extends to an element of Gal(®(a), ©)?
14
S: Fields
3.19	Determine which of the following extensions of Q are normal :
(а)
(b)	®(№);
(c)	®(№);
(d)	®(№).
3.20	Let E be a field of characteristic 0 and let К — E{Xj,..., Xn) be the field of rational functions in n indeterminates over E. Define the elementary symmetric polynomials in X1} ...,Xn by
а2 = ^^Х3, an = XiX2 •   Xn.
»=1	у у
Let F = E(cTian) and let
f(X) ^Xn- mX”"1 4- cr2Xn~2------4- (-l)non G F[X].
Prove that
(a)	If is a splitting field for f over F;
(b)	the Galois group of К over F is isomorphic to the symmetric group of degree n;
(c)	(K:F) = nL
15
4:	Modules
The easiest way to define a module is to say that it is an algebraic system that satisfies the same axioms as a vector space except that the scalars come from a ring R with a 1 instead of from a field F. This seemingly modest generalisation leads to an algebraic structure that is of the greatest importance. We use here the term Л-module, it being understood that the scalars are written on the left.
As with groups and lings, the substructures (submodules) and the structure-preserving mappings (Л-morphisms) are of especial interest, as are ‘new’ structures obtained from ‘old* such as quotient modules, cartesian (or direct) products, and direct sums. We assume that the reader is familiar with the correspondence theorem and the basic isomorphism theorems. If M, N are Л-modules then the set of Л-morphisms f : M —* N forms an additive abelian group (a 2-module) which we denote by Могк(М, TV), or End/i(Af) in the case where N = M. Sorafe authors write the former as Нотд(Л/, N).
Topics dealt with include exact sequences (the image of the input Л-morphism equals the kernel of the output Л-morphism), various diagrams that are commutative (all composite Л-morphisms defined from one given Л-module to another are equal), the chain conditions (as for rings in Chapter 1), and Jordan-Holder towers of submodules (as for composition series in groups). We also use the notion of a free Л-module (one that has a basis), and a projective Л-module (a direct summand of a free Л-module).
4.1	Let M be an abelian group and let End M be the ring of endomorphisms on M (i.e. group morphisms f : M —» M under addition and composition). Prove that M is an (End Af)-module under the external law
4: Modules
End M x M -» M described by (/, m) fm = f(m).
4.2	Let R be a ring with a 1, and M an abelian group. Prove that M is an R-module if and only if there is a 1-preserving ring morphism ц . R—> EndAf.
4.3	Prove that the ring of endomorphisms of the abelian group Z is isomorphic to the ring Z; and that the ring of endomorphisms of the abelian group Q is isomorphic to the field <D.
4.4	Let R be a commutative ring with a 1 and let f : R x R —♦ R be a mapping. Prove that f is an R-morphism if and only if there exist ct,ft G R such that
(Xfx,yER) f(x,y) = ax + fty.
4.5	Let — {a+b\/2 | a, b E Z} and consider the mapping f : Z[\/2] Z[\/2] given by
f(a + bx/2) — a + b.
Determine whether or not f is
(a)	a ring morphism;
(b)	a Z[x/2]-morphism;
(c)	a Z-morphism.
4.6	Let M be an R-module. For every R-morphism f : R —> M and every A E R let Xf : R —* M be given by (A/)(r) = f(rX)- Prove that Xf E Morfl(R,Af) and deduce that Morp(R,Af) is an R-module. Show also that the mapping & : Mor/e(R, Af) —» M given by $(/) ~ J[Ir) is an R-isomorphism.
4.7	Let f : M —» N be an R-morphism. If A is a submodule of Af let f ~*(A) = {/(a) | a G A}, and if В is a submodule of N let = {ж E M | /(sc) G B}. Prove that
(а)	/П-ГЧ-А)] = A+ Ker/;
(b)	/-[/-(B)l = BnIm/;
(с)	Г(ЛП/-И) = /-(Л)па
4.8	An R-monomorphism f : M —► N is said to be essential if, for every non-zero submodule A of N, is a non-zero submodule of Af. If Af is a submodule of N then TV is said to be an essential extension of Af if the canonical inclusion i : Af —> N is essential.
Prove that, as Z—modules,
(a)	Q is an essential extension of Z;
(b)	IR is not an essential extension of
17
Book 6
Rings, fields and modules
If M is a submodule of TV, use Zorn’s axiom to prove that there is a submodule A of N that is maximal with respect to the property MC\ A — 0. For such a submodule A prove that the composite morphism
M----i—>N—--Л/А
is an essential monomorphism.
4.9	Given the sequence
X—>Y-------i—>Z
of й-modules and й-morphisms, prove that
(a)	Im f П Kerg — f~* (Kerg о f);
(b)	Im / + Ker g— g*~ (Imgof).
4.10	Given the diagram
A-----—>C
'1
в
of й-modules and й-morphisms in which f is an epimorphism, prove that the following statements are equivalent :
(a)	there is an й-morphism h:B->C such that ho f = g‘
(b)	Ker/ C Kerg.
Prove further that such an й-morphism h, when it exists, is unique; and that it is a monomorphism if and only Kerf — Kerg.
If an й-morphism В : M —> TV can be expressed as a composite morphism
M—>A—>B 7 >N
where a is an epimorphism, (3 is an isomorphism, and 7 is a monomorphism, prove that A ~ М/ Kertf and В ~ Im0.
4.11	Establish the following statements concerning the various Z-modules involved :
(a)	Q is not finitely generated;
(b)	Q/Z is infinite;
(c)	0 is the only independent subset of Q/Z;
(d)	Morz(Z/2Z, Q) = 0;
(e)	Morz(®, Z) = 0.
18
4: Modules
4.12	An R-module M is said to be cyclic if it is generated by a singleton subset. Let M = Rx be a cyclic .R-module and let Аппк(ж) = {A € R | Аж = 0}. Prove that Аппк(ж) is a submodule of R and that M ~ R/Ann^ (ж).
Let m and n be integers, each greater than 1. Show that the prescription $(ж + m2) = nx + nmZ describes a Z-morphism «9 : 2/m2 —> 2/nm2. Prove that the Z-module Morz(Z/mZ, 2/nm2) is generated by {$} and deduce that
Mor^f^/mZ, Z/nmZ) ~ 2/m2.
4.13	If A, В are submodules of an .R-module M, establish a short exact sequence
0-----> А П В-------- A x В--->A + В------>0.
4.14	The diagram of R-modules and R-morphisms
О
0
is given to be commutative with the row and column exact. Prove that (a) ct and /? are zero morphisms;
(b) f and g are isomorphisms.
4.1Б Let M and N be R-modules and f : M —* N an R-morphism. If A is a submodule of M prove that the following statements are equivalent : (a) there is a unique R-morphism /* : М/A —* N such that	= f;
(b) A C Ker /.
19
Book 6	Rings, fields and modules
Show further that such an .R-morphism /* is a monomorphism if and only if A = Ker/.
If A, В are submodules of an R-module M, establish an exact sequence of the form
0----* M/(A П B)-* M/A x M/B---* M/(A + B)--♦ 0.
Hence deduce that
(Л + B)/( AB)/A x ( A + B)/B.
4.16 Suppose that the diagram of R-modules and R-morphisms
A—L—> В —C
<*	0	1
A1-----» B1----» C
Г	д'
is commutative and that ct, ft, 7 are isomorphisms. Prove that if the top row is exact then so is the bottom row.
4.17	[The 3x3 lemma.] Suppose that the diagram of R-modules and R-morphisms
is commutative, that all three columns are exact, and that the top two rows are exact. Prove that there exist unique R-morphisms ct" zC'—tC and (3" : C —♦ C" such that the resulting bottom row is exact and the completed diagram is commutative.
20
4: Modules
4.18	Let R be an integral domain. Given x G R with x / 0, show that there is a descending chain
R D Rx D Rx2 D • • • D Rxn D Rxn*1 D... of submodules of the Л-module R such that, for every n, Rxn/Rxn+I - R/Rx.
4.19	Determine which of the chain conditions, if any, are satisfied in each of the following modules.
(a)	Z as a Z- module;
(b)	Zm as a Z module;
(c)	as a Zm -module;
(d)	<D as a Q-module;
(e)	© as a Z-module;
(f)	Q[X] as a ©-module;
(g)	Q[X] as a Q[X]-module;
(h)	X\/M as a ©|X]-module where M is the submodule consisting
of those polynomials divisible by X6;
(i)	<D[X]/M as a G)[X]-module where M is the submodule consisting of those polynomials divisible by X2 + 1.
4.20	Let M be an Л-module of finite height. If N is a submodule of M prove that there is a Jordan-Holder tower of submodules
M - Mq D Mi D • • • D A/r_x D Mr = 0 such that, for some index k, Mk — N.
4.21	If M is an Л-module of finite height and if TV is a submodule of M prove that h(M) = h(N) + h(M/N). Deduce that Ar = M if and only if h(N) = h(M).
4.22	If M and N are Л-modules with M cf finite height and if f : M —» N is an Л-morphism, prove that Im f and Ker f are of finite height with
Л(1т/) + Л(Кег /) — h(M).
4.23	Let Mi,..., Mn be Л-modules each of finite height. If 0--------------------... Af„----------------------------.0
is an exact sequence prove that
£(-l)*fc(Jft) = 0. fc=l
4.24	Find Z-modules M, N with h(M) = h(N) = 2 and Mor^(Af, N) = 0.
21
Book 6	Rings, fields and modules
4.25	Let F be a field and let Mn be the ring of n x n matrices over F. For i = 1,..., n let E, G Mn be the matrix whose (г, i)~th entry is 1 and all other entries are 0. For i = 1,..., n define
Bi = Mn(E1+  + Ei).
Prove that
Mi = Bn D Bn_i Э   • Э Bi Э Bo = 0
is a Jordan-Holder tower for the Mn-module Mn.
4.26	Let Mi,...,Mn be submodules of an Я-module M such that Af — M,. For к — let JVfc be a submodule of Mk. If N = Х2Г-1 prove that N — Ni and that MjN = Mi/Ni.
4.27	If Ai,...,An are submodules of the Я-module M prove that M — ®Г=1 if and onty if f°r ®	there exist Я-morphisms /» :
Ai —> M and g,: M —> Ai such that
И E7=i Л°г. = Им-
Deduce that M is the direct sum of submodules Mi, M2 if and only if there is a split short exact sequence
0----_> Afj _——Af-----> д/2-----, о.
4.28	Let A be an Я-module. Use the fact that Я has an identity element to prove that every exact sequence of Я-modules and Я-morphisms of the form
0-----, x-----> A-----* Я----> 0
splits. In contrast, exhibit an exact sequence of Z-modules and 2-morphisms of the form
0----->Z------> A------------>0
that does not split.
4.29	An Я-morphism f : M —♦ N is said to be regular if there is an Я-morphism g : M —» N such that f о g о f = f.
By considering the combined canonical sequences
Ker	Ker f ~ Im f—^—.N—~^N/ Im f
and examining when each splits, prove that f : M -+ N is regular if and only if Ker / is a direct summand of M and Im f is a direct summand of N.
22
4: Modules
4.30	In the diagram of Я-modules and Я-morphisms
it is given that
(1)	the diagram is commutative;
(2)	each sequence Mt—-—»M—-—>M't is exact;
(3)	ki and k% are isomorphisms.
Prove that Ker я ft Ker j? — 0. If, for every x G M, the element z is defined by
® = (tx о Л”1 о ;2)(x) + (г2 о kf1 о я )(г),
deduce that z = x. Hence show that M = Im ii G Im i2 and that
Л.) о ki °	+ Л.2 ° fc2 о = 0.
4.31	Let (M)iEZ and (№)«€/ be families of Я-modules. If, for every i G /, fi : Mi —> Ni is an Я-morphism, define the direct sum of the family (fi)i$i to be the Я-morphism f :	M -» ®,6/ N> given by /((?n,)ie/) =
Determine Im f and Ker f.
If is also a family of Я-modules and if, for every i G I, gi  Li —> Mi is an Я-morphism, let g be the direct sum of the family (gi)i^i-Prove that
Ф i	ф M——. ф M
iei	iei	iei
is exact if and only if
Li—
is exact for eveiy г G /
4.32 Let Я be a commutative ring with a 1. Suppose that M is an Я-niodule and that A,,..., An are submodules of M with M — Ai. For j = 1,..., n let Lj be the set of Я-morphisms f : M -* N such that Ai C Ker f. Prove that Lj is an Я-module and that Lj ~ Могк(А,, JV).
23
Book 6	Rings, fields and modules
4.33 The diagram of й-modules and й-morphisms
. 1* I" Ь- Ь* ь
     
is given to be commutative with exact rows. If each q, is an isomorphism, establish the exact sequence
----->	e Bi—.
where tpi is described by
<Pi : at h-> (o,(ai),/,(a,))
and 0,- is described by
4.34	Give an example to show that a submodule of a free module need not be free.
4.35	Let f : M —» M be an й-morphism. Prove that if f is a monomorphism then f is not a left zero divisor in the ring End^fAf). If M is free, establish the converse.
4.36	Let f : M —* M be an й-morphism. Prove that if / is an epimorphism then f is not a right zero divisor in the ring End^(Af). Give an example of a free 2-module M and f G End^(-M) such that f is neither a right zero divisor nor an epimorphism.
4.37	If is a family of projective modules prove that Pi is also projective.
4.38	For й-modules M and N, let P(M, N) be the set of those й-moiphisms f : M —♦ N that ‘factor through projectives’m the sense that there is a commutative diagram
P k^ M——
in which P is projective. Prove that P(Af, TV) is a subgroup of the group Morj?(Af, TV).
24
4: Modules
4.39	Show that every diagram of R-modules and R-morphisms of the form
P'	P"
in which the row is exact and P\ P" are projective can be extended to a commutative diagram
0------*P'--*-—>P— —->P"-----------> 0
«	1	0
0------->£*----->E--------,E"------> 0
J »
in which the top row is also exact and P is also projective.
4.40	Let n be an integer greater than 1. For every divisor r of n consider the ideal r(Z/nZ) of the ring Z/nZ. Show how to construct an exact sequence
0------»"-(Z/nZ)----->Z/nZ------>r(Z/nZ)----->0.
T
Prove that the following statements are equivalent :
(1)	the above sequence splits;
(2)	h.c.f.{r,n/r} — 1;
(3)	the Z/nZ-module r(Z/nZ) is projective.
Hence provide an example of a projective module that is not free.
4.41	Let R be a commutative ring with a 1, and let X, Y be R-modules with X projective. If A, В are submodules of X, Y respectively, prove that
Дл,в = {/ 6 Могд(Х, Y) I Г(А) С B}
is a submodule of the R-module Mor#(X, У). Prove that Могд(Х/А, Y/B) ~ &a,bI&x,b-
4.42	Suppose that P is a projective R-module and that there is given a diagram of R-modules and R-morphisms of the form
P
о
X----------------
0
25
Book 6
Rings, fields and modules
in which the row is exact and /3 о & — 0. Prove that there is an R-morphism f : P —> X such that ct о f = Ф.
Let the diagram of R-modules and R-morphisms
---,p3	>A
kQ
• • -*Qs———>@2—-—*Qi-	>B-----—*0
fta	^2	hi
be such that both rows are exact and each is projective. Use an inductive argument to show that for every positive integer n there is an R-morphism kn : Pn -» Qn such that hn о kn — kn-i о gn.
4.43 Consider the diagram of R-modules and R-morphisma
0------->Ker g-1—>A ------------->0
p------>p/	o j}
6
in which the top row is exact, j is the canonical inclusion, i is a monomorphism, and I; is the canonical epimorphism. Prove that there is a unique R-morphism &: В —* P/lxn(ioj) such that &og = |] o«. Show also that Ф is a monomorphism.
An R-module P is said to be quasi-projective if, for every diagram
P
h
A------->B----->0
g
in which the row is exact and A is a submodule of P, there is a unique R-morphism 5 : P —» A such that g о f = h. Prove that P is quasi-projective if and only if, for every diagram
P
к
P-—-^P/Q---------->0
in which Q is a submodule of P, there is a unique R-morphism я : P —► P such that 1] о % — к.
26
Solutions to Chapter 1
1.1	(a) To say that R/I is an integral domain is equivalent to saying that
it has no zero divisors, which is equivalent to saying that, for all x, у G R, if (ж + l)(y + /) = 0 + / then either x + / = 0 +1 or у +I — 0 + /, which is equivalent to the condition xy € I implies x G I or у G /, i.e. to the condition that I is prime.
(b) Suppose that I is maximal and let x + I be a non-zero element of R(I. Then x^ I and so I4-(jc) is an ideal of R that properly contains I. Hence I 4- (ж) = R and so i + rx = 1 for some i G I and r G R. Passing to quotients (i.e. applying the natural morphism associated with I to this equation), we obtain (r + f)(x +I) — 1 +1 which shows that r +1 is an inverse of x +1. Thus R/I is a field. Conversely, if R/I is a field and J is an ideal of R with IC J, let b G J \ I- Since b + /^0 + 7it follows that for every r G R there exists x G R such that (b + /)(ж 4- 1} — r + I. It follows that bx — r = i G I and hence that r — bx — i G J and consequently J — R.
Since every field is an integral domain, it follows from the above that every maximal ideal is prime.
Clearly, the ideal (0) of 2 is prime but not maximal.
If every non-zero element of R/M is invertible then R/M has no proper non-zero ideals. For, if A were snch and a G A for some a 0 then a~l exists and a~ra G A gives the contradiction A ~ R/M. Hence R has no proper ideals that properly contain M.
Now
Mat2x2(2)/M Matsxsf^s)
and so, since Mat2x2(^2) contains no proper ideals, M is maximal in
Book 6
Rings, fields and modules
Mat2X2(Z). However, I 4- M is not invertible in Mat2x2(Z)/M since this would imply
1
1 + 2t I
which is impossible.
1.2	Suppose that (8) С IC 4Z. If I / (8) then there exists t = 4n E I with n odd, say n — 2m 4- 1. Then t — 8m + 4 which gives t 4- (8) = 4 4- (8) in Z/(8). Hence I — 4Z and 4Z/(8) has no non-trivial ideals. Clearly, 4Z/(8) is not a field since if x — 4 4- (8) then we have я2 = 0 in Z/(8).
The reason this can happen is that 4Z has no identity element.
1.3	(a) Since every element is idempotent, we have
x + X= (x + x)2 ~ X2 + X2 4- X2 + X2 — X+ x + X + X
whence x 4- x = 0 and the characteristic is 2.
(b) We have that
x 4- у = (ж 4- у)2 — ж2 4- xy 4- ух 4- у2 — х 4- ху 4- ух 4- у
so О = ху 4- ух. Using the fact that у 4- у = 0 implies that — у — у, we then obtain xy — —ух — (—у)я = yx.
(1)	=> (2) : If I is prime then A/I is an integral domain, so has no zero divisors. Now in A we have
4- у) = хух 4- xy2 — x2y 4- xy2 = xy + xy~O.
On passing to quotients modulo I this identity still holds, say яу(я-|-у) = 0. If then x / 6 and у £ 6 then we must have x 4- у = 6 whence x — —y = y. Thus A/I consists of two elements and so A/I ~ Z/2Z.
(2)	=> (3) : This follows by question 1.1.
(3)	=> (1) : This follows by question 1.1.
1 -4	J — X F[X, У] so J is principal Now we have that F[Xf Y]/J ~ Р[У]
which is an integral domain. Hence J is a prime ideal of F[X, У]. However, since У[У] is not a field it follows that J is not a maximal ideal of F[X,y]. In fact,
J E (J,Y2) e F[X,Y\.
28
Solutions to Chapter 1
1.5	(a) Since А С r(A) for every A we have
I + J C r(I) + r( J) C r[r(7) + r( J)].
Now let x G r[r(P) 4- r(J)]. Then xn G r(Z) 4- r{J) for some n G IN and so xn = у 4- z where ymi G I and л”*2 G J for some m\, m? G IN. Hence ж»(та*+то») _ £2 y^zv where p 4- v — mi 4- m^ and either p > mi or v > r^2- But if p > mi we have y,lzy G /, while if v > m2 we have y*zv g J. Hence sMmi+m,) G j 4. j so x e r(f + /).
(b) If In C J let x G r(I). Then x™ G I for some m. But then (sw)n G J since In C J, and consequently x G r(J).
The set of nilpotent elements of R/I is r(I/I) = r(I)/I. Hence R/I has no non-zero nilpotent elements if and only if I — r(I).
If P is prime and xn G P then x G P. Thus if x G r(P) we have x G P, and so r(P) = P.
1.6	(a) If x G I: К then xk G I for all к G К and so xk g J for all к G K,
whence x G J : K.
If now x G К : J then xa G К for all a G J and so xa G К for all a G I, whence x G К : I.
(b)	Observe first the general result that
(I: J) : К = I: JK.
In fact JK[(I : J) : К] C J(I : J) С I and so it follows that (I : J) : К С I : JK. Also, JK(I: JK) С I gives K(I: JK) C J: J whence I: J К C (f : J) : K. Now (b) follows on taking К — Jn.
(c)	If J С I and a G R then a J C J С I gives a G I: J. Conversely, if I: J = R then in particular we have 1 G I: J whence J — 1J С I.
(d)	I : (14- J) = (I: I) П (/ : J) = Rn(I: J) = I: J.
1.7	Clearly (pn) is a primary ideal of Z when p is prime. Conversely, suppose that m = pq with p,q coprime. Then pq G (m) and p (m) yet qn (m) for any n > 1. Therefore (m) is not primary.
It is straightforward to check that (4, X) is a primary ideal. For, if fg G (4, X) with f (4,X) then any term of g not divisible by 4 or X must be divisible by 2. Hence t?2 G (4, X), so (4, X) is primary.
If (4, X) is the power of a prime ideal P then P must contain (4, X). But Z[X]/(4, X) ~ Z4 and so the only possibility is P — (2,X) which corresponds to 2Z4, the only proper non-zero ideal of Z4. However, (2, X)n 7^ (4,X) for any n > 1.
Now R/P ~ "Z. so P is a prime ideal of R (since Z is an integral domain). However, ЗХ3 G P2 since 3X G P and X2 G P. But X3 P, so if P2 were primary we would have 3n G P2 for some n > 1. Since every element of P2 has constant term 0, this is a contradiction.
29
Book 6	Rings, fields and modules
1.8	Clearly (У2/2, XYZ) С (У) n (Z) г [X, У)2 n (X, Zfi. To obtain the reverse inclusion, suppose that /(X, Y, Z) is in the right hand side. Then since f e (T)n(Z) we have f(X,Y,Z} = YZg(X,Y,Z). But we also have f G (X, У)2 so
g(X, Y,Z)=Y a(X, Y,Z) + X b(X, Y,Z).
Now YZXl>(X, Y, Z) e (X, Zfi so we require Y‘2 a(X, У, Z) e (X, Z)2, i.e.
a(X, Y,Z) = Z h(X, Y, Z) + Xk(X, Y, Z).
Then / G (Y2Z2, XYZ) as required.
It is not true that (Y2Z2,XYZ) = (Y)(Z)(X,Y)2(X, Z)2. For, every polynomial in the right hand side has degree at least 6 and so in particular XY Z cannot belong to the right hand side.
1.9	Fj / 0 since clearly I G Fi. Let {Jp | /3 G B} be a totally ordered subset of Ff. Then U/зес J 0 / A, f°r otherwise we would have 1 G Jp for some fl whence the contradiction
A = 41 C AJp C Jp c A.
Suppose now that ж, у G U/jetf Then there exist a, G В such that x G Jo and G But either Ja C or C Ju. Suppose, without loss, that Jn C J7. Then x - у G C U/jec SimilarIV so, аж G J7 for every a G A. Thus we see that U/jetf ^0 *s an ^ea^ °f tbat is distinct from A. Thus Fj is inductively ordered.
Applying Zorn’s axiom, we obtain a maximal element M of Fj. It is clear that M is also a maximal idea! of A with IС M.
Let I g 1(A) with I / A. Then no element of I is invertible. For, suppose that x G I is invertible. Then 1 — ж-1 ж G 1 gives A = I, a contradiction. It follows that if a G A is invertible then a does not belong to any proper ideal, hence to any maximal ideal. Conversely, if a G A does not belong to any maximal ideal then, by the first part of the question, a does not belong to any proper ideal. Now Aa = a A is an ideal (by commutativity) of A with a — la G Aa. We deduce, therefore, that Aa = A whence there exists x G A such that xa = 1. Thus a is invertible.
1.10	MatnXn(F) is a finite-dimensional vector space over F. A left or right ideal of this is a subspace. Thus, if Li,L$ are left ideals with Li C L2 then dim Li < dim £2. It follows that every chain of left ideals is finite
30
Solutions to Chapter 1
and so both chain conditions are satisfied. The same holds for right ideals.
Let I be a non-zero proper right ideal of R. Suppose that I contains the non-zero elements
ci 0	di 0
c2 0 I’ I d2 0
Suppose in fact that Ci 0 and dt / 0. Then
and similarly
[ } °] [dfxda 0]
G/.
But then I contains
1
cf *C2
where d = cl1c2 — d11 d2. Now if d 0 then we have
whence
and therefore also
Consequently,
31
Book 6
Rings, fields and modules
It now readily follows that I — R, a contradiction. Hence we must have d = 0 and so dj eg = Ci ds which gives
Ci ol	Г di 0
Cg o]	I. ds 0
for some a € F. If Cj = 0 then di = 0, otherwise I = R. Hence the result follows.
No two proper non-zero right ideals can contain each other and so R has both chain conditions on right ideals.
If S is an additive subgroup of IR then S
Then
is a left ideal of R.
is an infinite descending chain of left ideals.
Taking Si — {t/in I t, n G Z} we obtain the infinite ascending chain of left ideals
1.11	Suppose that R — {r 4- xr | r g R}. Then x — r + xr for some r G R and so x 4- (— r) 4- z(—r) — 0, which shows that x is right quasi-regular. Conversely, if x 4- у 4- xy — 0 for some у G R then x = — у 4- s(—j/) G {r 4- жг | г E Й}. Since this is a right ideal of R, it then contains xr and hence also r 4- xr — xr = r so coincides with R.
Suppose now that x belongs to every maximal right ideal of R. If x is not right quasi-regular then A = {r+xr | r G -R} R. Let Af be a right ideal of R that is maximal with respect to containing the set A but not x. Then Af is a maximal right ideal of R and so x G Af, a contradiction. Hence x is right quasi-regular, so every element in the intersection of the maximal right ideals of R is right quasi-regular, which shows that xr is right quasi-regular for every r G R.
If Af is a maximal right ideal of R with x Af then {Af4-xr | r G R} is a right ideal of R with Af C {Af 4- xr | r G Й} so {Af 4- xr | r G R} = R whence we have — 1 == m 4- xr for some m g Af and r G R.
Suppose now tht xr is right quasi-regular for every r G R and that x M for some maximal right ideal Af. Then —1 — m 4- xr and xr is right quasi-regular, so there exists z E R with xr 4- z 4- xrz = 0. But -z — mz 4- xrz gives mz — xr. Hence xr G M and so —1 — m 4- xr G Af which shows that M = R, a contradiction. Hence the result follows.
32
Solutions to Chapter 1
1.12	The result is true (by hypothesis) for j = i. Suppose that j > i and that R3"1 = S3~l + R3. Then
R3 = R3~lR = (S3~l + R3)R
C SRjl + R3+l
= S(S3~l + R3) + R3+1 CS3 + R3+l.
Clearly, S3 + R3+l C R3 and so R3 = S3 + R3*1. Using induction and the above, we have
R3 = S3 + й^^1
= S3 + S3+k~l + R3+k
= S3 + R3+k.
Suppose now that R is nilpotent, say Rm = {0}.
(а)	Й’ — S* + й,+1 implies R3 = S3 + R3+k for j > i and any k. Thus R3 = S3 + R3+m = S3 for all j > i.
(b)	Let R = (a} + R2. Then by (a) we have R3 — {a}3 for all j > 1. Take j = 1 to obtain R = {a}.
(c)	If Й2 g. M then R = M + Й2 since M is maximal. Hence, again by (a), R = M which is impossible. Hence R? C M.
(d)	Since й2 С M the quotient R/М has zero multiplication, so every subgroup of the additive group M is an ideal of R. Hence R/M has a prime number of elements.
1.13	Let A be a subring of a miring Й. If a G A then a G R so a is nilpotent. Hence A is a nilring. Also, if b G R/I then b = x + I for some x G R, and since xn = 0 for some n > 1 we have bn — I. Thus Й/7 is a nilring.
Suppose now that A and BjA are nil. Let x G B. Since (x+ A)n = A for some n > 1 we have G A whence (xn)m = 0 since A is nil. Thus x is nilpotent and so В is nil.
If В is nil then so is B/(A CiB) ~ (A + B)/A. But A is nil. Hence so is A 4- B.
In IRp we have (01,02» «3» • • -)n = 0 if and only if a? = 0 for i > 1, which is the case if and only if every аг- is divisible by p. Hence, since there are only finitely many non-zero о», we have
(oi,a2,os,...)n = 0 <=> 0,-GpZ,,..
33
Book 6
Rings, fields and modules
Now it is easy to see that the set N of nilpotent elements is a two-sided ideal of Rp. However, N is not nilpotent. For if it were then we would have Nm = {0} for some m, giving (01,02, .. .)m = 0 for all (01, aa, ...) G N. However,
(0t0,...,0,p,0, ...)m # (0,0, ...).
ТП4-1
1.14	(a) J C Ai since J is the smallest ideal of A2 containing I. Thus
JI С I and IJ С I. Hence
J3 = J(I + A2I4- IA2 4- A2IA2)J
GI + JI+IJ + JIJ
C I.
(b)	If I = Ao C At c • •  C An = R, call I an n-step subideal. Let I be the smallest ideal of R containing I. We show that I3n С I by induction on n.
If J is the smallest ideal of A2 containing I then
J С A2 C A3 • - C An ~ R
and J is an (n—l)-step subideal. By the induction hypothesis, J3w-3 C J where J is the smallest ideal of R containing J. But since IC Ai C A2 and J is the smallest ideal of A2 containing I we have J3 С I by (a). Consequently, J3n C J.J3 C JI С I. However, I C J С I so J — I and then C /.
(c)	If Im — {0} we have (f3n)m = {0} and I is nilpotent.
1.15	Suppose that Q is semiprime and that An C Q. We show that AC Q. The result clearly holds for n = 2. Suppose by way of induction that if A™ C Q with m < n then A C Q. If n is even then А»пАзп C Q implies A®" C Q whence A C Q by the inductive hypothesis. If n is odd then An+1 C Q and n 4-1 is even, whence again A C Q.
Let t] : R R/Q be the natural morphism. Suppose that Q is semiprime and that X is nilpotent in R/Q, say Xn — 0. Then, denoting the inverse image of a subset T of R/Q underj} by T, we have X C Xn — Q. Since Q is semiprime it follows that X C Q and so X — 0.
Conversely, suppose that R/Q contains no non-zero nilpotent ideals. Let A be an ideal with A2 C Q. Then [11(A)]2 = tj(A2) “ 0 whence h(A) = 0 and A C Q.
34
Solutions to Chapter 2
e.l (а) z]y <=> у = xr <=> у G (z) <=> (у) c (z).
(b)	If x and у are associates then (z) == (p) follows from (a). Conversely, if (z) = (y) then there exist r, s G R such that x = ry, у = sx whence z — rsx giving rs — 1 so that r, s are units and z, у are associates.
(c)	Using (b) we have
и is a unit <=> и ~ 1 <=> (w) = (1) = R.
(d)	Suppose that x = yr where y,r are not units. Then certainly (z) С (p) C R. Now (z)	(p) since otherwise x, у are associates (by
(b)) and then r would be a unit; and (y) R since otherwise у would be a unit (by (c)). The converse is clear.
(e)	If (z) is not maximal among the principal ideals then for some у we have (z) С (p) C R- By (d), у is then a proper factor of z and so z is not irreducible. Conversely, if z is not irreducible then it has a proper factor у and the result follows from (d).
(a)	True. If a|6 implies 6|a then every pair of non-zero elements are associates (for then z xy ~ p). Hence x ~ 1 for every x G R* and so R is a field.
(b)	True, as observed in the proof of (a).
(c)	a = bu and c = dv give ac = bd.uv and so ac ~ bd.
(d)	False. We have 1 ~ 1 and 1 4- 2t ~ —2 4- i in Z[t] but 1 4- (1 4- 2i] is not an associate of (—2 4- i) 4-1.
(e)	True. If every element of R* is a unit or a prime then every element of R* must be a unit. For, suppose that p G R* is prime and consider p2. Clearly, p2 is not prime, and it cannot be a unit since p2 G (p) R. Thus every element of R* is a unit and so J? is a field.
Book 6	Rings, fields and modules
£.S In Z, 5 is irreducible since it is prime.
In Z[X], 5 is irreducible since 5 = f(X)g(X) implies that deg/(X) = deg<?(X) — 0 so f(X) and <?(X) are constant polynomials. Then one of f,g is ±1.
In Z[i] we have 5 = (2 4- i)(2 — i) where neither 2 4- i nor 2 — i is a unit (since £(2 + i) — £(2 — ») = 5), so 5 is not irreducible.
In	5 is irreducible. In fact, if 5 = сф then 25 = £(5) =
£(a)£(/?). But if a = a 4- b\fi^2 then £(ft) = a1 2 4- 262 / ±5.
Й-4 We have that £(43i — 19) = 2210 = 2 • 5 13 17. Consider now a = a 4- ib. We have that <x is of length
(1)	2 if and only if a2 4- b2 = 2, which is the case if and only if ot is an associate of 1 4- i;
(2)	5 if and only if a2 4- b2 — 5, which is the case if and only if a is an associate of 2 4- i or of 1 4- 2г;
(3)	13 if and only if a2 4- 62 — 13, which is the case if and only if a is an associate of 3 4- 2» or of 2 4- 3»;
(4)	17 if and only if a2 4- b2 — 17, which is the case if and only if a is an associate of 4 4- i or of 1 4- 4i.
It is now as easy matter to check that
43» - 19 = (2 4- 3i)(l 4- »)(2 4- »)(4 - i).
£.5	(a) 114- 7i has length 170 — 2 • 5 ♦ 17. Now 1 4- i has length 2, and
(11 + 7i)= |(11 - Hi + 7i + 7) = 9 - 2i
(which has length 5 • 17) and so 11 4- 7» = (14-»)(9 — 2t). Now 14-2» has length 5 and
1 — 2i (9 - 2») —— = |(9 - 2i - 18» - 4) = 1 - 4»
(which has length 17) and so 9 — 2г = (14- 2»)(1 - 4»). Thus
11 4- 7» = (14- »)(1 4- 2»)(1 - 4г).
(b) 4 4-7\/2 = х/2(74-2ч/2).
(c) 4 - has length 19 which is prime so 4 - \/—3 is irreducible in Z{\Z=3].
36
Solutions to Chapter 2
2.6
Unique factorisation is not violated since
6 = 2 3 = |(-2 + ч/б)(2 + ч/б)][(3 - x/c)(3 + ч/б)] = [(-2 + x/6)(3 + ч/б)][(2 + ч/б)(3 - ч/б)] = ч/б ч/б.
In Z[v^lO] we have
6 = 2 • 3 = (4 + ч/10)(4 - v^O)
which gives two distinct factorisations of 6 into irreducibles.
2.7	In Z^y/—6] the only units are ±1 since a 4- by/—6 is a unit if and only if
£(a 4- ЬуЛ-б) = a2 4- 662 = ±1, i.e. if b — 0 and a = ±1. Consider now the decompositions
10 = 2 - 5 = (2 + \ЛГ6)(2 -
No element of	can have length 5 since «2 4- 662 = 5 is clearly
impossible. Hence 2 4- y/-6 and 2 — y/—6 are irreducible since each has length 10 and any decomposition would involve an element of length 5. Similarly, 5 has length 25 and is irreducible. Also, since a2 + 662 = 2 is also impossible, no element can have length 2 and hence 2 (of length 4) is irreducible. We thus have two distinct factorisations of 10 into products of irreducibles and so Z[\/—6] is not a unique factorisation domain.
(a)	2 + y/^6 is irreducible but not prime, for it divides 10 but does not divide either 2 or 5.
(b)	The greatest common divisor of 10 and 2(2 4- y/—6) does not exist. In fact, 2 and 2 4- y/—6 each divide 10 but 2(2 4- y/~6) does not divide 10 since £[2(2 4- y/—6)] — 40 which does not divide £(10) — 100.
(c)	Since 5 and 24-уЛ-6 are irreducible their greatest common divisor exists and is 1. Suppose that 5a 4- (2 4- y/—6)/3 — 1. Then 10a 4- 2(2 4-y/=¥)/3 = 2 which is impossible since 2 4- y/~6 divides the left hand side but does not divide the right hand side.
2.8
We have
8=22-2 = (14- >/—7)(1 - y/^7).
Now each of 2,14- y/—7,1 — \Z—7 is irreducible. For these elements have lengths 4,8,8 respectively so if any were reducible then there would exist a 4- by/^7 with a2 4- 7 b2 = 2 which is clearly impossible.
37
Book 6
Rings, fields and modules
Now 8k G 7] can be written as
8fc = (1 -	4- \/^7)fe
= (1 - v/=7)fc“1(l 4- x/^7)fc“123
= (1....У=7)*-‘(1 4- У^7)л-1’23*
_ 23fc
2k irreducibles
2k + 1 irreducibles
2k+ i irreducibles
3k irreducibles
whence the result follows on taking t — 2k.
An element that is irreducible but not prime is 2; for 2 divides 8 = (14- \/^7)(1 — \f—7) but divides neither of the factors.
Since 2 and 14-\Л-7 are irreducible they have greatest common divisor 1. Suppose that 72 4- 5(1 4- \/^7) = 1. Then on multiplying each side by 1 —	7 we obtain
72(1 - Vе?) 4- 58 = 1 - y/^7.
But 2 divides the left hand side and does not divide the right hand side, so we have the required contradiction.
2.9	Let I — (r). Then given a G Z[t] we have a = /9r4-7 where N(q) < N(r). Since 7/1 = a 11 and there are only finitely many 7 with < N(r) it follows that Z[i]/f is finite.
2.10	Let a — a + ib be a prime in Z[iJ. Suppose first that €(a) is even. Then a2 4-b2 is even. Since a, b cannot both be even (otherwise we can extract a factor 2 and a would not be prime), it follows that both must be odd. But then
a + ib = [|(a 4- 5) 4- |(6 - a)a](1 4- i) whence ex ~ 14- i since |(a 4- b) 4-1(b — a)i must be a unit (otherwise a 4- ib would not be prime).	'
Now suppose that £(a) is odd. Then if a, b are both non-zero we have £(a) s= a2 4- b2 is prime in Z; for otherwise (a 4- ib)(a — ib) = a2 4- b2 and a decomposition of a2 -r b2 into primes in Z gives a contradiction to a 4- ib being prime. Now a+ib is a factor of p — a2 4- b2 and p = 1 mod 4, which is (c).
If b = 0 then ex — a G Z and a must be prime. If a = 1 mod 4 then a = c2 + d2 for some c, d G Z so again case (c) occurs. If a = 3 mod 4 then case (b) occurs.
If a — 0 then a is an associate of b G Z and the same argument works.
38
Solutions to Chapter 2
2.11	(a) If n < —1, say n — — p, then Pell’s equation is a2 + bp2 — 1. The
only solutions are a = ±1,6 = 0 and so the group of units is {—1,1}.
(b)	If n > 1 then Pell’s equation is a2 — nb2 = ±1. И we write a — a + by/n G IR and a = a — by/n E IR then this becomes act = ±1. It follows that if a is a unit then so also is ak for every integer к > 1, for then we have akak — ±1.
(c)	By (b) we see that the group of units of Z[\/2] contains the set {±(1 + x/2)fe | к > 1}. Suppose now that a is a positive unit of Z[y/2]. Then for some к > 1 we have
(I + >/i)k < и < (1 + V2)fc+l
whence we deduce that
1 < «(1 +V2)~k <1+^2.
Let 1 ± y/2} k ~ a+ b\/2, so that
1 < a + bx/2 < 1 + x/2.
Since a,b E Z it follows that a = 1 and 6 = 0. Thus we have that w(l + \/2)~fc = 1 and so и = (1 + >/2)fc. Thus we see that the group of units coincides with the set {±(1 + \/2)fe | к > 1}.
2.12	Let A be an ideal of R. Then the set
M = Im G Z — E A for some — G R} n	n
forms an ideal of Z. For, if mi,m2 EM then —, — G A whence 72-1 Г12 mi ~
mi,m2 E A since A is an ideal, and consequently -------j----G A, giving
m< — m? E M. Also, if a G Z then ~ ~ E A shows that ami G M.
1 Ti-i
Now every ideal of 7L is principal, so we have M = (m), say. Thus if m = pkt where t is coprime to p then we must have A = —•'j, showing
that A is also principal. However,
contains A and so
proper ideal of R.
is a maximal ideal of R which
is the unique maximal ideal that contains every
39
Book 6
Rings, fields and modules
2.1S	(a) If x and 3/1 • • • yn are not relatively prime then there is an irre-
ducible d 6 D such that d\x and d|j/i  • -yn. Since, in a principal ideal domain, irreducible and prime are the same we have й|ж and d|pt- for some i. This contradicts the fact that ж, у, are given to be relatively prime.
(b) If pr> and py' are not relatively prime then there is an irreducible (=prime) p G D such that p\pr> and p|py’ whence the contradiction p, ~ p ~ pj.
Let b — Pi1 - •  p^ be the unique factorisation of b as a product of irreducibles. By (b) the elements Pi*,...,p„n are pairwise relatively prime. Hence, by (a), for every i the elements pp and	are
relatively prime. It follows that the elements
ПК*. Пк*> TI
t/l
are relatively prime. Thus there exist ai,..., an such that
1 = tti П й* +    + П iytn
Multiplying by a and dividing by b — П£=х Pi* we obtain
a acti	aan
~ b ~ p\l	prnn '
2,14	(1) => (2) : Let 7 be a family of ideals of A. Choose Iq G 7. Either Iq
is maximal in 7 or there exists /j G 7 with Iq G Ii • Either Д is maximal in 7 or there exists I? G 7 with Io C R C /2- Since A is noetherian, this argument can be repeated only finitely many times, ending with a maximal ideal of 7.
(2) => (3) : Let I be an ideal of A and let C be the collection of all ideals of A that are contained in I and are finitely generated. We have that C / 0 since clearly (0) G C. By (2), C has a maximal element J. For ж G / consider the ideal J + (ж). This is finitely generated (since J is) and is an ideal of A contained in I, so belongs to C. The maximality of J then gives J — J4- (ж) whence x G J. Since ж was chosen arbitrarily in I it follows that I = J G C and hence that I is finitely generated.
(3) => (1) : Let R С /2 C /3 C ... be an increasing chain of ideals of A. By the hypothesis (3), every ideal of A is finitely generated. Let I = Ua>i -(?• Then I is an ideal and so is finitely generated, say by
40
Solutions to Chapter 2
{si,...,zn}. Now every ж,- belongs to some I3i so let к be the greatest such j encountered. Then we have each ж,- G Д and it follows that
/ = (zi,...,z„) C Ik
whence I — Ik. Thus Ik = 4+l = /fc+2 = • • • and the chain terminates finitely.
(a) => (/?) : If A is a principal ideal domain then every ideal is principal so is finitely generated (in fact, generated by a singleton) so, from the implication (3) => (1) above, A is noetherian. Also, the sum of two principal ideal is principal (since every ideal is principal).
(/?) => (a) ; Suppose now that (/?) holds and let I be an ideal of A. By (1) => (3), I is finitely generated, say
1=	= (34) + • • + (»,).
A simple inductive argument shows that (by (/?)) every finite sum of principal ideals is principal. Hence I is principal.
2.15	(a) a\b implies that b G I = (a). But a is such that N(a} is minimal
for all elements of I. Thus, since N{b) — N(a) and b G I we have (a) = I = (6) and so a, b are associates.
(b) Let a, b G R be such that neither divides the other. Consider the ideal I = (a, 6) generated by a and b. Since b (a) and a (6) we see that I / (a) and I (6). Now choose d G I with N(d) minimal. Then I = (d) and N(d) < N(a),N(d) < N(b). From (a, b) = (d) we deduce that d G (a, 6) so there exist a, fl G R with d — aa 4- fib as required.
2.16 Suppose that there exist a, b G R \ {0} such that
6(a + b) > max(6(a),6(6)).
Then from
b = 0(a 4- b) 4- b, b = l(a 4- 6) — a
with 6(—a) = 6(a) < 6(a 4- b) and 6(6) < 6(a 4- 6) we see the lack of uniqueness of quotient and remainder.
Conversely, suppose that 6(a 4- b) < max(6(a), 6(6)) and that a G R has two representations, say
a — qb 4- r	(r — 0 or 6(r) < 6(6));
a — q'b 4- r' (И — 0 or 6(И) < 6(6)),
41
Book 6	Rings, fields and modules
with r^r' and q q*. Then we have the contradiction
6(6) < 6[(g — g*)4 =	- r) < тах(6(И), 6(-?-)) < 6(b).
Thus r — r' or q = q'. Since each of these implies the other, uniqueness follows.
In Z[i] take a = 1 + 4г and 6 = 5 + 3®. Then we have
1 + 4i = (5 + 3®)(1 + i) + (-1 - 4i)
= (5 + 3i)i + (4 — ®).
S.17	(a) We have -1 +3® = (1 +®)(1+2i). Since £( 1+i) = 2 and £( 1+2г) = 5
we see that 1 + i and 1 + 2i are irreducible.
(b) Since g.c.d.(a0,...,an) = 1 we cannot write f(X) = afi(X) with a irreducible. Hence if f(X) is reducible we can suppose that f(X) = g(X)h(X) where
<?(X)=6O + 61X + • + brxr,
6-(X) — cq + C\X + •  • + ceX8,
with r,s < n. Now a0 = 6oco and so, since p|a0, we have p|6o or p|cq. Also, since p2 does not divide feoCo we can assume that p|feo and p does not divide co. If p|6, for all i then p\an which is a contradiction. Hence there is a first coefficient, bt say, such that p does not divide bt. But
= (boCt +  • + 6*-iCi) + btC0
and p divides the bracketed part since p|6t- for 0 < i < t. Hence p does not divide at since it does not divide btco- Since t < n, this is a contradiction.
(c) Since — 1 + 3® = (1 + ®)(1 + 2г) we take p — 1 + i and note that p[(— 1 + 3i) but p2 does not divide —1 + 3®'. Also, since (1 + ®)(1 — •) = 2 we see that p|6 and p|8®‘ but p does not divide 1. Hence, by (b), the polynomial
X3 + BiX2 - 6X - 1 + 3®
is irreducible in 2[®'ЦХ].
2.18 Z[v/3] is a euclidean domain so every pair of non-zero elements has a greatest common divisor.
We have, as products of irreducibles,
13 = (4 х/з)(4 + \/з),
7 + 5\/з = (1 + \/з)(4 + х/з).
Hence the greatest common divisor is 4 + \/3-
42
Solutions to Chapter 2
2.19	(a) a24-5fc2 = ±1 if and only if a — ±1 and 6 — 0. So if (a+6\/^5)(c4~
dx/--5) — 1 then
(a2 + 562)(c2 + 5d2) = £(a 4- b\^S)£(c + dy/^5) = £(1) = 1 gives a2 4- 562 = 1 = c3 4- 5d2 and the result follows.
(b)	£(3) — £(2 4- x/~5) = ^(2 — V—5) = 9, so if z is a factor of 3,2 4- yf—5,2 — л/^5 then we must have £(a) G {1,3,9}. Now if £(л) = 3 we have z = a+by/^5 where a,b G IL are such that а2 4-562 — 3. Clearly, there are no solutions. Thus if a is a factor then we must have either t(z) = 1 or t(z) — 9. In either case, one of the two factors is of length 1 showing that it is a unit.
(c)	We have that
9 = 3 • 3 = (2 + T=5)(2 - 7=5).
To show that these are different factorisations it suffices to observe that 3z 2 -1- \T~5 for any unit г, for ±1 are the only units.
(d)	To show that P\ = (3,2 4- \f—5) is prime, consider Pi Г) Z. We have 3 G Pi HZ, and if any integer not divisible by 3 lies in Pi П 7L then, by the euclidean algorithm, 1 G Fi о Z and we have the contradiction Pi = Z[x/^5], Hence we see that Pi П Z = 3Z.
Given x, у G Z{>/—5} let w, v be integers such that
x — « G Pi, у — v G Pi.
Suppose that xy G Pi- Then uv G Pi and, since uv G Z, we have uv G Pi П Z — 3Z. Thus и G 3Z or v G 3Z, so either x G Pi or у G Pi-Hence Pi is prime. Similarly, so is Pg = (3,2 — Vr-5).
(e)	With the above notation, we have
P1P2 =(3,2 + 7=5)(3,2 - 7=5) = (3)(3,2 + 7=5,2 - 7=5) = (3),
since 1 = -3 + (2 + 7=5) + (2 - 7=5) e (3,2 + 7=5,2 - 7=5).
2.20	First note that
m + nx/10 G Pi tn + пл/10 G Fa m 4- тгл/10 G P3
2|m,
m — n = 0 mod 3, tn 4- n zzz 0 mod 3.
43
Book 6	Rings, fields and modules
Using this we can show that each of Pi, P?t P3 is prime. For example, if
(mi + niVlo)(m2 + 712'/10) € Pa
then we have
31 [(mi — ni)(m2 — 7i2) + 9711П2]
and so 3|(mi — tii) or 3|(тп2 — ng).
Now (6) = (2)(3) = (4 + v/10)(4 - x/10) and
(2)=P?, (3)=P2P3, (4 +\/10) = P,P2, (4 - \/10) = P,P3.
Hence both decompositions of the element 6 lead to the decomposition
(6) = P?P2P3
for the ideal (6) as a product of prime ideals.
44
Solutions to Chapter 3
S.l
The binomial theorem (which holds in any commutative ring with a 1) gives
(a + bf = £ 0)a^br, (a - (,)» = £ (₽)
For 0 < r < p the coefficient
0_ p!
r!(p - r)!
is an integer. Since p|p!
and p does not divide either r! or (p — r)!, we see that Г j is an integer
that is divisible by p. Hence we have that
(a + b)p =^ар + Ьр
and, if p is odd,
(a - by = ap +	= ap- bp.
If p is even then we must have p = 2, in which case
(a — fe)2 — a2 -t- b2 — a2 - b2 + 2b2 ~ a2 — b2.
For the second part, proceed by induction. The result has just been established for n = 1. For the inductive step, suppose that
(a±t)p ±6₽
Book 6	Rings, fields and modules
where к > 1. Then we have
(a ± Vf" = [(a ± b)'’"f = (apk ± i"')''
= (</?' ± ((/)'’ = <Z+1±(/+‘.
S.2 If F is of characteristic 0 then clearly
Df = 0 <=> ах = a2 = • • • — ап — 0 <=> / — ao.
If now F is of characteristic p then
(1) if p\k we have
как — (kl)ak = Oak ~ 0;
(2) if p does not divide к we have
как — 0 > (kl)ak ~ 0 => ak = 0.
Thus Df — 0 if and only if f is of the form
a0 + арХР + a2pX2₽ + • - - 4- orpXr₽.
S.S Clearly, every non-zero constant polynomial is a unit in F[X). That these are all the units follows from the fact that if f, g are non-constant polynomials then degfg > deg / > 1 so that fg is non-constant. Hence no polynomial of degree greater than zero is a unit.
If <p :	—» F[X| is an automorphism then tp sends units to units
(since the units are the invertible elements). Hence the image under <p of a non-zero constant polynomial is also a non-zero constant polynomial.
Define •& : F -* F as follows : let $(0) = 0 and for every a / 0 let #(a) be the constant polynomial <p(a). Then clearly & is an automorphism on F and the diagram
F-------—>F[X]
•9
F----—> F[X]
is commutative.
For the monomial X G F[X], consider <p(X) E F{X]. If degy?(X) < 1 then <p(X) is a constant polynomial so
<p(a0 +atX +	+anXn) ~ d(a0) + tf(ai)p(-X) + " + ^(«n)(v>(-X))n
46
Solutions to Chapter 3
is also a constant polynomial, which is impossible since <p is an automorphism. If now degy?(X) = к > 1 then the image tp(f) of a polynomial of degree n will have degree nk. Thus only polynomials of degrees k,2kt3k,... can appear as images under <p. Again this is impossible since ip is an automorphism. We must therefore have deg <^(X) = 1 so that <p(X) — aX 4- b for some a ± 0.
S.j We have that f, g are equivalent if and only if
(VaGF) (/~ff)(a)=0,
which is the case if and only if X— a divides f — g for all ex G F. Suppose that deg(f — g) = n.
If F is infinite and /, g are equivalent then taking n + 1 distinct elements oio.Oi,... in F we have that f - g is divisible by
(X-tt0)(X-ttl)-(X-ft,)
which is of degree n+1. Since a non-zero polynomial of degree n has at most n distinct roots, we conclude that f — g = 0 and so / — <?.
If F is finite, say F = {<h, a2,..., an}, then the above argument shows that f — g is divisible by
ro(X) = (X-ai)(X-a2)---(X - a„).
Conversely, if f — g is divisible by m(X) then it is clear that / and g are equivalent.
In the case where F — GF(p) = Z/pZ we have
m(X) = X(X - T)(X - 2) • -  (X - p^T).
Since, for a = б, I,..., p — 1 we have (xv — a [recall Fermat’s theorem that a?-”1 = 1 mod p], we have that X — cx divides X1’ — X. Hence m(X) divides X₽ - X, say Xp - X — m(X)h(X). Equating degrees gives p = p 4- deg h and so h(X) is a non-zero constant polynomial. Equating leading coefficients now gives h(X) = 1 and so m(X) — Xp — X.
3.5	Consider X2 + У2 — 1 as an element of F[F][X] = F[X, У], i.e. as a polynomial in X with coefficients in F[F].
Now У 4- 1 is irreducible in F[y], and У + 1 divides У2 — 1. Also, (У 4-1)2 does not divide У2 — 1; for otherwise, since each is monic, we would have (У 4-1)2 = У2 — 1 whence 2У 4-2.1 = 0 which is possible only if F is of characteristic 2. We can therefore apply the result of question 2.17(b) to see that X2 4- У2 — 1 is irreducible in F[X, У].
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Book 6	Rings, fields and modules
3.6	The natural morphism Z —»Zn extends to a morphism Z[X] Zn[X]; simply reduce the coefficients modulo n.
Given f G Z[X], choose n such that n does not divide the leading coefficient of /. Then if f factorises over Z the image of f under the morphism factorises over ZR. Thus if the image of f is irreducible over Zn then f must be irreducible over Z.
The whole point behind this useful test for irreducibility is that Zn is finite and so there are only finitely many possibilities to check.
In Z5[X] we have X4 + 15X3 + 7 — X4 + 2. This has no linear factor since no t G Z5 satisfies t4 + 2 — 0. Suppose that
X4 + 2 = (X2 + aX + t>)(X2 + cX + d).
Then we have
a + c = 0, ac + b + d = 0, ad + be = 0, bd = 2.
From a — —ewe have a(b — d) — 0. If a — 0 then b — —d so t>2 = — 2 = 3, while if b — d then fe2 = 2. However, only 0,1,4 are squares in Z5, so neither fc2 = 3 nor b2 = 2 is possible. Hence we see that X4 + 2 is irreducible over Z5. By the above, it is therefore irreducible over Z. But if it were reducible over Q then it would be so over Z.
3.7	{l,\/2} is a basis for Q(\/2) over Q, and {1, \/2, (^)2} is a basis for
©(?2)(л/2) over Q(72). Thus
{1, \/2, ^2,2&l6,22/3,27l6 }
is a basis for Q(\/2, v^) over Q.
Now 27/e = 2 - 21/6. Hence 21/6 G Q( A ^2) and so
® с ®(№) c ®(V2, №).
Now \/2 is a zero of X6 — 2 which by Eisenstein’s criterion is irreducible over Q. This is therefore the minimum polynomial of \f2 over Q. Hence we have
6 = (®(\/2, #2) : ®) = (®(x/2, $2) : ®(№))(©(№) : ®) = (®(T2, #2) :®(№)) -6
so that (®(\/2, \/2] : ®(v^)) = 1 and hence ®(\/2,	= ®(у/2), a
simple extension.
48
Solutions to Chapter 3
3.8	Suppose that Z were algebraic over F. Then there would exist a polynomial
/(X) = a0 + a1X+  + anXn	0)
over F for which f(Z) = 0, i.e.
(X3 \	/ X3 \2	/ X3 \n
x+i;+o4v+t) + +0”Vx4i) =0'
from which we have
c„X=” + a^iX3n~3(X + 1) + • • • + a„(X + l)n = 0.
Since the left hand side is a non-zero polynomial of degree 3n, we have a contradiction. Thus Z must be transcendental over F.
Now in (F(Z))[y] we have that X satisfies У3 — ZY - Z — 0. But Y3 — ZY — Z is irreducible in (F(Z))[y]; for, if not, it has a linear factor У — л where a € F(Z) and a3 — Za — Z ~ 0, and this would imply the existence of a non-zero polynomial f for which f(Z) = 0 which, as we have just seen, is impossible. Thus we see that У3 — ZY — Z is the mimimum polynomial of X over F(Z). Thus F(X) is a simple algebraic extension of F(Z).
3.9	Both are irreducible by the Eisenstein criterion. We have
x2 - з = (x - V3)(x + 7з)
X2 — 2X — 2 = (X — 1 —\/3)(X - 1 + VS)
and the splitting field in each case is Q(\/3).
3.10	Over C we have
(X2 - 2X - 2)(X2 + 1) = (X - 1 + V3)(X - 1 - \/3)(X - i)(X 4-1).
Hence a splitting field is <Q(\/3, i).
We also have
X'1 - ЗХ3 + X2 - 3
= (X2 - 3)(X3 + 1)
= (x - 7з)(х + 73)(X + 1)(X - i(-i + i\/3))(x -1(-1 - tV3)) and a splitting field is
И-1 + ^)) = A*)-
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Book 6
Rings, fields and modules
Let К = Q(x/3,i). Now x/3 has minimum polynomial X2 —3 over Q so (Q(x/3) : Q) = 2. Since i $ Q(x/3) and since i has minimum polynomial X2 4-1 over Q(\/3) we have (Q(\/3, i) : Q(x/3)) = 2. Hence
(K : Q) = (K : Q(\/3))(Q(x/3) : Q) = 2 - 2 = 4.
3.11	It is readily seen that
X3 - 3abX + a3 + b3 = (X 4- a 4- tifiX2 4- (a 4- 6)X 4- (a + 6)2 - 3a6).
Since XB — 6X3 + 8 = (X3 — 2)(X3 — 4), a splitting field over Q is Q(v^2,	But (v^2)2 = so Q( Й, \^4, n/З) = Q(</£,iv/3).
Hence S = Q(^2, г>/3) and
(S : Q) = (Q(№n/3) : Q(^2))(Q( ^2) : Q) = 2 -3 = 6.
Since S contains xfiz and x/Л. we have </2 4- y^4 G S. Now take a ~ — \fii,b — — \/2 in the first part of the question to see that
X - (</4 + </2) divides X3 - 6X - 6.
Since \/4+ xf2 G S, which is of degree 3 over Q, and since \/4+ x/2 £ Q, we see that the minimum polynomial of xfk 4- xf2 over Q is a cubic. By the above observation, it must therefore be X3 — 6X — 6.
Since \/4 + xfl G S we have Q( 4- Q S- But Q( y^4 4- ^2) has degree 3 over Q while S has degree 6, so we conclude that Q(	\/2) Ф
S.
3.12 In Q[X] we have
/(X) = (X2 + 4X + 1)(X2 - 2X - 1)
and in IR[X] we have
/(X) = (X 4- 2 - V3)(X 4-2 4- V3)(X - 1 - v/2)(X - 1 + V5).
Thus К ~ Q(v/2, x/3) is a splitting field for f.
Now if we had \/3 G Q(\/2) then there would exist atb G Q with х/З = a 4- b\/2 whence
3 = a2 4- 2b2 4- 2abx/2.
50
Solutions to Chapter 3
3.13
If a, b 0 we have the contradiction \/2 GO. If a = 0 then \/3 = b\/2 so \/6 = 2b G Q, another contradiction. If b = 0 then \/3 — a G Q, also a contradiction. Thus we see that \/3 £ 0(ч/2) and so
(Q(v/2, \/3) : 0) = (Q(v/2, \/3) : О(ч/2))(О(\/2) : Q) = 2  2 = 4.
A basis for Q(x/2,V3) over Q is
{1, ч/^,л/з, ч/б}.
Now the minimum polynomial of ч/З — V2 has degree 2 or 4. If it were of degree 2 then for some a, b G 0 we would have
(ч/З - 4/2)2 + a(v/3 - v/2) + Ь = 0
which gives
b + 5 + ач/з - aV2 - 2\/б = 0,
which contradicts the fact that {1, >/2, ч/З, \/б} is a basis. Hence we see that the minimum polynomial has degree 4 and Q(\/2,4/З) = Q(\/3 — 4/2).
The minimum polynomial of 4/З— >/2 over Q is X4 — 10X2 +1 whereas over ®(\/3) it is X'2 - 2v/3X + 1.
We have that
X4 - X2 - 2 = (X2 - 2)(X2 + 1)
with X2 — 2 and X2 + 1 irreducible over Q. Now over (I
x4 - X2 - 2 = (X - \/2)(X + ч/2)(Х - i)(X + i)
so f splits completely over Q(\/2,i). Since clearly f cannot split in any smaller subfield of C, we have that К — Q(x/2,i) is a splitting field.
Now 4/2 has minimum polynomial X2 — 2 over Q and so we have Q(\/2) : Q) = 2. Since i Q(\/2) and since i has minimum polynomial X2 + 1 over Q(\/2) we also have that (0(ч/2,г) : Q(\/2)) = 2. Hence
(K : Q) = (K : Q(^))(Q(v/2) : Q) - 2 • 2 = 4.
A basis for К over Q is {l,\/2,i,i\/2}.
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Book 6
Rings, fields and modules
Since Q(i 4- V2) C Q(i_, x/2) we have that (Q(i 4- \/2) : Q) is either 2 or 4. Now (i 4- x/2)2 = 14- 2i\/2, and if the minimum polynomial of i 4- x/2 were of degree 2 we could find a, b € Q such that
0 = (t + x/2)2 + a(i + x/2) + b = 64-14- axf2 4- ai 4- 2ixf2,
which contradicts the fact that {l,\/2,i,i\/2} is a basis. Thus the minimum polynomial of i 4- x/2 is of degree 4 and so (Q(t 4- \/2) : Q) = 4 and consequently Q(i 4- x/2) —	\/2).
Since clearly (14- x/2)2 = 14- 2ix/2 we have (i 4- x/2)4 = (14- 2ix/2)2 — —7 4- 4ix/2 and so
(i+\^)4-2(i + ^)2 + 9 = 0,
whence the minimum polynomial of i 4- x/2 is X4 — 2X2 4- 0.
(a)	Since (i 4- x/2)2 = 14- 2i\/2 =14- 2i(i 4- x/2) 4- 2 the minimum polynomial of i 4- x/2 over Q(i) is X2 — 2iX ~ 3.
(b)	Since (i 4- x/2)2 = 14- 2ix/2 =14- 2x/2(i 4- x/2) — 4 the minimum polynomial of i 4- x/2 over Q(\/2) is X2 — 2x/2X 4- 3.
(c)	Since (i4- x/2)2 = 14- 2ix/2 the minimum polynomial of i 4- x/2 over Q(ix/2) is X2 - 1 - 2ix/2.
З.14	(1) => (2) : If f is reducible then it has a linear or a quadratic factor.
If it has a linear factor X — a then о is a root, whence so is —a, whence f has the factor A'4o. Thus f has the quadratic factor (X-|-o)(X —a).
(2) => (3) : If X4 4- r = (X2 4-0X4- /?)(X2 4- o*X 4- /?') then
a 4- a‘ = 0, fl 4- fl' 4- ос/ — 0, txfl’ 4- a*fl — 0, fl fl' = r.
If a / 0 then a1 ~ —ot so a{fl' — fl) = 0 so fl’ = fl, which gives r — fl2 with 2fl = ~aa' = o2, i.e. r = ^o4.
If a = 0 then a* = 0, fl' = — fl and r — —fl2.
(3) => (1) : If r = -p2 then X4 + r = X4 - p2 = (X2 4- p)(X2 - p); and if r — ^g4 then
X4 + т = X* +	= (X2 + qX +	- qX + |g2).
Every element of К = is of the form у a 4- b^ 4- c^2 4-
52
Solutions to Chapter 8
where a,b,c, d G Q and £4 = —r G 0- Now (Q(£) : Q) = 4 so if Q C L c 0(f) we must have (L : 0) = 2. Thus L is necessarily of the form Q(??) where j/2 G Q and tj Q. Now
jy2 - a2 + fe2£2 + c2^4 4- d2^6 + 2ab£ + 2ac£2
4- 2ad£3 4- 2bc£3 4- 2fed£4 + 2cd£5
— a2 4- b2? — rc2 — rd2£2 4- 2ab£ 4- 2ac£2
4- 2adf3 4- 2bc£3 — 2bdr — 2cdr£
— (a2 — rc2 — 2rbd) + 2(ab — rcd)£
4- (fe2 - rd2 4- 2ac)£2 4- 2(ad 4- fec)f3,
so f]2 G 0, rj £ Q is equivalent to
ad 4- be = 0, ab — red — C, b2 — rd2 2ac = 0, 6, c, d not all 0.
Now if c 0 then since r / —b2/dP we must have a = b = d = 0 in which case T] = c£2 and Q(?j) — Q(c£2) = Q(£2). On the other hand, if c = 0 then
ad •= 0, ab = 0,
6, d not both 0.
Thus if \/r Q there are no further solutions. If, however, \fr G Q then these conditions give
b a — 0, - — ±r.
d
In this case we can take d — 1 and b = ±^/r so that there are two other solutions, namely
4 ~ Vr( + (3 and 4 = -y/rf + f3.
3.15	(a) The Galois group has order 4 since the minimum polynomial of
e2iri/5 is X4 4- X3 4- X2 4- X 4- 1.
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Book 6
Rings, fields and modules
(b) The order of the Galois group is 1 (i.e. every Q-auto morphism of Q(w) is the identity). For, suppose that a is a Q-automorphism. Then since w = \/2 we have
(o(w)]3 = or(w3) = a(2) — 2.
But since Q(w) C IR we have a(w) G IR and so o(tu) — w. Hence a is the identity map.
3.16 In C[X’ we have
X4 - 2 = (X + r)(X - r)(X + ir)(X - tr)
where r = \^2. Hence a splitting field for X4 — 2 is ©(-^2,$). Also,
(Q(\/2,i) : Q(\/2))(Q(</2) :Q) = 2-4==8.
A Q-automorphism of Q( \/2, i) is completely determined by its effect on x^2 and i. Now the conjugates of i are i, —i since its minimum polynomial is X2 4-1. Also, the conjugates of are
</2, -xfi, ix/2,
since its minimum polynomial is X4 — 2. Hence there are eight elements in the Galois group
Gal(Q(№i),Q).
The elements a,P given by a(r) = —r, ct(i) ~ —i and /3(r) = ir,/3(i) = i do not commute, so the Galois group is not abelian.
3.17 Let о = ^2. Then
X3 - 2 = (X - л)(Х2 + лХ + a2)
= (X a)(X - a(-l + *Я)(Х - a(-l - ^)).
By the uniqueness theorem, any splitting field of X2 — 2 is isomorphic to the subfield of (E generated over Q by
{°. (4 + ^)“. (4-¥)“}•
All these elements belong to Q(a,t\/3); and since
sV3=q-‘[(4 + ¥)“-(-!- ¥)“)
54
Solutions to Chapter 3
it follows that a splitting field is Q(
Now (Q(	: Q) is a multiple of (Q(^2) : Q) and (Q(t\/3 : Q),
so is a multiple of 6. But X3 — 2 is of degree 3, so the splitting field has degree at most 3! = 6. Hence (Q( -ffi, t\?3) : Q) = 6. Thus the Galois group of X3 — 2 over Q is of order 6.
Now let К = Q(v/2,®\/3). A Q-automorphism of К is completely determined by its action on x/2 and on i\/3. The conjugates of a = v^2
are
(~5 + ^)а.(-^)“
and those of i\/3 are 2\/3, So the six elements of Gal(K, Q) are represented by
The group table is
b b
9 9 h h к к
о e
a b g h к a b g h к b e h к g e a к g h к h e b a g к a e b h g b a e
The subgroups are
{e}, {e,c, b}, {е,^},	{e,k},
and the associated Hasse diagram is
55
Book 6
Rings, fields and modules
By the fundamental theorem of Galois theory, the subfields of the splitting field have a lattice diagram that is the dual of the above subgroup lattice diagram. As К ~ Q(^2,i\/3) it must have four subfields (other than К and Q). These are
Q(\/2) of degree 3 —the fixed field of {e, g);
of degree 2 —the fixed field of {e,a,b};
Q((—| — *^)a) °f degree 3 —the fixed field of {e, h}\
Q((—| + M^)a) of degree 3 —the fixed field of {e, к}.
3.18 First note that since a is a zero of f we have a3 = За — 1. Now
(X - a)(X - a2 + 2)(X + a2 + a - 2)
= (X2 + (-a2 - a 4- 2)X + a3 - 2a)(X + a2 + a - 2)
= (X2 4- (-a2 - a + 2)X + a - 1)(X + a2 + a - 2)
= X3 + (-a2 - a + 24-a24-a - 2)X2
4-	[[—o2 - a 4- 2)(a2 + a - 2) + a — 1]X + (a — l)(a2 + a — 2).
But the coefficient of X2 is clearly 0, whereas that of X is
- a4 - a3 + 2a2 - a3 - a2 4- 2a + 2az 4- 2a - 4 + a - 1
— — a(3a — 1) — 2(3a —1)4- 3a2 4- 5a — 5
= -3,
and the constant term is
a3 4- a2 — 2a ~ ex2 — ct 4- 2 = 1.
Since /(X) = (X — a) (X — a2 4- 2) (X 4- a2 4- a — 2) we see that Q(a) is a splitting field for f over Q and that {l,a,a2} is a basis. Hence we have (Q(a) : Q) = 3.
Since i?(a) = a2 — 2 we must have
0(a2) = [0(a)]2
- a4 - 4a2 4- 4
= —a2 — a 4- 4
56
Solutions to Chapter 3
since а3 = За — 1. Hence 0 extends to a unique ©-automorphism defined ЬУ
0(a + ba + ca2) = a + b(a2 — 2) + c(-a2 - a 4- 4)
- (a — 2b + 4c) + (—c)a + (6 — c)a2.
Therefore Gal(©(a), ©) = {1,$,02} is the cyclic group of order 3.
Note that any ©-automorphism that sends a to a2 — 2 sends a2 — 2 to (—a2 — a + 4) — 2 = —a2 — a 4- 2, so no ©-automorphism interchanges a and a2 — 2.
3.19 Only (a) is normal, a splitting field of X"2 — 2. All of the others have minimum polynomials with complex roots.
3.20	(a) We have f(X) = {X-XV )(X - X2) - - - (X - Xn) and f does not
split in any proper subfield of К.
(b) An F-automorphism of К maps conjugates to conjugates, so permutes the X^ Conversely, if $ G Sn, the symmetric group of degree n, then the mapping : К —» К such that
$*(a) = a (af~E)
^(Xi} = X^ (i = l,...,n)
defines an F-automorphism on К since the symmetric polynomials are unaltered. Thus we see that $»-+$* describes an isomorphism from Sn onto Gal(X,F).
(c) (K:F)=|Gal(K,F)|=|Snl-n!.
57
Solutions to Chapter 4
4.1	This involves simply a trivial verification of the axioms; for example, (f + g)™ =(f + s)(m) = /(m) + g(m) - fm + gm.
4.2	=►: Define : M —+ Af by цг{т) = rm. Then r i-> цг is a 1-
preserving ring morphism д : R —♦ EndAf.
<=: Define an external law by (r, m) rm — [д(г)](т). Then since д is a ring morphism we have
r(m1 4- m2) = rmi + rm2,
(r + s)m = rm + sm,
[rs)m = r(sm),
Im — m,
and so M is an Я-module.
4.3	Take R — Af = IL in question 4.2. If f G EndZ then / is completely determined by /(1) since /(m) = /(ml) = m/(l) for every m G Z. Now observe that Д/р) = / so that ц is surjective :
(Vm e Z) A7(i)(m) = fWm = f(m).
If now r G Ker д then ц(г) = 0 gives rm = 0 for all m G Z whence r — 0. Thus д is a ring isomorphism.
The result for Q is similar : here we use the fact that if q -= — G Q m then mq = n so mf(q} = f{mq) = f(n) = n/(l) whence
/(«)= -/(!) = «/(!) m
4-4	If л,/? G Я exist such that f(x, y) — ax+fly for all xty G R then clearly f is an Я-morphism.
Solutions to Chapter 4
Conversely, if f : R x R' —* R is an й-morphism then for all я, у G R we have
+ J/(0,l)] = s/(l,O) + y/(0,1)
whence, since R is commutative, the result follows with a = /(1,0) and /? = /(0,l).
4.5	(a) Not a ring morphism; for example,
/(72 • 72) = f(2) = 2 and /(72)/(72) = 1-1=1.
(b)	Not a 2[72]-morphism; for example,
72/(1) = 72 and /(T2-1) = /(T2)= 1.
(c)	/ is, however, a Z-morphism since
/[(с + bx/2) 4- (c 4- d\/2)] — e4-b-*-c4-d = /(g 4- b\/2) + /(c 4- dx/2), f[n(a 4- b\/2)] — na 4- nb — n(a 4- b) — nf(a 4- bx/i).
4.6	That А/ G Могл(й, Af) and that Могд(й,М) is an й-module is routine. That 0 : Могд(й,АГ) —» M given by 0(/) = /(Ik) is an R-morphism follows from the fact that
W + g)^(f + p)(i*) = /(M + p(i«) = ад +
0(А/) = (А/)(1Д) = /(1яА) - /(А) = А/(1Я) = А0(/).
That$ is injective follows from the observation that if &(f) ~ «9(g) then /(1д) == <7(1r) and so ftg agree on the basis {1r} of the й-module Й. Thus we have f = g. Finally, given m G Af let fm : Й —* Af be defined by /m(r) = rm- Clearly, we have /w G Могд(й,АГ) and in = 1цт = fm(i/г) —	Consequently & is also surjective.
4.7	Observe first that f~* and f*~ are inclusion-preserving. Since Ker / = f*~ {0} and Im / = /"* (Af) it is then clear that
A + Ker/C/-[/-(A)], /-[/‘-(B)]CBnIm/.
(a)	If now x G /*“(/“*(Л)| then we have /(ж) = /(c) for some a G Л whence x — a G Ker/ and x G A 4- Ker/. This establishes the first equality.
(b)	If x G В П Im / then for some у G Af we have f(y) = s 6 B. This gives у G /*“(5) and so x = f(y) G f~*[f‘~(B)], which establishes the second equality.
(c)	Since f~* is inclusion-preserving we have f~*(AC\f^~(B)) C f~*(A) and f~*(AC\ f*~(B)) C (В)] С B. Thus it suffices to prove that if x G /“’’(А) П В then x = /(g) for some g G An /’“(B); and this is immediately seen from x = /(a) G B.
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Book 6	Rings, fields and modules
4-8	(a) Let i : Z -» 0 be the canonical inclusion and let M be a non-
zero submodule of Q. Then there is a non-zero rational ~ G M. Thus m	n
m — n - — G TVf and so Л1 nZ {0}. Consequently	{0} and
i is essential.
(b) Let j : Q —> IR be the canonical inclusion. Let a G IR be irrational. Then oZ is a non-zero submodule of IR. Clearly, oZ n Q = {0} so j'*“(otZ) = {0} and j is not essential.
Let A be the collection of submodules T of N such that MC\T = {0}. Then A 0 since clearly {0} G A. Now A is inductively ordered; for, let Ni С Л2 С Л3 C ... be a chain of submodules in A. Then Uf>i № =	№ is a submodule of TV, and
M n U Ni = U(MnM) = J{0} = {0} »>1 t>l	i>l
so Ut>i № G A- Applying Zorn’s axiom we can therefore assert the existence of a submodule A of TV that is maximal in A.
Consider now the composite morphism
M-------—> N---!—> N/A.
That this is a monomorphism follows from the fact that if x G Kert] о i then == ° and 80 ® = i(x) G Kerlj = A. Thus x G M П A == {0} and so x = 0.
To see that |j о i is essential, let G be a non-zero submodule of TV/A. Then by the correspondence theorem we have G — X/A where X is a non-zero submodule of TV with А С X; and, indeed, А С X since G is non-zero. Now if we had (h °	= {0} then we would have
{0} = «*_[I]*_(X/A)] — X ГУ M whence X G A in contradiction to the maximality of A. Thus (t] о i\~(G)	{0} and so tj о i is essential.
4-Q	(a) We have that
x G /“"(Ker (?of) <=> ж = /(у) where ₽[/(«/)] = 0
4=» x = /(t/) and д(к) = 0
<=> x G Im / П Ker g.
(b) Likewise,
f) <=> (3y) jfx) = s[f(y)]
<=> (3y) z-f(y)6Kery
<=> x G Im/ + Kerg.
60
Solutions to Chapter 4
4-10	(a)=>(b) : If h о f — g then
x G Ker f => g(x) — fe[f (ж)] = h(G) — 0 => x G Ker g.
(b)=»(a) : If Kerf C Kerg then for x,у G A we have
f(p) ~ f(y) => x “ У G Kerf C Kerg => е(ж) = g(y).
Since f is surjective, we can therefore define a mapping h : В —♦ C by the prescription h\f(ж)] = Clearly then ho f — g, and h is an .R-morphism since
M/fa) + r(sdl = МЛ® + »)1 = <ЛХ + »)
= g(x) + s(y) = МЛ®)] + fel/(sOL
= Л[/(Аз=)] = <z(A3c) =	= А/ЦЛ®)]-
If such an R-morphism h exists and if к : В —> C is also an R— morphism with ко f = g then ко f = ho f whence к — h (since f, being surjective, is right cancellable). Thus h is unique. Also,
Ker h — {0} <=> (g(sc) = 0 => f (ж) = 0)
<=> Kerg? C Kerf
<=> Ker g = Ker f.
Consider now the diagram
in which i о f о t is the canonical decomposition of f and a,/7,7 are respectively an epimorphism, an isomorphism, and a monomorphism-
Since f = 7 о p о a and since 7,/? are each monomorphisms we have Дж) =0 if and only if с>!(ж) = 0 so Ker a = Kerf — Kertj. By the previous part of the question it follows that there is a unique monomorphism h : A —* М/ Ker f such that h о a = I}. Since t] is surjective it follows that so also is h. Thus h is an isomorphism.
Arguing dually, we see that Im 7 = Im f and there is a unique epimorphism к : Im f —» В such that 70 k — 1. Since 1 is injective so is k. Thus к is an isomorphism.
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Book 6	Rings, fields and modules
4-11	(a) Let {-7-,...,^-} be a finite subset of Q and let b = ГП=1 ai-
al an
Let p be a prime such that p does not divide b. Then — is not in P
the submodule generated by	^act« suppose that it
were : then we would have, for some ®i,..., xn G Z,
1 n
1	G>i Z
p “X* a-i b
for some z€ Z, so that b ~ pz and we have the contradiction p\b. Thus Q is not finitely generated.
(b)	If p, g are distinct primes then - + Z - + Z, since otherwise 11	VI
- — and so q — p = pqn for some n G Z whence, for example, q — p(l + gn) and the contradiction p\g. Thus we have that if pi,P2,P3, • •• is the sequence of distinct primes then the elements
— + Z, —+ Z, —+ Z, ...
Pi P2 Рз
are distinct elements of Q/Z, so Q/Z is infinite.
(c)	For every x G Q/Z there exists n G Z such that n / 0 and nx — 0.
In fact, let x = - 4- Z; then q 0 and
Q
gs: = g^£ + z) = ^ + Z = p+ Z = 0 + Z = 0.
Thus no singleton subset of Q/Z is independent. No element of Q/Z can therefore belong to an independent subset. The only independent subset is then (by definition) 0.
(d)	If 19 G Mor2(Z/2Z,Q) let x — «9(1 4 2Z). We have
2x = 2i9(l + 2Z) = «9(2 + 2Z) = i?(0 + 2Z) = 0
whence x = 0 and consequently 19 — 0.
(e)	Let i9 G Morz(Q, Z) and suppose that «9(1) / 0. Then for every non-zero r G Z we have «9(1) = n9(~) and so r|i9(l). Since «9(1) has only finitely many divisors, we deduce that we must have «9(1) = 0. Consequently, for all p,q G Z with p,g / 0 we have
° = p?(i) =	= w’ Q) -
whence 19(~) =0 and so i9 is the zero map.
62
Solutions to Chapter 4
4-12 Consider f t R -* M — Rx given by /(A) — Xx. Clearly, f is an R-morphism with Ker f = Аппд(г). By the first isomorphism theorem, M — Im f ~ R/ Kerf = R/knnji(x).
That «9 is well defined follows from the fact that
x 4- mZ. — у 4- mZ => x — у G mZ
=> nx - - ny С nmZ
=> t9(z 4- m2) = j9(y + m2).
Clearly, «9 is a Z-morphism.
Let g G Mor2(Z/?nZ,Z/nmZ) and let g(l 4 mZ) — t 4 nmZ. Then 0 4 nmZ = g(0 4- mZ) = g(m + mZ)
= mg(l 4- mZ)
= m(t 4- nmZ)
= mt 4 nmZ
so mt G nmZ and consequently t G nZ, say t — nr. Thus g(l 4 nZ) = nr 4- nmZ and so, for every x G Z,
g(x 4 mZ) — xg(l 4- mZ) = xnr 4- nmZ
= r(nx 4- nmZ)
= rs9(z 4- mZ)
whence we have g — n9. Thus Morz(Z/mZ,Z/nmZ) is generated by {$}. Now
A G Annjp(i9) <==> At9 ~ 0 <=> (Aj9)( 1 4- mZ) =04 nmZ
<=>• An 4 nmZ = 04 nmZ
<=> An G nmZ
<=> A G mZ,
so Annz(tf) — rnZ. By the first part of the question we deduce that Mor^(Z/mZ, Z/nmZ) ~ Z/Ann^(«9) = Z/mZ.
4-1S Define & : An В —» A x В by &{x) = (x, z), and тг:АхВ~»А4ВЬу ir(z, y) = x — y. Clearly, «9 is a monomorphism, and тг is an epimorphism since тг(а, — b) = a-t-b. Now Im «9 = {(z, z) | z G А П B} and
Kerjr = {(z,y) | zGA, у G B,x — y = 0} = {(z, z) | z G А П B}, so the sequence
0---->A П B~A x B- r >A + В-------->0
is exact.
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Book 6	Rings, fields and modules
4.14	(a) By commutativity we have
a — 7га о q!2 = 03 о 5г2 о i3 = g о 0 = 0.
Similarly, /3=0.
(b) Since Tri о i2 = a = 0 we have Imta Q Kerjri = Im и so there is a unique Я-morphism к : A2 —* A\ such that о к = 1%. It follows that ii о к о f = t2 о f = ц whence, it being injective (and hence left cancellable), к о f = id^'.
Since ТГ2 о = p = 0 we have similarly a unique Я-morphism к' : Aj —* A2 such that t,2 о к1 = tj. Then кок' = id*'.
Now t2 о fc' о Л = о fc = z2 so, t2 being left cancellable, к1 о к — id^». Thus we see that к is an isomorphism with A-1 = kf. Since ко f — id a1 we deduce that f = k~l, whence f is an isomorphism. Similarly, so is g-
4-15 The first part is a special case of question 4.10, but we include the details.
(a)=>(b) : If /* о Ьл = f then since Ker Ijx = Л we have
(Va G A) f(a) ~	= A(0) = 0
so АСКег/.
(b)=>(a) : If A C Ker / then we have
e4-A — s/4-A=>e-0GAC Ker f => f(x) = f(y)
so we can define a mapping /* : M/A —» N by setting /*(e 4- A) = /(к). Clearly, /* is an Я-morphism, and /* о 1ц = f.
That /* is unique follows from the fact that 1ц is surjective and hence right cancellable.
Now /* is a monomorphism if and only if
f(e) = 0 => e4-A = 04~A—A
i.e. if and only if Ker f C A, whence the result follows since the reverse inclusion holds by hypothesis.
Consider the sequence
0------.M/(A П B)—^—>M/A x MIB—^~->MI(A + B)---------->0.
Let p be given by P(x + A,y + В) = e — у + A + B. Then P is an Я-epimorphism; for example, P(z 4- A,0 4- B) — z 4- A 4- B.
64
Solutions to Chapter 4
As for a, consider first the mapping f : M —» M/А x M/B given by /(ж) = (x + A,x + B). This is an .R-morphism with Ker / = А П В so, by the first part of the question, there is a (unique) monomorphism a : Af/(A П B) -> M/A x M/B such that а о |]лпе = /.
To establish exactness, we must show that Im a = Ker 0. Now
Im ex = {(ж 4- A, x 4- B) j x€M}\
and
Ker/3 = {(ж 4- A, у 4- В) | x — yeA + B}.
But we have that
ж — у G A 4- В <=> (За G A)(36 E В) x — у — a + b
<=> (За G A)(36 EB'jx-a —y + b = z say
<=> (3z G M) x 4- A = z 4- A, y+ В ~ z + B.
Hence we see that Ima = Ker 0.
For the last part, take M — A 4- В to get the sequence
0----> (A + B)/(A П B)—»(A 4- B)/A x (A 4- B)/B-------> 0----* 0
the exactness of which shows that a is an isomorphism.
4.16	Suppose that the top row is exact. Then
gf of* oa — •yogo f = 7 о 0 = 0
so,	a being right cancellable, д' о /' = 0. It now follows that Im f* C Ker д'.
To obtain the reverse inclusion, let b' G Ker</. Since 0 is an isomorphism, there is a unique b G В such that b' — /5(6). Then
O = s'(t') = </[/3(6)1 = *(!>)]
so,	7 being an isomorphism, p(6) — 0 whence b G Kerg = Im/ so b = /(a) for some a G A. Then
6' = /3(6) = /?[/(“)] = /'!“(“)! e Im/'.
Thus the bottom row is exact.
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Book 6	Rings, fields and modules
4.17	gootof' — go foa' = Oca* = 0 so Kerg' == Im/1 C Kerg°a. Since д' is surjective it follows that there is a unique .R-morphism a" : C' —» C such that й"о / = jo a. Similarly, g"оfi© / — 0 so Ker g = Im f C Ker g” © (3 and there is a unique R-morphism /3" : C —» C" such that PHog = g"°P-To show that the thus completed bottom row is exact, we have to show that
(1)	is injective;
(2)	/3" is surjective;
(3)	Imo" — Ker/3".
This we do by the usual process of ‘diagram chasing’.
(1)	Let c' G Ker a". Since д' is surjective there exists b' G B' with g*(b') = c". Then
о = а"(г) = ""1»'(б')1 = 9[«(И)
so a(fe') G Ker g = Im f so a(fc') = /(a) for some a G A. Since then
O = P[«(t')]=W(a)] = rm)
we have /?'(o) G Ker/" = {0} so a G Ker/?' = Imo', say a = a’(a*). Consequently
a(fc’) = /(a) = /{a'(a')] = a[/'(a')L
giving b‘ — f'(a') since a is manic. It now follows that
c' = /(&') = /[/'(«')] = 0
and so Kero" = {0} as required.
(2)	Since fi"о g = g" о/3 and since gn,/3 are each surjective we see that /3" © g is surjective, whence so also is /3".
(3)	/3" © о" о д' — g" о (3 о о = д" о 0 = 0 so, д' being surjective (and hence right cancellable), /?" © a” = 0 whence Im о" C Ker/3".
To obtain the reverse inclusion, let c G Ker/3". Then, since g is surjective, c = g(b) for some b G В so
о = p“(c) = p"[g(t>)] = <nw\
whence /?((.) e Ker = Im/”. Thus /3(1;) = /"(a") for some a" e A" and, since (3' is surjective, a" = /Зг(а) for some a G A. Thus
p(b) = /"№)] = /’№)!
ее
Solutions to Chapter 4
giving b — f(a) G Ker/? = Imo. Thus, for some b* G B*, we have b — f(a) = ct(b') and it follows that
C = g(b) = 9|/(a) + a(t')] =	+ p[a((>'))
= д[л(&')] since g о f — 0
= «"У(И]
G Imo"
so Ker/?" C Ima".
4.18	Clearly, = Rx.xn C Rxn so we have the descending chain
R D Rx D Rx2 D • • • D Rxn D Rxn+l Э • ••
of submodules of R.
Consider the mapping f : R —» Rxn//tan+l given by the prescription /(r) = тхп 4- Rxn+l. It is clear that f is an Л-epimorphism. Now
Kerf = {re R | rxn G Лзс"+1}
= {reR | (3t G R) rxn = txn+l}
= {r G R I (3t G R) (r - tx)xn = 0}
- {r G R | (St G R) r - lx = 0}
— Rxt
the penultimate equality resulting from the fact that x / 0 and R has no zero divisors. It now follows by the first isomorphism theorem that
Rxn/Rxn+1 = Im/ ~ R/ Ker / = R/Rx.
4-19	(a) The 2-module 2 satisfies the ascending chain condition but not
the descending chain condition. For, every submodule of 2 is finitely generated (in fact by a singleton); and an infinite descending chain of submodules is
22 D 42 D 82 D 162 D ....
(b)	The 2-module 2m satisfies both chain conditions since it is finite.
(c)	This is similar to (b).
(d)	The Q-module Q is a vector space of dimension 1 over Q. Any submodule of this is therefore a subspace and so is either {0} or Q. Thus both chain conditions are satisfied.
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Book 6
Rings, fields and modules
(e)	As a Z-moduIe, © satisfies neither chain condition. To see this, let the submodule generated by f G © be given by (t) — {nt | n G Z}. Then	*
(2) Э (4) D (8) D (16) D  
is an infinite descending chain of submodules, and (i)c(|)c(|)c(^)c--
is an infinite ascending chain of submodules.
(f)	As a ©-module, ©[X] satisfies neither chain condition. To see this, let (t) — f© be the submodule generated by t G ©[X]. Then
(X,X2,X3,X4, ...)d (X2,X3,X4, ...) D(X3,X4, ...)□•
is an infinite descending chain of submodules, and
(X) c (X, X2) c (X, X2, Xs) c (X, X2, X3, X4) c  • -
is an infinite ascending chain of submodules.
(g)	As a ©[X]-module, ©[X] satisfies the ascending chain condition but not the descending chain condition. In fact
(X)d(X2)d(X3)d(X4)d...
is an infinite descending chain of submodules, and every submodule is finitely generated (since a submodule of ©[X] as a ©[X]-module is an ideal which, by the division algorithm, is generated by the monic polynomial of least degree that it contains).
(h)	By the correspondence theorem the submodules of ©[X]/M are of the form T/M where T is a submodule of ©[X] that contains M. But the only such submodules T are of the form (/(X)) where /(X) divides X6, and there are finitely many of these submodules. Hence ©[X]/M as a ©[X]-module satisfies both chain conditions.
(i)	As in (h).
4-20 Every tower of submodules can be refined to a Jordan-Holder tower. In particular, the tower {0} C N С M can be so refined.
4.21 By the previous question there is a Jordan-Holder tower that ‘passes through’ N, say
M = M D M D   D Mn A D Mn+J D • • • D Mn+t = 0
68
Solutions to Chapter 4
which is of height n + t. Since
M/N = M0/JV D MJN D - Э Mn/N = N/N = 0 is then a Jordan-Holder tower (of height n) for M/N and
TV = Mi D Mi+i D • • • Э Mn+t = 0
is a Jordan-Holder tower (of height t) for TV, we deduce that h(M) = n + t^ h(M/N) + h(N).
For the last part observe that
N = M <=> M/TV = 0 <=> h(M/N) = 0 <--> h(M) = h(N).
4-22 Since M is of finite height and Ker f is a submodule of M it follows that Ker f is of finite height. Since Im/ ~ M/l&ex f and since M/Kerf is also of finite height (for quotient modules inherit chain conditions), it follows that Im / is of finite height. It now follows by the previous question that
Л(1т/) + Л(Кег/) = Л(М/Кег/) + A(Ker/) = h(M).
4-23 The proof is by induction. If n = 1 then the sequence
o--->m--->o
can only be exact if M = 0, in which case the result is trivial. When n = 2 the exactness of
0-----------------------> 0
requires / to be an isomorphism, in which case ft(Mi) — Л(М)« Suppose then that n > 2 and that the result holds for all exact sequences of the given type of length less than or equal to n — 1. Consider the diagram
0—*	—	—
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Book 6
Rings, fields and modules
in which the row is exact, К = Ker = Im /n_2, тг is the epimorphism induced by fn-2, and i is the canonical inclusion.
The sequence
0-----*	----*-------♦ Mn_2—К----------* 0
is exact and so, by the inductive hypothesis,
n-2
+ (-1)’-‘A(K) = 0.
fc=l
Also, the sequence
0----->k--------------------------.0
is exact and so, again by the inductive hypothesis,
-h(K) +	- h(Mn) = 0.
It follows from this that
(-1)"- *h(K) = (-1)’>-1[Л(М,_1) - Л(М„)] = 0.
Substituting for (—1)П-1Л(К) in the previous equation now gives the result for n and completes the induction.
4-%4 Consider M = Z2 © Z2 and N = Z3 © 7L3. Clearly, h(M) — h(N) = 2. Also, every element of M is of order 2. If now £ Morz(M, N) then
(Vz GM) 0 = 0(0) = 0(z + z) = 0(z) + 0(z).
Since N has no elements of order 2 we deduce that 0(z) = 0 for all z G M whence 0 = 0.
4-85 It is readily seen that MnE, is the submodule of Mn consisting of those matrices of the form
0	...	0	an	0	...	0
0	...	0	ац	0	. .	0
0	...	0	a3i	0	...	0
0	...	0	ani	0	...	0
70
Solutions to Chapter 4
and that Bi is the submodule of Mn consisting of those matrices of the form
an Ojs ...	0 ... 0
О21	e22	•••	fl2t	0	- 0
031	a32	• • •	a3i	0	- 0
onj	aft2	...	oni-	0	...	0
We thus have the tower
Mn = Bn D Bn^ Э dB1DB0 = {0}.
4.26
Now there are no submodules A such that Bi-i С A C Bi and so, by the correspondence theorem, Bi/Bi~i is simple. This is therefore a Jordan-Holder tower.
For the first part it suffices to prove that
for i = 1,..., n. This follows from the fact that Ni C Mi for every i and M = ф*1-! Mi, so that
Ni П Nj C JU П = J#»
Consider now the mapping
Й : M = Ф	—* Ф M‘/N‘
i=l	1=1
given by the prescription
t?(?n1>...,7nn) — (mt +	+ Nn).
Clearly, 6 is an R-morphism with
Ker 6 = {(mi,..., mn) | тп,- E TV,} — JVf — N.
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Book 6
Rings, fields and modules
Then, by the first isomorphism theorem,
ф Mt/Ni = Im Й ~ Ml Ker0 = M/N. i-1
4-27 Suppose that M — ф"=1 Л» and let pr£ : M —> and in, : Mi —♦ Af be the ith projection and injection respectively, i.e.
pr£(ai,...,on) = ait ini(aj) = (0,...,0,a,-,0,
Then it is clear that рг,- and in,- satisfy (1) and (2).
Conversely, suppose that /,- : M —» Л,- and р, : А,- —» M satisfy (1) and (2). Let X be any Я-module and for each t G I let hi: Ai —* X be an Я-morphism. Consider the diagram
M
Define h : M —* X by
Л=52А*°«>‘-
For every i we have
ho fi = 52 hj о g3- о R =? hi.
If, moreover, t: M —> X is such that t о /,- = hi for every « then
t = t O idw (= t ° 52 A ° & = 52* ° A ° & = 52h3 ° 9з = h-
Thus M is the direct sum of Ai,..., An-
We have that Af = Afj © M?, if and only if there exist Я-morphisms
M: J1~ *Af. *a 'Af.
72
Solutions to Chapter 4
such that
(a)	91 ° A = «Ur,!	(b) 91 °/2= °;
(c)	92 о h = idlH,;	(d) 92 0 /1 = °;
(e)	A ' 91 + /2 ° 92 = idM .
Now if these conditions are satisfied then the sequence
0----------------------------------> 0
is exact; for ft is injective by (a), g2 is surjective by (c), Im ft C Ker g2 by (d), and by (e) if g2(x) = 0 then ® = (ft о <7ц)(зс) so that Kerg2 C Im ft. Moreover, the sequence splits by (a) and (c).
4-28 Note that R is free with basis {1}. If
0-------,x------.a—----->R------> 0
is exact then f is surjective so there exists a G A such that /(a) = 1. Choosing any such a G A define g R АЪу g[l) — a and extend this to the whole of R by linearity. Then
(A» 9>(1) = /(<*) = 1
so f о g and id/? coincide on the basis {!}. Hence f о g = id/? and g is a splitting morphism.
The sequence
0------>Z —Z------------!—> Z/2Z-----► 0
is exact, but cannot	be split	since,	as can be seen	in a similar	manner
to part (d) of question 4.11, we have
Mor2(Z/2Z,Z)=0.
4-29 To say that Ker f is a direct summand of M and that Ini / is a direct summand of N is equivalent to saying that in the following diagram (in which »2 о b о tji is the canonical decomposition of /) both short exact sequences split:
0	0
Kerf—М/ Ker ^Im	——>K/ hnf
 I ......... 1
о	0
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Book 6
Rings, fields and modules
Let g : N —* M be given by
g — Pi ° b~l ° ^2-
Then we have
f ° 9 ° f ~ «2 ° 6 ° tli ° Pi ° b~* о 7Г2 о i2 о Ь ° (h
id	id
= ®2 ° b о b~l о b о tj]
= «2 О 6 О |J1
= f,
and so f is regular.
Conversely, suppose that f is regular. Then there exists g : N —» M such that f°g°f — f- Using the canonical decomposition f = 130601]! this translates into
i2 о b о th о g о i2 о b о th = i2 0 6 0 hi •
Using the fact that «2 is left cancellable, that t|i is right cancellable, and that b is cancellable on both sides, we deduce from this that, on the one hand,
fcoth ojot2 =idIm/
and, on the other,
hi о goi2 ob = idM/Ker/-
Thus 7f2 = b о hi о g and pi — g a i2 о b are splitting morphisms for the above short exact sequences and the result follows.
4-SO If ж G Ker ji П Kerj2 then ji(ar) = 0 == j2(x) and, by the exactness, x G Imiinlmi2 so that x = ii(mi) — «2(^2)- Then, using juxtaposition to denote composites for convenience, we have
0 = Ji(z) =	= *i(m2)
from which we deduce, since hi is an isomorphism, that m2 — 0. Consequently,
x =	= 2'2(0) — 0
and so Kerj! П Ker/2 = {0}-
74
Solutions to Chapter 4
If we define £ =	then
Ji(®) = лй k^ji^x) 4- 3\i2 kf^x) = ji(x) о	k,
Ь(®) = Ja«i k^lj2(x) + 32*2 ki^jxtx} = j2(x) kz	0
and hence £ — x £ Ker 3'1 пКег/з* It now follows from the first part that £= x.
The fact that x — x can be translated into
+ iafcrb'i = idM .
Thus we see that M — Imij 4- Imi2. Since also
Imii П Im i2 = Kerji n Ker/2 = {0}>
it follows that M = Im«i ® Im is-
For the last part we observe that
h\kylli -rhsk^1^ = ;с®2^Г1Я«о 4-joh^a‘ja^o
= Уо(»2ЛГХЛ 4-— Jo ° >dw °«c — Jo’o = 0.
4-31 We have that
(^a)acl £ bn f <=> (Va £ /)(3m« £ Ma) fa(ma) - na
<=> (Va e Z) na e Im fo [ua)iei £ ^)Im/a;
»€/
(ma)a6/ e Ker/ <=> (Va e 1) /«(ma) — 0
<=> (Va e I) ma e Ker/a
<=4> (ma)a€I e Ker fa, a<=I
so that Im f = ®a€/ Im/a and Ker / = фле/ Ker fa.
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Book 6
Rings, fields and modules
If every sequence La—>MU—^—>Na is exact then for every a € I we have Impa = Ker fa and hence
Imp = фТтра = фКегД = Ker/. aEI	atf
Thus the sequence
«€/	a£I	aEI
is also exact.
Conversely, suppose that the direct sum sequence is exact. Then we have the commutative diagrams (one for each a)
La '°  Ma ----» Na
ф L a     >®Ma---♦ ф Na
f	я
in which the bottom row is exact and 2*1,12,1*3 are the canonical injections. Now g о f = 0 gives g о f о =0 and so, by commutativity,
° 9a ° fa = 0. Since г'з is monic, and so left cancellable, we deduce that ga о fa = 0 and hence that Im/„ C Kerpa.
In order to establish the reverse inclusion, given ma 6 Ma let denote the element of	whose а-component is ma and all other
components are 0. Then we have
ma & Ker ga => ma e Ker g => ma e Im /
(RH square commutative) (bottom row exact)
=> ma (E Im fa (LH square commutative)
and so Ker ga C Im fa-
Thus we see that the top row is also exact.
j.S2 Note first that, since R is commutative, Могд(М, N) is an R-module. Also, Lj ф 0 since it clearly contains the zero morphism from M to TV. To show that L3- is an jR-module, it therefore suffices to show that it is a submodule of Могд(М, N)- and for this it suffices to prove that if
76
Solutions to Chapter 4
f,g G L3 then f + g G L3, and if f G Lj, A G R then Xf G by. For the first of these, observe that
Ker f П Kerg С Кег(/ + g), since /(sc) = 0 — g(x) implies (/ + g)(sc) = f(x) + g(x) — 0; and for the second, observe that
Ker /С Ker Xf, since /(sc) = 0 implies (А/)(ж) = A/(sc) = AO = 0.
As for the stated isomorphism, note that by applying the first isomorphism theorem to pry : M = ф”=1 At -+ A3 we obtain
Aj ~ М/ ф,уу Ai.
Let •Oj : Aj —» М/ ф,уу Ai be the Д-isomorphism given by
0y(%) = ay + ф,/3 Ai.
If / e £y then ф,#3 Л; C Ker / so there is a unique Д-morphism /* : М/ ф,у у Л,- —» JV such that /* ° 1] — / in the diagram
MI®wAt—
A
Then clearly /* о 0j G Могя(Лу,2У). Consider the mapping described by / •—> /* ° $y- This is injective, since &3 is an isomorphism and /* is unique; and it is an Д-morphism, since the uniqueness of /* gives (/ + £?)* — /* + £?* and (A/)* = A/*. It remains, therefore, to show that this map is surjective. Now given any h G Могд(Ау, N) we have h о pry G Мог/г(М, N). Since clearly
(h о pr3)- (	Л,) = 0,
we have he pr? g Lj. Moreover, for all ay G Aj we have
|(h о pry). O t»y](ay) = (h O pry), (ay + ф,уу A,)
= (Л ° pry)(“r)
= h(ay)
whence we see that h — (h о pry)* e> i?y.
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Book 6	Rings, fields and modules
£.33 First we show that the sequence is semi-exact m the sense that the composite of successive morphisms is zero; this will show that the image of the ‘input’ morphism at each stage is contained in the kernel of the ‘output’ morphism, for
f о g = 0 <=> Im g C Ker f.
We shall find it convenient to mix notation and use juxtaposition to denote composition.
(1)	<pi о	— 0 follows from the given exactness, commu-
tativity, and
a.fc.-ii.7!1,?!-! = M-i7*-i7il‘is!-i =	= °;
/.л.-пйй-! =o.
(2)	о (pi = 0 since
# . !₽<(“.)! = #.(“•(“<)> Mo.)) = Z?[o.(o.)l - ftl/.(o.)l = 0-
(3)	Ы^1д\ о = 0 since
(МГ‘лЧ)(<4Л) = hi^r1 g'JKa'i) - h.'ii' s'/W.) = 0 - h.lpl.S.G.)
= 0.
To show that the sequence is exact at B’it it now suffices to observe that
Ь[ e Ker =* hi'lF* Siffli) ~ ®
=> 7?М(Ю e Kerftf = Im ft
=* 7,”1 sJG!) =
=> »!(*><) = ngitM =
=> b'i - A(M e Ker 9' = Im Д
4 = №) +	= M“I) - fiM.) e Im#.-.
78
Solutions to Chapter 4
As for exactness at A{ ® Bi, first observe that
(aj, M € Ker
=>/a«:-)=№)
=> о = A-(af) = Mi) = ngi(bi)
=> 0 = ffi(bi)
=> bi e Ker gi = Im £
(*) =>	= fi(ai)
=> fi(ai) = A(M = PifiM = f&M
=> а[ — oti(ai) G Ker Д- = ImJ£_j
=> a, — at(at) =	= Af-i7i-i(c»-i) = a»^i-i(ci-i)
=> a'- = a*(a» +
Using (*) and the fact that — 0, we also see that
bi — fi[a,i + Л<-1(с;_1)].
Hence (a'*,6«) G Im y?,.
Finally, for the exactness at Ai, if fl* G Ker<pi then ot- G Kero,- and «* G Ker fa. If «» G Ker fa — Im h^i then we have
a, — At-ifci-i) =
Now if ai G Ker a,- then
о = a,(a,) = a.ftf-tlil'jM-l) =	= AI-x(c!-i)
which gives G KerAJ-j = so
c'i-i =sj-i(4-i)-
Thus we have that
a* G Ker$p, Of =
4'34 Let В = Zx 2 and consider the ring R as an f?-module. It is free, with basis {(1,1)}. Let M = 2x {0}. Then M is a submodule of R. But M is not free, since from
(O,l)(sc,O) =(0,0)
we see that no non-empty subset of M can be linearly independent.
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Book 6	Rings, fields and modules
4.35	If f is a monomorphism then since f is left cancellable we cannot have f о g = 0 with p / 0. Hence f is not a left zero divisor in Endfl(M).
Suppose now that M is free, with basis {m, j i € I}. We obtain the converse by showing that Ker f / {0} implies that f is not a left zero divisor. Suppose then that Ker/ / {0}. Let (п^),е/ be a family of nonzero elements of Ker / and let g : M —> M be the (unique) Д-morphism such that = щ for eveiy i G I. Since {m,- | i G Z} is a basis of M it is clear that
{0} 7^ Imp C Kerf
whence f о g — 0 and / is not a left zero divisor.
4.36	If / is an epimorphism then since / is right cancellable we cannot have go f = 0 with g / 0. Hence / is not a right zero divisor in End/?(Af).
The Z-module Z is free. Multiplication by 2 is in Endz(Z) and is clearly neither a right zero divisor nor an epimorphism.
4.37	Suppose that each P, in the family (Z4)»er is projective. To prove that *s projective, consider the diagram
0 (exact)
Since Р, is projective there exists an Д-morphism hi : Pi —* A such that go hi = f о in,-. By the definition of coproduct, there is a unique Д-morphism h :	Pi —♦ A such that h о in,- = h,. We show that
goh = f whence ф,«е/Pi is projective.
Now g о h о in,- = g о hi = f о in,- and so, since every x E ®,-е/ Л can be written uniquely in the form
* = 52 inJм
jeJ
80
Solutions to Chapter 4
where J is a finite subset of I and p3- e P3 for all j e J, we have
(»> hXz) = (g о h) (	“/(₽/))
S'GJ '
= E(?oA°in.>)W
= Ew-yw
ae J
'/GJ '
= M
whence go h = f.
4-38 If Pi and F2 are projective then (as seen in the previous question) so is Pi Ф P2, We first observe that F(M, TV) 0 since it clearly contains the zero morphism from M to N. Now given f,g G P(M, N) we have to show that f — g e P(M,N). For this purpose, consider the following diagram in which f factors through the projective module Pi and g factors through the projective module P2. Here Ki, тг2 are the projections of Fi © P2 onto Pi, P2 respectively.
It follows from the definition of coproduct that there is a unique R-morphism t? : M —» Pi ©P2 such that щ of? = and 7Г2 о & = fc2. Since then
hi о 7п о $ = hi о ki = f
h2 о тг2 о t? = h2 о k2 = g we see that
f - g — (hiTTi — h27T2) о $
i.e. f — g factors through the projective module Pi © P2. Thus we have that f-ge P(M, N).
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Book 6
Rings, fields and modules
j.39 Since Р’,РМ are projective so also is P1 ® P". The sequence
0--> P1--—> P' © P" —»P"-> 0
where i(p'} = (p',0) and n(p,,plt) — p" is then split exact. If we let j : P' © P" —► P' be a left hand splitting morphism and /? a projective lifting of (3 then the given diagram can be extended to the following :
Consider the mapping 7 : P' © P" —» E given by
7 = faj + /Зтг.
It is clear that 7 is an P-morphism. That it makes the diagram commutative follows from the observations
7г = (faj 4-	= f о ct о j о i -|-/3 о тг о i = fa-
91 — 9(f<*3 +	— gofoaof + go/3on = {3v-
0
4.4O	(1) <=> (2) : If h.c.f.{r, -} — 1 then there exist integers x,y such that
xr + y— — 1 whence, on passing to quotients,
r(x + nZ) -I- ~(y + nZ) — 1 + nZ
and therefore, for all t © Z,
t+ nZ— r(xt + nZ) + ^(yt + nZ).
82
Solutions to Chapter 4
441
Also, we have that
“(sc + nZ) = r{y + nZ) <=> -x - ry £ nZ
n
<	=> — x — ry = an
<	==> -}ry (since -|n) r	r
<	=> (since h.c.f.{~,r} = 1)
<	=> ry e nZ
<	=> r(y + nZ) = 0 + nZ.
Thus if (2) holds we have that
z/nz ~ ”(z/nz} er(z/nz)
and the sequence splits.
Conversely, if the sequence splits then we have the above isomorphism and so there exist x, у G Z such that
г(г + nZ) + —(y + nZ) = 14- nZ.
It follows that rx + ~y — 1 + an for some a gZ. This can be written
.	. П „
rx + {y — ar) — — 1
r
and shows that h.c.f.{r, —} = 1.
(2) <=> (3) : If (2) holds then r(Z/nZ) is a direct summand of the free Z/nZ-module Z/nZ (a basis of which is {14-nZ}); hence (3) holds. Conversely, if (3) holds then, by the definition of projective module, the sequence splits.
Taking n = 6 and r — 3 we have that (2) is satisfied and so it follows that Z/2Z ~ 3(Z/6Z) is a projective Z/6Z-module. However, it cannot be free; for a free Z/6Z-module is a direct sum of copies of Z/6Z and so has at least six elements, whereas Z/2Z has only two.
That Дд.в = {/ € Могд(Х, Y) | /“*(A) С B} is a submodule of the Я-module Могд(Х, У) follows immediately from the observations
(1)	ДА1в 0 since it contains the zero morphism;
(2)	f,g G Да,е => (/ 4- £?)”* (А) С В => f 4- g G &a,b ;
(3)	f G &a,b => А/ G &A,B-
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Book 6
Rings, fields and modules
Suppose now that / G &a,b- Then since /*(Л) С В there is a unique Я-morphism f* : X/A -*Y/B such that the diagram
X/A—^Y/B
is commutative. We can therefore define a mapping
f  &a,b Motr(X/A,Y/B)
by the prescription f(/) = f*. It is readily verified that f is an R-morphism; in fact, by the uniqueness of /* we have (/ + p)* = /* + #* and (A/)* = Af*.
To see that f is surjective, let t G Могд(Х/Л,У/В) and consider the diagram
X/A—^Y/B
Since X is projective there exists an R-morphism g : X —♦ Y such that hsoSf = i°tlAjin fact, 9 is a projective lifting of t о tjA. It follows that <Г*(Л) С В and so $(g) = t.
Now examine Ker f. We have
f e Kerf <=> f* = 0
(VzG X) f(z) + B-O + B
<=> (VxGXjf^jGB
<==> Г(Х) G В
<=> f G &x,b-
Thus, by the first isomorphism theorem,
MorR(X/A,Y/B) = Imf ~ Дл,B/Kerf = &a,b/&x,b-
84
Solutions to Chapter 4
4-4% The diagram
P
1'
X--------->Y--------->Z
a	0
is given to be such that the row is exact and /3 о $ — 0. It follows that
Im $ C Ker /3 — Im a.
Let a+ : X —* hna and : P —* Imo be the Isomorphisms induced by a and Then q+ is an epimorphism and we have the diagram
P
|tf+
X-----»Im a---------->0
in which the row is exact. Since P is projective, there exists f : P —♦ X such that q+ of = #+. Now let i : Im a —» Y be the canonical inclusion. Then we have
at о f = i о a+ о f = i о = &.
To apply induction in the second part of the question, note first that the existence of ki : Pt —* Q\ such that hi о ki = ко a gi follows from the fact that hi is an epimorphism and Pi is projective.
Suppose, by way of induction, that for t = 1,..., n — 1 we have constructed kt : Pt -» Qt such that ht okt = kt_i ogt. Consider the diagram
E> 0"	. E>	. n
rn	1------2
I*"-3
Qn	~ *Qn-1 7 *Qn—2
fin	fin — 1
in which each of the rows is exact. We have
hn-i о о gn) - (hn_i о fcn_i) о gn
= kn-2 0 9n~i ° gn
= кп-г О О
= 0
and so, Pn being projective, the first part of the question yields the existence of an й-morphism kn : Pn —» Qn such that hnokn — kn-iogn. This then completes the inductive argument.
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Book 6
Rings, fields and modules
4.4S In the given diagram it is clear that tj о i о j = 0 and so
Ker 0 = Im/ C Kerlj о i.
Since g is an epimorphism there therefore exists a unique P-morphism
•&-. В Р/Imt о j such that t 01 = i^cg.
To prove that t? is a monomorphism we use the fact that g is surjective.
Now
l?(si(a)| = t>[s(6)] => ft о »)(a) = ft о t)(b)
=> i(a) —1’(6) e Im« о j
=> i(a — b) = «[/(sc)] for some x 6 Ker g
=> a - b — j(x) since i is monk
=> s(“) - glV) = »(“ - 0 = g[j (z)l = 0
and the result follows.
If P is quasi-projective then the last part follows from the above on taking A — P, В = P/N and g — t]/v.
For the converse, consider the diagram
By the first part of the question there exists a unique monomorphism & such that -д^д = |]o». By hypothesis there exists a unique Я-morphism % such that t] о jr = t? © A. Let p e Im тг. Then p = ir(q) so we can chase to get
h(p) = tik(«)l = й1л(«)1 = <>[»(“)] =
giving
p — i(a) 6 Ker t] = Imt о j
whence p G Im i. Thus we have that Im тг C Im i. Since i is monk there therefore exists a unique Я-morphism f : P —* Л such that i о f = jr. Then
tJo^of = l|oiof = t]oTr = tfoh
and, since t? is monic (by the first part of the question), g о f = h.
86
Test paper 1
Time allowed : 3 hours
(Allocate 20 marks for each question)
1	Let R be a ring and let N be the smallest subring of R that contains every nil two-sided ideal of R. Show that TV is a nil two-sided ideal of R.
Suppose that I is a right ideal of R. Prove that I + RI is a two-sided ideal of R. Prove also that, for any integer m > 1,
(I 4- Я/)т C Im + RIm.
Deduce that if I is a nilpotent right ideal of R then I C N.
Does TV contain every nilpotent left ideal of Й?
2	Let R be an integral domain in which it is impossible to find an infinite sequence (an) with the property that Лп+i is a proper factor of an for every n. Prove that Я is a unique factorisation domain if and only if all the irreducible elements of R are prime.
Express 9 + 19i as a product of irreducibles in Z[»J.
Show that in the quadratic domain Z[>/14] the equations
14 == >/14 • >/14 = 2-7
do not violate unique factorisation.
3	Let F be a field and let f be a non-linear irreducible polynomial in F[X]. Show that E ~ F[X]/(f) is an extension of F and that f has a linear factor X — ct in where a — X + (/) G E. Show also that E — F(pt).
Now let F = GF(3) — Z/(3) be the field with three elements and let /(X) = X3-X+1.
(a)	Show that f is irreducible over F.
(b)	How many elements are there in F(a)?
(c)	Determine the inverse of a in F(ct).
(d)	Show that f splits completely into linear factors in F(o)[X] and find these factors.
(e)	Describe the Galois group Gal(F(o),F).
4	Let M and N be Л-modules and f : M -♦ N an Я-morphism. If A, В are submodules of M, N respectively prove that the following statements are equivalent :
(1)	f~(A) С B;
(2)	there is a unique Я-morphism f* ; M/A —♦ N/B such that the diagram
M —» N
M/A—---->N/B
is commutative.
Show further that such an Я-morphism /*, when it exists, is
(a)	a monomorphism if and only if A =
(b)	an epimorphism if and only if N = В + Im f.
Given the diagram M —-—♦ TV -—~—> P of Я-modules and Я-morphisms show that, with appropriate definitions of the morphisms involved, the sequence
0 —» Kerf —> Kergof	Kerg —> 7V/Imf —» P/Tmgof —» P/hag —» 0
is exact.
5	Let Ai,..., Am and Blt.. ,,Bn be Я-modules. For each Я-morphism / : ФГ=1 Ak ez=i define
fii = pr° ° f ° in;1
where prj :	Bfc —> Bj is the canonical epimorphism and inf :
is the canonical monomorphism. If M denotes the set
89
of n X m matrices |0y,-] where each tfy,- : At- —> B3 is an Й-morphism, show that the mapping
f:MorB(®A,©B,
— M
described by f(f) = [fyf] is an abelian group isomorphism, so that f is uniquely determined by the n x m matrix [/j-,-].
Show also that the composite Й-morphism
m	n	p
Ф A	ф By —ф Ck
i=l	7=1	fc=l
is represented by the matrix product (<Zfcj][/>»]-
Hence establish, for every Й-module A, a ring isomorphism
Могк(Лпч,Лт) Matmxm[Mor«(Z,Z)].
90
Test paper 2
Time allowed : 3 hours
(Allocate 20 marks for each question)
1	Let R be a commutative ring with a 1 and let A be an ideal which is maximal in the set of ideals of R that are not finitely generated. Suppose, if possible, that xy G A with x,y $ A. Show that (A, x) and C — {c G R | ex G A} are finitely generated ideals of R. E
(A, x) = (dj + x, a2 + b2x,...,an + bnx)
and C ~ (ci, c2,..., cm) show that
A =	• j Owj cix,c2x,.. .tcmx).
Deduce that A is a prime ideal.
Hence show, using Zorn’s axiom, that if every prime ideal of a commutative ring with a 1 is finitely generated then every ideal is finitely generated.
2	Let w — |(—1 + \/-3) he a primitive cube root of unity. Define
Z[w] = {a -t- bw | a,b G Z).
Show that Z[cu] becomes a euclidean domain on defining a norm by
N(a + fccu) ~ a2 — ab + ft2.
[Hint. The proof is similar to the usual argument which shows that Z[^/n] is a euclidean domain for — 2 < n < 3.]
Prove that Z[\/^3] is a subring of Z[w|. Deduce that a subring of a unique factorisation domain need not be a unique factorisation domain.
3	Let К be a finite extension of the field F. Explain what is meant by saying that К is a normal extension of F.
Is a normal extension of Q? Is Q(\/2) a normal extension of Q?
Find the Galois group of the extension Q(\/2, \/5) of Q. Find the subgroups of this Galois group and determine the corresponding fixed fields.
4	If TV is an Я-module and *8 a family of Я-modules, establish the following isomorphisms of abelian groups (in which denotes cartesian products):
(1) Могд ( фм-,Л j ~ П Mor/?(M-, ЛЭ;
' «€/	' tei
(2) МиЦЛГ.ПмЛ ~ MorB(JV, M.j-' *€/	'	»е/
Deduce that if (M,),er and (IV3)y€j are families of Я-modules then there is an abelian group isomorphism
Могя^фм, пФ п Могд(М,, ЛГу).
' iei	'	(i,y)
g/xj
Establish the abelian group isomorphism
Morz(Z, Z) ~ Z.
If, for every positive integer n, the cartesian product of n copies of Z is denoted by Zn, deduce that, for positive integers n and m,
Morz(Z",Z™) = Znm.
5 If P is a projective Я-module and if the diagram
P
Z
93
is such that the row is exact and /3 о $ = 0, prove that there is an .R-morphism f : P —♦ X such that a о f = &.
An exact sequence of the form
(*)	. —> p„ —p„_,	.
Pl
Fo—>0
is said to split if there exist jR-morphisms gt-: Pt- —♦ P,'+i such that
(1) /i°9o = idp„;
(2) (Vi > 1) gi-ifi + fi+igi - idP1.
Prove by induction that if each Pi is projective then the sequence (*) splits.
94
Test paper 3
Time allowed : 3 hours
(Allocate 20 marks for each question)
1 Explain what is meant by a euclidean domain and prove that every euclidean domain is a principal ideal domain.
Which of the following are euclidean domains? Give a proof, or a counter-example, to justify your assertions.
(a)	Z[X];
(b)	©И;
(c)	Z[V5]i
(d)	Z[i],
By showing that (2,14- \Z—3) is not principal in Z[V^3] prove that Z[\/—3] is not euclidean.
2 Let ao — 2 and for n = 0,1,2,... let an+i be the positive square root of an. Define Kn = Q(an).
(a) Show that Kn C Kn+t.
(b) Show that Kn+i — Kn(<Wi) and that (^»+i : Kn) < 2.
(c) Show by induction that an+i Kn.
(d) Deduce the value of (Kn : Q) for n > 1.
3 Find the irreducible factors in <D[X] of
f(X) = X4 - 2X2 3 - 3.
Show that Q(i,\/3) is a splitting field for f over Q. Show also that : Q) ~ 4 and write down a basis for Q(£, \/3) over Q. Find
the minimum polynomial of i + %/3 over Q. Deduce that Q(t,x/3) = Q(i + V3).
Find the minimum polynomial of i 4- \/3 over
(a)	Q(i);
(Ъ)	в(Л);
(c)	Q(iV3).
4 The diagram of Я-modules and Я-morphisms
0------> A —в —«_ c —--^1)
О	Д	7
0--------A‘------>B*----->C'
Р д'
is given to be commutative with exact rows. If 7 is an isomorphism, establish the exact sequence
0------> A A’® В —-—> B'------------‘—> D
where f ~ h о 7-1 о д' and are given by
6) = y'(a') - Р(Ъ), p(a) = (a(a), /(a)).
5 If (jVfa)ag/ is a family of Я-modules, explain what is meant by a product of this family. If (P, (pa)ae/) and (Q, (qa)aei) are products of (Ma)ae/ prove that there is a unique Я-morphism & : P —» Q such that qa о $ = pa for every a G I.
Let (naG/ (pra)ae/) be the cartesian product of (Ма)ае/. A submodule M of Пае/ is said to be a subdirect product of (Ма)ае/ if, for every a G I, the restriction pr£* : M —» Ma of the canonical projection pra is an Я-epimorphism.
If N is an Я-module and there is a family (/о)ае/ of Я-epimorphisms fa:N—> Ma such that
П Ker/,, = {0},
prove that N is isomorphic to a sub direct product of
Deduce that Z is isomorphic to a subdirect product of (Z/nZ)n>i.
97
Test paper 4
Time allowed : 3 hours
(Allocate 20 marks for each question)
1 Let Ai,..., An be two-sided ideals of a ring R. We say that R is the direct sum of Aj,..., An, and write R =	if
(i)	я = БГ=1Л;
(2)	(i = 1,..., n) A. n Ej¥, Ay = {0}.
If Si,..., Sm are two-sided ideals of a ring R such that
(а)	Я = £^5.-;
(b)	no proper subset of {Si,... ,Sm} generates Й;
(c)	each S, is a ring with a 1 containing no ideals other than {0} and
show that R = Si.
Now suppose that
m	n
R= ф Si = ф Ti i-1	i-l
are two representations of R as direct sums of two-sided ideals. For i = 1,..., m show that SiR = Si and deduce that there is exactly one j G {1,..., n} such that S,Tj — S,. Hence show that Si = T3- and deduce that the direct sum decompositions are essentially identical.
2 Let К be a subfield of C that is the splitting field of a polynomial f E Q[X]. By considering the Galois group of К over Q, show that there are only finitely many distinct subfields of К containing <Q.
Let D = {(Q(C) •• Q) j C € A} be the set of degrees of simple extensions of Q that are contained in K. Explain why D has a greatest
element, d say. If a G К is such that (<Q(a) : Q) = d, show that Q(a) = K.
[flint. If Q(a) / К choose /3 G K\Q(ot) and consider the fields Q(tt +q/3) where q G Q.|
3 Explain what is meant by saying that a finite extension of a field F is normal. Show that if f E F[X] and if К is the splitting field for f over F then К is a normal extension of F.
Let F = GF(p) = Zp and let К = GF(pn), the splitting field for Xp" — X over F. Show that every irreducible polynomial of degree n in F[X] splits completely over K.
Show that if a G К then deg ma divides n, where ma is the minimum polynomial of a over F. Deduce that no irreducible factor of Xp ~ X in F[X] has degree exceeding n.
Find the irreducible factors of X9 — X in GF(3)[X].
4 Prove that an Я-module is noetherian (i.e. satisfies the ascending chain condition on submodules) if and only if every submodule of M is finitely generated.
Let M and N be noetherian Я-modules and let P be an Я-module such that there is a short exact sequence
0----->M—1->P—!_>N-----------.0.
If A is a submodule of P show that А П Im f is finitely generated, say by {xi,..., xr}- Now show that there exist yr,..., yn G A such that
(yij-^yn^i^.^Xr}
generates A. Deduce that P is also noetherian.
Hence show that if P is an Я-module and M is a submodule of P then P is noetherian if and only if M and P/Af are noetherian.
5 Prove that the following statements concerning an Я-module P are equivalent :
(1)	P is projective;
(2)	every exact sequence M--->P----->0 splits;
(3)	P is a direct summand of a free Я-module.
Let An be the ring of lower triangular n x n matrices X = [ачу] over a field F (so that хц = 0 if i < j). Let A, В G An be given respectively by
a =P = J +	= j =
15	(0 otherwise, *3	10 otherwise.
99
If &n ~ {X G Дп I (» = 1,..., n)	= 0} prove that
0--->ДЯВ—^Дп—^->0n--------------->0,
where f is the canonical inclusion and g is given by g(X) — XA, is a split exact sequence of An-modules.
Deduce that 0n is An-projective.
100
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Algebra through practice
A collection of problems in algebra, with solutions
Problem solving is an art that is central to understanding and ability m mathematics. With this senes of books the authors have provided a selection of problems with complete solutions and test papers designed to be used with or instead of standard textbooks on algebra. For the convenience of the reader, a key explaining how the present books may be used in conjunction with some of the major textbooks is included. Each book of problems is divided into chapters that begin with some notes on notation and prerequisites. The majority of the matenal is aimed at the student of average ability but there are some more challenging problems. By working through the books, the student will gain a deeper understanding of the fundamental concepts involved, and practice in the formulation, and so solution, of other algebraic problems. Later books in the series cover matenal at a more advanced level than the earlier titles, although each is, within its own limits, self-contained.
Students in schools, colleges, polytechnics and universities will find these books invaluable as an accompaniment to their courses, and of great assistance m preparing for examinations.
Book 1 Sets, relations and mappings
Book 2 Matrices and vector spaces
Book 3 Groups, rmgs and fields
Book 4 Linear algebra
Book5 Groups
Book 6 Rings, fields and modules
CAMBRIDGE UNIVERSITY PRESS