solutions in groups and rings

Problems and Solutions in

GROUPS & RINGS

William J. DeMeo

November 2, 2010

Abstract

This document contains solutions to some of the problems appearing on comprehensive exams given by theMathematics Department at the University of Hawaii over the past two decades. In solving many of these problems,

I benefited greatly from the wisdom and guidance of Professor William Lampe. A number of other professorsprovided additional expert advice and assistance, for which I am very grateful. In particular, I thank Professor Ron

Brown for leading an enlightening ring theory seminar in the fall of 2009, and Professor Tom Craven for taking overthis job in the spring of 2010.

Some typographical and mathematical errors have surely found their way into the pages that follow. Nonetheless, Ihope this document will be of some use to you as you explore two of the most beautiful areas of mathematics.

Finally, its no secret that mathematics is best learned by solving problems. The purpose of this document, then, is togive you some problems to work on in order to locate any gaps in your understanding. The solutions are there toassist you in filling these gaps, and to provide cautionary notes about typical oversights and points of confusion.

Please send comments, suggestions, and corrections to:[email protected].

Contents1 Group Theory 3

1.1 1993 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 1999 March . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 2002 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 2003 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5 2004 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 2006 April . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.7 2006 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.8 2007 April . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.9 2008 January . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.10 2008 April . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 Ring Theory 222.1 1992 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.2 1995 April . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.3 1995 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.4 1996 April . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.5 1999 March . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1

CONTENTS CONTENTS

2.6 2003 April . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.7 2003 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.8 2004 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.9 2008 January . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.10 2008 April . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.11 2008 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

A Basic Definitions and Theorems 47A.1 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47A.2 Congruence Relations and Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48A.3 Group Theory Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50A.4 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52A.5 Ring Theory Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53A.6 Factorization in Commutative Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54A.7 Rings of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54A.8 Nakayamas Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55A.9 Miscellaneous Results and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

2

1 GROUP THEORY

1 Group Theory

1.1 1993 November1. Prove that there is no non-abelian simple group of order 36.

Solution: Let G be a group of order |G| = 36 = 2232. Then the Sylow theorem implies that G has a subgroup Hof order |H| = 9. Define G/H = {gH : g G}, the set of left cosets of H in G. This is a group if and only ifH P G, but let us assume that H R G (otherwise were done), in which case G/H is merely a set of subsets of G.Note that the set G/H contains exactly [G : H] = |G|/|H| = 4 elements. Denote by Sym(G/H) the groupof permutations of elements of G/H , and consider the map : G Sym(G/H), defined for each g G by(g)(xH) = gxH . The usual trivial calculation shows that this map is a homomorphism of G into Sym(G/H). Inthis case, it is obvious that cannot possibly be a monomorphism (embedding, or faithful representation) of G inSym(G/H) since the order of Sym(G/H) = S4 is 4! = 24, while |G| = 36. Therefore, must have a non-trivialkernel, ker = {g G | (g) = idG/H} = {g G | gxH = xH for all x G}. Thus, (e) 6= ker P G (sincethe kernel of a homomorphism is a normal subgroup), so G is not simple.Remark: A5 (which has order 60) is the smallest non-abelian simple group. uunionsq

2. Prove that for all n > 3, the commutator subgroup of Sn is An.

3. a. State, without proof, the Sylow Theorems.b. Prove that every group of order 255 is cyclic.

Solution:Theorem. [L. Sylow (1872)] Let G be a finite group with |G| = pmr, where m is a non-negative integer and r is apositive integer such that p does not divide r. Then

(i) G has a subgroup of order pm. Such a subgroup is called a Sylow p-subgroup of G.(ii) If H and J are Sylow p-subgroups of G, then J 6 gHg1 for some g G. In particular, the Sylow

p-subgroups of G form a single conjugacy class.(iii) Let np denote the number of Sylow p-subgroups of G and let H be any Sylow p-subgroup of G. Then

np 1 (mod p), np | [G : H], and np = |G : NG(H)|.(Note that [G : H] = r so the second condition says that np divides r.)

b. Let G be a group of order |G| = 255 = 3 5 17. We will show that G must be abelian. Then, the fundamentaltheorem of finitely generated abelian groups gives G = Z3 Z5 Z17 = Z255 (since 3, 5 and 17 are pairwiserelatively prime), whence G is cyclic.For each p {3, 5, 17}, let np denote the number of Sylow p-subgroups. By the Sylow theorems,

n3 1 (mod 3) and n3 | 85 n3 {1, 85},n5 1 (mod 5) and n5 | 51 n5 {1, 51},

n17 1 (mod 17) and n17 | 15 n17 = 1.In particular, the Sylow 17-subgroup is normal.1

1Recall, if P is a Sylow p-subgroup, then so is g1Pg (g G). Thus, if np = 1, then P = g1Pg (g G), so P is normal.

3

1.1 1993 November 1 GROUP THEORY

For each p {3, 5, 17}, let Pp denote an arbitrary Sylow p-subgroup. Note that Pp Pq = (e) for p 6= q {3, 5, 17} since each non-identity element of Pq generates Pq . Suppose both n3 = 85 and n5 = 51. Then, sinceP3 P5 = (e), a simple counting argument (2 85 + 4 51 = 374 > 255) shows that there are too many distinctgroup elements. Therefore, either n3 = 1 or n5 = 1 (or both).

Now note that P17 P G, so G/P17 is a group of order 3 5, hence cyclic.2 Similarly, if P3 P G, then G/P3 is agroup of order 5 17, hence cyclic, while if P5 P G, then G/P5 is a group of order 3 17, hence cyclic.From here there are (at least) two ways to show thatG is abelian and complete the proof. One uses the commutator,and the other uses the following easily proved

Lemma. If H P G and K P G, then H K P G and the natural map : G/(H K) G/H G/K givenby (g(H K)) = (gH, gK) is a monomorphism.(For the proof see Problem 3 of November 2004.)

By the lemma, we can embed G/(H K) (as a subgroup) in the group G/H G/K. Thus, take H to be P17, andK to be either P3 or P5 (whichever is normal). Then H K = (e), so G itself is (isomorphic to) a subgroup of theabelian group G/H G/K, and is therefore abelian.Alternatively, recall that the commutator3 G P G is the unique minimal normal subgroup of G such that G/Gis abelian. (In particular, G is abelian iff G = (e).) Therefore, since G/P17 is abelian, G 6 P17. For the samereason, we have at least one of the following: G 6 P3 or G 6 P5. In any case, G 6 P17 Pq = (e) for someq {3, 5}. Thus, G is abelian.

uunionsq

4. Let G be a group and let X be a set. We say that G acts on X (on the left) if there is a multiplication GX Xsuch that 1x = x and (gh)x = g(hx) for all g, h G and x X .a. Let H be a subgroup of G. Prove that G acts on the space of left cosets G/H .b. Let x X and let H be the subgroup of G fixing x. Define the orbit of x under G and prove that there is aone-to-one correspondence between this orbit and the space of left cosets G/H .In what sense can this one-to-one correspondence be called an isomorphism?

c. Let g G and let x = gx. Let H be the subgroup of G fixing x and let H be the subgroup fixing x. Provethat H and H are conjugate.

Solution: a. Let H be a subgroup of G, and let X = G/H denote the set of left cosets of H in G. Define(g, xH) 7 gxH = (gx)H for each g G and xH X . This is clearly well-defined. (Proof: If (g1, xH) =(g2, yH), then (g1x)H = (g2y)H because (g2y)1(g1x) = y1x H .) Also, this map takes G X into Xand satisfies 1xH = xH . Finally, by associativity of multiplication in G, we have (gh)(xH) = ((gh)x)H =(g(hx))H = g(hx)H = g(hxH). This proves that (g, xH) 7 gxH = (gx)H defines a group action by G on thespace of left cosets G/H .

b. Let x X and let H be the subgroup of G fixing x. (This is called the stabilizer of x, and often denoted byGx.) That is, H := Gx := {g G : gx = x}. The orbit of x under G is the set Gx := {gx : g G}. We provethat there is a one-to-one correspondence between this orbit and the set of left cosets G/H . Indeed, consider themap G 3 g 7 gx Gx, which is clearly surjective, and its kernel is given by the relation

{(a, b) GG : ax = bx} = {(a, b) : b1ax = x} = {(a, b) : b1a H} = {(a, b) : aH = bH}.2Recall, if |G| = pq with p < q, both odd primes, and p does not divide q 1, then G is cyclic.3The commutator, G, is the subgroup generated by the set {aba1b1 : a, b G}.

4


In other words, a and b are in the same leftH-coset iff they take x to the same element of the orbit Gx. Equivalently,there is a one-to-one correspondence between the orbit of x and the set of left cosets G/H .

The bijection is an isomorphism in the following sense: the algebra Gx;G is a transitive G-set, as is the algebraG/H;G. The map : Gx;G G/H;G given by (ax) = aH is a well-defined bijection (by the argumentabove) and it respects the (unary) operations of the algebras; that is, for all operation symbols a G, and allbx Gx, we have (a(bx)) = ((ab)x) = (ab)H = a(bH) = a(bx). Thus, is a (G-set) isomorphism.c. Let g G and let x = gx. Let H be the subgroup of G fixing x and let H be the subgroup fixing x. Then Hand H are conjugate, since

H = {a G : ax = x} = {a G : agx = gx} = {a G : g1agx = x}= {gbg1 G : bx = x} = g{b G : bx = x}g1 = gHg1.

uunionsq

5. Prove that a finite group is nilpotent if and only if it is a product of p-groups.

Solution: (See March 1999 Problem 4.)

5

1.2 1999 March 1 GROUP THEORY

1.2 1999 March1. There is no simple group of order 328.

2. Every finite p-group has a nontrivial center.

3. Suppose G is a group and H is a subgroup of G and N is a normal subgroup of G.

(a) If G is solvable, then H is solvable.(b) If G is solvable, then G/N is solvable.(c) If N and G/N are solvable, then G is solvable

4. Suppose H and K are subgroups of the group G. Then [H,K] is the subgroup generated by {hkh1k1 | h H & k K}. We recursively define Gn by G0 = G and Gn+1 = [Gn, G]. A group is nilpotent iff for some k,Gk = {1}. Use this definition of nilpotent group in what follows. For the rest of the problem, G is assumed to bea finite group.

(a) If P is a Sylow subgroup of G, then N(N(P )) = N(P ), where N(P ) denotes the normalizer of P .(b) If G is a nilpotent group and H is a subgroup of G not equal to G, then N(H) 6= H .(c) If G is nilpotent and P is a Sylow subgroup of G, then P is a normal subgroup of G.(d) If each of the Sylow subgroups of G are normal in G, then G is the direct product of its Sylow subgroups.(e) Every finite, nilpotent group is the direct product of its Sylow subgroups.

Solution: (a) Let P be a Sylow p-subgroup of G. Of course, for any H 6 G, we have H P N(H). Inparticular, N(P ) P N(N(P )). To prove the reverse inclusion, fix g N(N(P )). That is, g G andgN(P )g1 = N(P ). Then, since P 6 N(P ), it follows that gPg1 6 N(P ). Also, P and gPg1 are bothSylow p-subgroups of G, so they are both Sylow p-subgroups of N(P ). But P P N(P ), so P is the uniqueSylow p-subgroup of N(P ). Thus, gPg1 = P . That is, g N(P ), which proves N(N(P )) 6 N(P ), asdesired.

(b) G is nilpotent iff there exists n N such that

G = G0 Q G1 Q Q Gn1 Q Gn = (e).Now Gn = (e) 6 H and, since H is a proper subgroup, G0 = G H , so there is some 0 6 k < n such thatGk+1 6 H and Gk H . Take any x Gk \H . We complete the proof by showing that x N(H). Indeed,x Gk implies that, for any h H ,

xhx1h1 [Gk, G] = Gk+1 6 H.

so xhx1h1 H . Therefore, xhx1 H , for all h H . That is, x N(H).(c) Assume G is nilpotent and P is a Sylow subgroup of G. We show P P G. By (a), N(N(P )) = N(P ), and

so (b) (with H = N(P )) implies that N(P ) cannot be a proper subgroup of G, so it must be all of G. That is,N(P ) = G, which means P P G.

6

1.2 1999 March 1 GROUP THEORY

(d) Let P1, . . . , Pr be the Sylow subgroups of G with orders |Pi| = peii , where p1, . . . , pr are distinct primes ande1, . . . , er are positive integers. Since Pi P G for each 1 6 i 6 r, P1 Pr is a subgroup of G. If a Pi hasorder o(a), then o(a)|peii , by Lagranges theorem. Since p1, . . . , pr are distinct primes, o(a) - pejj , for j 6= i.Therefore, again by Lagranges theorem, a / P1 Pi1Pi+1 Pr. This proves that, for each 1 6 i 6 r,Pi (P1 Pi1Pi+1 Pr) = (e). Finally, P1 Pr < G and |P1 Pr| = pe11 perr = |G| togetherimply that G = P1 Pr. Therefore, by Corollary D.6. of the appendix, G = P1 Pr. uunionsq

7


1.3 2002 November1. Let A be a finite abelian group. Let n be the order of A and let m be the exponent of A (i.e., m is the least positive

integer such that ma = 0 for all a A).a. Show that m | n (i.e., m is a factor of n).b. Show that A is cyclic if and only if m = n.

2. Let G be a finite group. Let H 6 G be a subgroup.a. Show that the number of distinct conjugates of H in G is |G|/|N(H)|, where N(H) is the normalizer of H inG and | | indicates order.b. Show that G =

gG gHg

1 if and only if H = G.

c. Deduce that G is generated by a complete set of representatives of the conjugacy classes of elements of G.

Solution: a. Let G act on Sub[G], the set of subgroups of G, by conjugation. That is, for each g G andK Sub[G], let Kg = gKg1. The orbit and stabilizer of the subgroup H under this action are, respectively,

OH = {Hg : g G} and Stab(H) = {g G : Hg = H} = N(H).

I claim that the index of N(H) in G is [G : N(H)] = |OH |. Indeed, there is a one-to-one correspondence betweenthe left cosets of N(H) and the elements of OH . This is verified as follows: for any x, y G,

xN(H) = yN(H) x1y N(H) x1yH = Hx1y yHy1 = xHx1.

This proves that the map xN(H) 7 xHx1, from the cosets G/N(H) to the orbit OH , is well-defined and one-one. It is clearly onto. Therefore, |OH | = [G : N(H)] = |G|/|N(H)|. uunionsq

b. We must show: G =gG gHg

1 iff H = G.

For every g G, |H| = |gHg1|, so, for any x, y G, |(xHx1) (yHy1)| 6 (|H| 1) + (|H| 1) + 1 (atworst they intersect at the identity). More generally,

gG

gHg1

6 |G|(|H| 1) + 1 (1)But, instead of using |G| as a crude bound on the number of distinct conjugates of H , we can use the result frompart a. That is,

gG

gHg1

6 |G||N(H)| (|H| 1) + 1To finish the proof, we verify that

|G||N(H)| (|H| 1) + 1 6 |G|

with equality if and only if H = G.

8


Heres a more concise proof which can be found, e.g., in Dixon [1] problem 1.9:

Let [G : H] = h and |G| = n. Since H 6 N(H), therefore, by 1.T.3, H has at most h conjugates inG. Since all subgroups have the identity element in common, the number of distinct elements in theseconjugates of H is at most 1 + (|H| 1)h = n h+ 1 < n.

c. Deduce that G is generated by a complete set of representatives of the conjugacy classes of elements of G.

3. Let p, q and r be distinct prime numbers.a. List, up to isomorphism, all abelian groups of order p4.b. Up to isomorphism, how many abelian groups of order p4q4r3 are there?

4. Let G be a nonsolvable group of least order among nonsolvable groups. Show that G is simple.

Solution: LetG be a group satisfying the hypothesis and supposeG is not simple. Then there is a proper nontrivialnormal subgroup N P G. Now |N | < |G| and |G/N | < |G|, so N and G/N are both solvable (by the minimalityhypothesis). But this implies that G is solvable. (By Problem 3 of March 1999, G is solvable iff N and G/N aresolvable). This contradiction completes the proof. uunionsq

5. Let G be a group of order pnm, where p is a prime number, n > 1 and p is not a factor of m. Let r be the numberof Sylow p-subgroups of G and let E = {K1, . . . ,Kr} be the set of Sylow p-subgroups of G. Let N(Ki) be thenormalizer of Ki in G, and let G act on E by conjugation.

a. Show thatri=1N(Ki) is a normal subgroup of G.

b. Show that |ri=1N(Ki)| > pn/(r 1)!c. Deduce that a group of order 48 is not simple.

Solution: a. By definition,N(Ki) = {g G : Kgi = Ki}

where Kgi = gKig1 = g(Ki) are various common notations for the action (on E) of conjugation by G. (The

subgroup N(Ki) is sometimes called the stabilizer of Ki under this action.) Let : G Sym(E) be the mapwhich sends each g G to the action g Sym(E), the group of permutations on the set E. It is trivial to checkthat Hom(G,Sym(E)), with kernel

ker = {g G : Kgi = Ki for all Ki E} =ri=1

N(Ki).

As kernels of homomorphisms are normal subgroups,4ri=1N(Ki) P G. uunionsq

b. By part a. and the first isomorphism theorem,

G/N(Ki) = im 6 Sym(E)

4In fact, the set of all kernels of homomorphisms of a group is precisely the set of all normal subgroups of the group.

9


Therefore, |G|/|N(Ki)| 6 |Sym(E)| = r!. That is,pnm

r!6 |N(Ki)|.

To finish the proof of part b., then, we need only show pn/(r 1)! 6 pnm/r!; equivalently, r 6 m. In otherwords, we must show that the number r of Sylow p-subgroups is at most m. The Sylow theorem implies that rdivides [G : K], for any Sylow p-subgroup K. In the present case |K| = pn, so [G : K] = m. Thus r divides m,so r 6 m, as desired. uunionsq

c. To deduce that a group of order 48 is not simple, suppose |G| = 48 = 243, and, as above, let r denote the numberof Sylow 2-subgroups of G, and denote these subgroups by {K1, . . . ,Kr}. Then the subgroup

ri=1N(Ki)

is normal in G (by a.) and has order at least 24/(r 1)! (by b.). The Sylow theorem implies that r divides[G : Ki] = |G|/|Ki| = 48/24 = 3, so r {1, 3}. In either case, |

N(Ki)| > 24/(r 1)! > 1, which proves

thatN(Ki) is a nontrivial normal subgroup of G, so G is not simple. uunionsq

6. Show that a finite group G is nilpotent if and only if G is isomorphic to the product of its Sylow p-subgroups.

10


1.4 2003 November1. Let H be a finite index subgroup of G. Show that there exists a finite index subgroup K of G such that K H

and K is normal in G.

2. Let G be a group of order 84. Show that G is not simple.

3. Let G be a group and let A be an abelian normal subgroup of G. Show that there is a nontrivial homomorphismfrom G/A to the automorphism group of A.

4. State what it means for a group to be solvable, and show that any group of order 280 is solvable.

5. Prove that a group of order 343 has a nontrivial center.

11


1.5 2004 November1. a.5 Carefully state (without proof) the three Sylow theorems.

b. Prove that every group of order p2q, where p and q are primes with p < q and p does not divide q1, is abelian.c. Prove that every group of order 12 has a normal Sylow subgroup.

2. a. Among finite groups, define nilpotent group, in terms of a particular kind of normal series. State two conditionson G which are equivalent to the condition that G is nilpotent.

b. Prove that the center Z(G) of a nilpotent group is non-trivial.c. Give an example which shows thatN P Gwith bothN andG/N nilpotent is not sufficient forG to be nilpotent.

3. a. IfK,L are normal subgroups ofG prove thatG/KL is isomorphic to a subgroup ofG/KG/L (the externaldirect product). What is the index of this subgroup in G/K G/L, in terms of [G : K], [G : L] and [G : KL]?

b. Prove either direction of: If N P G, then G is solvable if and only if both N and G/N are solvable.Solution: a. Define the natural mapping

G/K L 3 g(K L) 7 (gK, gL) G/K G/L.

We must check that this mapping is well-defined and one-to-one, as follows:

x(K L) = y(K L) x1y K L x1y K and x1y L xK = yK and xL = yL (xK, xL) = (yK, yL).

To be clear, the forward implications () prove well-definedness, the reverse implications () one-to-oneness.Also, it is important to note that these calculations take G/K, G/L, and G/K L to be groups, with multipli-cations given by xKyK = xyK, etc. This requires that the subgroups K and L, and hence K L, be normal.If, for example, K were not normal, then G/K would not be a group.It remains to check that the mapping is a homomorphism. Let denote the mapping, and fix two cosets x(KL)and y(K L) in G/(K L). Then

(x(K L)y(K L)) = (xy(K L)) = (xyK, xyL) = (xKyK, xLyL)= (xK, xL) (yK, yL) = (x(K L))(y(K L)).

By exhibiting a monomorphism : G/KL G/KG/L, we have proved thatG/KL can be embeddedas a subgroup in G/K G/L.The index of G/K L in G/K G/L, in terms of [G : K], [G : L] and [G : K L], is

|(G/K G/L)|/|G/K L| = (|G|/|K|)(|G|/|L|)|G|/|K L| =[G : K][G : L][G : K L] .

uunionsq5cf. problem 3, November 2006, and others.

12


b. (See Problem 3 of March 1999 exam.)

4. a. If H is a subgroup of G and [G : H] = n > 2, prove that there exists a homomorphism from G into Sn, thegroup of all permutations of an n-element set. Show that the kernel of is contained in H , and the image of is a transitive subgroup of Sn.

b. If [G : H] = n andG is simple, thenG is isomorphic to a subgroup ofAn, the subgroup of all even permutations.c. Every group of order 23 32 112 is solvable.

5. Let G be a finite group, and suppose the automorphism group of G, Aut(G), acts transitively on G \ {1}. Thatis, whenever x, y G \ {1} there exists an automorphism Aut(G) such that (x) = y. Prove that G is anelementary abelian p-group for some prime p, by proving that:

a. All non-identity elements of G have order p, for some prime p.b. Here Z(G) 6= 1, and the center of every group is a characteristic subgroup.c. Thus G 6= Z(G), i.e., G is abelian.

6. Suppose N is a normal subgroup of G. CG(N) denotes {g G | g1ng = n for each n N}.a. Prove that CG(N) P G, and if CG(N) = {1} then |G| divides |N |!.b. If N is also cyclic, prove that G/CG(N) is abelian, and hence that G 6 CG(N), where G is the commutator

subgroup of G.

13

1.6 2006 April 1 GROUP THEORY

1.6 2006 AprilInstructions: Do as many problems as you can. You are not expected to do all of the problems. You may use earlierparts of a problem to solve later parts, even if you cannot solve the earlier part; however, complete solutions arepreferred. Most importantly, give careful solutions.

1. Show that there is no simple group of order 992.

Solution: Suppose G is a group of order |G| = 992 = 25 31. Our goal is to show that G contains a normalsubgroup. By the Sylow theorems, the number np of Sylow p-subgroups satisfies6

np 1 (mod p) and np | [G : H]

where H is any Sylow p-subgroup. Therefore, n31 1 (mod 31) and n31 | 32, so n31 {1, 32}. If n31 = 1, thismeans there is a unique Sylow 31-subgroup, which must be normal in G (since conjugates of Sylow p-subgroupsare also Sylow p-subgroups).

Suppose n31 = 32. Now, any element of a group of prime order generates the whole group. Therefore, the Sylow31-subgroups must intersect at the identity. Thus, there are 30 32 = 960 elements of order 31 in G. This leavesonly 32 elements in G that are not of order 31. Also, G must contain a Sylow 2-subgroup of order 25 = 32.Apparently, only one of these can fit inside G, so n2 = 1 (since this results in exactly 992 = 960 + 32 elements). Inthis case, it is the Sylow 2-subgroup that is unique, hence normal in G. uunionsq

2. Let G be a nonabelian simple group. Let Sn be the symmetric group of all permutations on an n-element set, andlet An be the alternating group.

a. Show that if G is a subgroup of Sn, n finite, then G is a subgroup of An.b. LetH be a proper subgroup ofG, and, for g G, let g be the map of the set of left cosets ofH onto themselvesdefined by g(xH) = gxH . Show that the map g 7 g is a monomorphism (injective homomorphism) of G intothe group of permutations of the set of left cosets of H .

c. Let H be a subgroup of G of finite index n and assume n > 1 (so H 6= G). Show that G can be embedded inAn.

d. If G is infinite, it has no proper subgroup of finite index.e. There is no simple group of order 112.

Solution: a. Suppose G is a subgroup of Sn, n finite, and assume G is nonabelian and simple. Clearly G 6= Sn,since Sn is not simple (for example,An P Sn). ConsiderGAn. This is easily seen to be a normal subgroup ofG.(Normality: if G An and g G, then gg1 G and gg1 An, since An P Sn, so gg1 G An.)Therefore, as G is simple, either G An = (e) or G An = G. In fact, G An = (e) never occurs under thegiven hypotheses, and two proofs of this fact are given below. In case GAn = G, of course, G 6 An and we aredone.

Proof 1: Suppose GAn = (e). Then G contains only e and odd permutations. Let G be an odd permutation.Then 2 is an even permutation in G, so it must be e. Thus, every nonidentity element of G has order 2. Suppose is another odd permutation in G. Then is an even permutation in G, so = e. Therefore, = e = , so = . This shows G has only two elements e and . But then G = Z2 is abelian, contradicting our hypothesis.

6The notation a | b means a evenly divides b; i.e. ac = b for some c.

14


Proof 2: If G were not a subgroup of An, then the group GAn 6 Sn would have order larger than |An| = |Sn|/2.By Lagranges theorem, then, it would have order |Sn|. Therefore, GAn = Sn, and, by the second isomorphismtheorem,

G/(G An) = GAn/An = Sn/An.This rules out G An = (e), since that would give G = Sn/An = Z2 (abelian), contradicting the hypothesis. uunionsq

Remark: Although the first proof above is completely elementary, the second is worth noting since it reveals usefulinformation even when we dont assume G is simple and nonabelian. For example (Rose [6], page 77), if G is asubgroup of Sn containing an odd permutation, then we can show that exactly half of the elements of G are evenand half are odd. Indeed, under these conditions we easily get that GAn = Sn (as in the second proof above),and then the second isomorphism theorem implies |G/(G An)| = |Sn/An| = 2. Therefore, |G An| = |G|/2,which says that half the elements of G belong to An, as claimed.

b. Let H G, and let S denote the set of left cosets of G. Let Sym(S) denote the group of permutations of S.Then the map : G Sym(S) defined by (g) = g , where g(xH) = gxH is a monomorphism of Ginto Sym(S). To prove this, we first show that (g) = g is indeed a bijection of S, then we show that is ahomomorphism of G, and finally we show that ker is (e).To see that g is injective, observe

g(xH) = g(yH) gxH = gyH (gy)1gx H y1x H xH = yH.To see that g is surjective, fix xH S, and notice that g(g1xH) = gg1xH = xH .To see that is a homomorphism, note that, for any xH S,

g1g2(xH) = g1g2xH = g1(g2xH) = g1 g2(xH).That is, (g1g2) = (g1)(g2), which proves that is a homomorphism.Finally, note that ker is a normal subgroup of G. Therefore, as G is simple, either ker = (e), or ker = G. But

ker = {g G : x G,g(xH) = xH} = {g G : x G, x1gx H} 6 H(since e G). Therefore, H 6= G, so ker 6= G, so ker = (e). This proves that is a monomorphism. uunionsq

c. Let H be a subgroup of G of finite index n and assume n > 1 (so H 6= G). We must show that G can beembedded into An.

Part b. showed that embeds G into Sym(S), where S is the set of left cosets of H in G. Now [G : H] = |S| = n,which implies that Sym(S) = Sn, the group of permutations of n elements. Therefore, G is isomorphic to asubgroup of Sn. By part a., then, G is isomorphic to a subgroup of An. uunionsq

d. Suppose G is infinite. We must show it has no proper subgroup of finite index. If, on the contrary, H 6 G with[G : H] = n, then, G would be isomorphic to a subgroup of the finite group An. Impossible. uunionsq

e. Suppose, by way of contradiction, that G is a simple subgroup with |G| = 112 = 24 7. Clearly, G isnonabelian. (Proof: |G| is not prime, so G has nontrivial proper subgroups. Also, G is simple, while if G wereabelian, all subgroups would be normal.) Now, the Sylow theorems imply that G has a Sylow 2-subgroup of order24, call it H . Then [G : H] = 7, so, by part c., : G A7. Thus G is (isomorphic to) a subgroup of A7 and|A7| = 7!/2 = 7 5 32 23. By Lagranges theorem, |G| divides |An|. But, 24 7 does not divide 23 32 5 7.This contradiction proves that a group of order 112 cannot be simple. uunionsq

15


3. a. Let be an element of the symmetric group Sn and let (i1i2 . . . ik) be a cycle in Sn. Prove that1(i1i2 . . . ik) =(i1i2 . . . ik). [Note that it is assumed that permutations act on the right, so maps i to i.]

b. Show that A4 is not simple.c. Show that any five-cycle S5 and any two-cycle S5 together generate S5.

4. If A and B are subgroups of a group G, let A B be the smallest subgroup containing both, and let AB = {ab :a A, b B}.a. Show that if A is a normal subgroup then AB = A B and that, if both A and B are normal, then A B is

normal.

b. If A,B and C are normal subgroups of G and C A, prove Dedekinds modular law:

A (B C) = (A B) C

5. Let G be a group and let Z be its center.

a. Show that if G/Z is cyclic then G is abelian.b. Show that any group of order p2, where p is a prime, is abelian.c. Give an example of a non-abelian group G where G/Z is abelian.d. Let be a homomorphism from G onto K, where K is an abelian group. Let N be the kernel of and suppose

N is contained in Z. Suppose that there is an abelian subgroup H of G such that (H) = K. Show G isabelian.

6. Let G be a finite group and let H P G be a normal subgroup. Show that G is solvable if and only if H and G/Hare solvable.

16


1.7 2006 November1. Find (up to isomorphism) all groups with at most fifteen elements.

2. Let G be a finitely generate group.

a. Prove that every proper subgroup H < G is contained in a maximal proper subgroup.b. Show that the intersection of the maximal proper subgroups of G is a normal subgroup.

3.7 a. Carefully state (without proof) the three Sylow theorems.b. Prove that every group of order p2q, where p and q are primes with p < q and p does not divide q 1, is abelian.

4. Prove that the center of a finite abelian p-group is nontrivial.

5. Let G be a group and H a subgroup of G, N a normal subgroup of G. Show that HN is a subgroup of G, thatH N is a normal subgroup of H , and that HN/N = H/H N .

7cf. problem 1, November 2004, and others.

17


1.8 2007 April1. Let H and K be (not necessarily normal) subgroups of a group G. For g an element of G the set HgK is called a

double coset. Show that any two double cosets are either identical or do not intersect.

2. State (without proof) the three Sylow theorems. Show that every group of order 56 contains a proper normalsubgroup.

3. Define G, the commutator subgroup of G. Show that G is a normal subgroup of G. Show that every homomor-phism from G to an abelian group A factors through G/G (i.e., given : G A there exists : G/G A suchthat = pi where pi : G G/G is natural).

4. Short answers.

a. Give examples of groups K,N , and G with K a normal subgroup of N , N a normal subgroup of G, but K nota normal subgroup of G.

b. Give an example of a simple group that is not cyclic.c. Give an example of a solvable group that is not nilpotent.d. Give an example of a group G that is not abelian but G/C(G) is abelian where C(G) denotes the center of G.e. Make a list of abelian groups of order 72 such that every abelian group of order 72 is isomorphic to exactly one

group on your list.

f. Give a solvable series for S4, the symmetric group on four elements.

5. Let H be a finite index subgroup of G. Show there is a normal subgroup of G that is contained in H and has finiteindex in G.

18

1.9 2008 January 1 GROUP THEORY

1.9 2008 JanuaryInstructions: Do as many problems as you can. You may use earlier parts of a problem to solve later parts, even ifyou cannot solve the earlier part; however, complete solutions are preferred. Most importantly, give careful solutions.

1. Prove that if H is a subgroup of a finite group G, then |H| divides |G| and use this to show that the order of eachelement of a finite group divides the order of the group.

2. a. State Sylows theorems.b. Show that there is no simple group of order 56.c. Show that every group of order 35 has a normal subgroup of order 5 and one of order 7.d. Is every group of order 35 abelian?

3. How many nilpotent groups of order 360 are there? Justify your answer. Your justification should indicate clearlywhich theorems you are applying. If you are not able to find the number of nilpotent groups, at least find the numberof Abelian groups of order 360.

4. Show that every group of order 1000 is solvable. You can use the fact that if G has a normal subgroup N such thatboth N and G/N are solvable, then G is solvable.

19


1.10 2008 April1. Prove that S4 is solvable but not nilpotent.

Preliminaries:8 The symmetric group on four letters, S4, contains the following permutations (here we use {0, 1, 2, 3}to denote the four letters that the elements of S4 permute):

permutations type

(01), (02), (03), (12), (13), (23) 2-cycles

(01)(23), (02)(13), (03)(12) product of 2-cycles

(012), (013), (021), (023), (031), (032), (123), (132) 3-cycles

(0123), (0132), (0213), (0231), (0312), (0321) 4-cycles

The set of subgroups of S4, denoted by Sub[S4], contains 30 elements. In particular, there is the alternating group,A4, which has 12 elements and is generated by the 3-cycles in the table above. The alternating group consists of theeven permutations, which are the permutations that can be written as a product of an even number of 2-cycles ortranspositions. For example, (012) = (02)(01), e = (01)(01), are even, while (01) and (0123) = (03)(02)(01)are odd. Thus, the 12 elements of A4 are (from the table above): the eight 3-cycles, the three products of disjoint2-cycles, and the identity.

Another important subgroup is the Klein four group V4. This is the subgroup of A4 generated by the three productsof disjoint 2-cycles; that is, V4 = {e, (01)(23), (02)(13), (03)(12)}. It is useful to note that V4 = Z2 Z2, andthe subgroup lattice Sub[V4] is the five element modular lattice M3. That is, V4 has three subgroups {e, (01)(23)},{e, (02)(13)}, and {e, (03)(12)}, and each pair meets at the identity and joins to V4.

Solution: We know that S4 is solvable simply because we can write down a normal series, (e) P {e, (12)} PV4 P A4 P S4, with abelian factors:

{e, (12)}/(e) = Z2, V4/{e, (12)} = Z2, A4/V4 = Z3, S4/A4 = Z2.In fact, the factors of the subnormal series are cyclic, which is more than we need.9

We now prove that S4 is not nilpotent. Let C0 = (e), C1 = Z(G) (=the center of G), and let C2 = pi11 (Z(G/C1))where pi1 : G G/C1 is the canonical projection epimorphism. That is, C2 is defined to be the (unique) normalsubgroup of G such that C2/C1 = Z(G/C1). Given Cn, define Cn+1 = pi1n (Z(G/Cn)). Then, by definition, Gis nilpotent if and only if there is an n N such that Cn = G. The series (e) < C1 < C2 < < Cn = G iscalled the ascending central series.

We claim that the center of S4 is trivial. That is, Z(S4) = (e). Once we prove this, then it is clear that S4 is notnilpotent. For the ascending central series never gets off the ground in this case, so it has no chance of ever reachingS4.

Recall, if N P G and G/N is cyclic and N Z(G), then G is abelian. Note that |A4| = 12 = 22 3. If thereexists N P A4...Finish it!

8This problem appears, for example, in Hungerford [2], p. 107.9Note that we could have used the series (e) P V4 P A4 P S4 to prove that S4 is solvable, and in this case the factor groups are not all cyclic,

since V4/(e) = Z2 Z2 is not one-generated.

20


2. List all abelian groups of order 100. Prove that your list is complete.

Solution: Since 100 = 2252, the (isomorphism classes of) abelian groups of order 100 are

Z22 Z52 = Z100Z22 Z5 Z5 = Z20 Z5Z2 Z2 Z52 = Z2 Z50

Z2 Z2 Z5 Z5 = Z10 Z10.

This list is complete by the fundamental theorem of finitely generated abelian groups.

Elaborate!

3. How man elements of order 7 are there in a group of order 168?

Solution: Let G be a group of order |G| = 168 = 23 3 7. If a G and |a| = 7 then (a) is a subgroup of order 7,and since 7 is the highest power of 7 dividing dividing 168, (a) is a Sylow 7-subgroup of G. We count the numbern7 of Sylow 7-subgroups using the Sylow theorems, which imply

n7 1 (mod 7) and n7 | [G : (a)].

Since [G : (a)] = 24, then n7 {1, 8}. If n7 = 1 , then there are only 6 elements of order 7 in G. If n7 = 8, then,since the Sylow 7-subgroups must intersect at (e), there are 6 8 = 48 elements of order 7...

Finish it!

4. Let G < G be the commutator subgroup of a finite group G. Let Z be the center of G. Let p be a prime number.Suppose that p divides |Z| but does not divide |Z G|. Show that G has a subgroup of index p.

5. Assume G is a finite group, p the smallest prime dividing the order of G, and H a subgroup of index p in G. Provethat H is a normal subgroup of G.

21

2 RING THEORY

2 Ring Theory

2.1 1992 November1.10a. Define injective module.

b. Prove: The module Q is injective if every diagram with exact row

0 // P

// P

Q

with projective P is embeddable in a commutative diagram

0 // P

// P

Q

2. Let R be a commutative ring with identity. Recall that an ideal P R is said to be prime if P 6= R and ab Pimplies a P or b P . Let J R be an ideal.a. Show that J is prime if and only if R/J is an integral domain.b. Show that J is maximal if and only if R/J is a field.

Solution: This is problem 2 of the April 2008 exam. (See section 2.10 below.)

3. Let R be a principal ideal domain. Let J1, J2, . . . be ideals in R such that

J1 J2 Ji Ji+1

Show that there is an integer n > 1 such that Ji = Jn for all i > n.

Solution: This is simply the statement that a PID is Noetherian. It is proved as follows: Let J =Ji and check

that J is an ideal of R. Since R is a PID, J is principal, or one-generated. That is, J = Ra for some a R.Now a Ra = Ji implies a Jn for some n N. Therefore, as Ra is the smallest ideal containing a, we haveRa Jn

Ji = Ra, and similarly for all Ji with i > n. uunionsq

4. Let R be a ring and M a (left) R-module. Suppose M = K L and is an R-endomorphism of M with theproperty that (L) K. Prove that M = K (1M + )(L).

10Part b. of this problem makes no sense to me. If it makes sense to you, please email me! [email protected]

22

2.1 1992 November 2 RING THEORY

Solution: In fact, if M = K L, then every direct sum decomposition M = K J is of this form. That is,11J = (1M )(L), for some HomR(L,K). (The proof of this fact is easy, but it is not needed for thisproblem.)12

Fix m M . Then m = k + l for some k K and l L, and (l) K. Therefore,m = k + l = k (l) + l + (l) = k + (1M + )(l).

This shows that we can write any m M as a sum m = u + v with u K and v (1M + )(L). Thus M =K+(1M+)(L). To show that the sum is direct, supposem belongs toK(1M+)(L). Thenm = k = l+(l),for some k K and l L. Thus, l = k (l) K, which puts l K L = {0}, so (l) = (0) = 0, since is a homomorphism. Therefore, m = l + (l) = 0, and we have shown M = K (1M + )(L). uunionsq

5. Let R be any ring, M,N,K submodules of some R-module such that N 6M . Prove that the sequence

0 M KN K

M

N M +K

N +K 0

is exact where the maps are the natural (obvious) ones.

Solution: The natural maps are

f :M KN K 3 x 7 x+N M/N

andg : M/N 3 x+N 7 x+ (N +K) M +K

N +K.

It is easy to check that these are well-defined homomorphisms. To see that f is injective, consider

ker(f) = {x+ (N K)|x M K and x+N = 0} = {x+ (N K)|x N K} = {0},where the first 0 denotesN (the zero ofM/N ), while the second 0 denotesN K (the zero of MKNK ). To see that gis surjective, consider im(g) = {x+(N+K)|x M} = M/(N+K), and note that x+(N+K) = x+y+(N+K)for any y K, so M/(N +K) = (M +K)/(N +K).It remains to show that im(f) = ker(g), and, as we see below, this turns out to be equivalent to

The Modular Law13 If N,M,K are submodules of some module, and N 6M , then

M (N +K) = N + (M K). (2)This law is easy to prove by checking both inclusions; that is, consider an element of each side of (2) and show itis contained in the other side.

Finally, observe,

ker(g) = {x+N M/N |x N +K} = {x+N |x M and x N +K}= {x+N |x M (N +K)} = {x+N |x N + (M K)} (the modular law)= {x+N |x M K} = im(f).

uunionsq11I write 1M instead of 1M + because the former seems a bit more natural to me, but it makes no real difference.12See Kasch and Mader [4], page 4, or better yet, work out the details over dinner.13In lattice theory, we say that a lattice is modular if it satisfies the modular law: for elements a, b, c of the lattice, if a 6 b, then b (a c) =

a (b c). Not coincidentally, the lattice of submodules of a module satisfies the modular law (where join is + and meet is ). Alas, mathematicalterminology is not always perverse!

23


Remark: The modular law is sometimes called the Dedekind identity. This identity, and its consequence, im(f) =ker(g), is the crux of problem 5. In fact, Kasch and Mader ([4], page 5) call the exact sequence

0 M KN K

M

N M +K

N +K 0

a useful and attractive version of the Dedekind identity.

6. Let F be a field and let R = F (X,Y ] be the polynomial ring over F in indeterminates X,Y . Let , F , andlet : R F be defined by (f(X,Y )) = f(, ) for all polynomials f(X,Y ) R. Show that ker() =(x , y ).

7. Let R be a ring which is generated (as a (left) R-module) by its minimal left ideals (i.e., R is semi-simple).

a. Prove that R is the direct sum of a finite number of minimal left ideals.b. Prove that every simple R-module is isomorphic to a left ideal of R.

Solution: a. Assume the left regular R-module, RR, is semi-simple, by which we mean here that

R =iI

Ai (3)

for some collection {Ai|i I} of minimal left ideals of R. We assume that all rings are unital, so 1 R. Notethat, although the direct sum (3) may be infinite, each r R is a sum of finitely many elements from the set{Ai|i I}. In particular, 1 = ai1 + + ain for some finite collection of indices {i1, . . . , in} I andaij Aij . Therefore,

R = R1 = R(ai1 + + ain) nj=1

Aij iI

Ai = R,

which proves that R is the finite direct sumn

j=1Aij . uunionsq

8. Let R be a commutative ring. Prove that an element of R is nilpotent if and only if it belongs to every prime idealof R.

Solution: If s R is nilpotent, then sn = 0 for some n N. In particular, sn = 0 P for any ideal P of R. IfP is prime, then sn P implies s P .To prove the converse, recall the following useful theorem:14

Theorem: If S is a multiplicative subset of a ring R which is disjoint from an ideal I of R, then there exists an idealP which is maximal in the set of all ideals R disjoint from S and containing I . Furthermore, any such ideal P isprime.

If s R is not nilpotent, consider the set S = {s, s2, s3, . . . }. Clearly this is a multiplicative set which is disjointfrom the ideal (0). The theorem then gives a prime ideal P disjoint from S. In particular, s / P . uunionsq(See also April 08 (2d).)

14This is proved in the appendix; see Theorem A.2.

24

2.2 1995 April 2 RING THEORY

2.2 1995 April1.15 Let R be a commutative ring. Recall that the annihilator of an element x in an R-module is the ideal ann(x) ={s R : sx = 0}. Let Rx and Ry be cyclic R-modules.a. Show that RxR Ry = R(x y).b. Show that R(x y) = R/(ann(x) + ann(y)).c. Show that ann(x y) = ann(x) + ann(y).

2.16 Let R be a ring and suppose that I, J are left ideals of R such that I + J = R. Prove that

R

I J= RI RJ.

3.17 Recall that an ideal I of a ring R is nilpotent if In = 0 for some positive integer n.

a. Show: If I and J are nilpotent ideals, then so is I + J .b. Let R = Q[X,Y ]/(X2, Y 2). Show that R contains a unique largest nilpotent ideal and find it.

4. The injectivity class of an R-module C is defined to be the class Inj(C) consisting of all modules B such thatfor every submodule A of B, every R-homomorphism h : A C extends to a homomorphism h : B C.Prove that the injectivity class of C is closed under epimorphic images; i.e., if B Inj(C) and f : B B is anepimorphism, then B Inj(C).

5.18 Let M and N be finitely generated modules over the polynomial ring Q[x]. Show that M and N are isomorphicif M is isomorphic to a direct summand of N and N is isomorphic to a direct summand of M .

15See also November 99, problem 4.16See also November 95, problem 4.17See also November 97, problem 5.18See also November 97, problem 2.

25


2.3 1995 NovemberInstructions: Ph.D. students should attempt at least three problems, M.A. students should attempt at least two.It goes without saying that all rings have an identity. Integral domains are commutative, other rings are not necessarilycommutative unless so specified.

1. Recall that an element r in an integral domain R is called irreducible if whenever r = ab for a, b R, then eithera or b must have an inverse.19

a. Prove that an element r is irreducible if and only if (r) is maximal among principal ideals; i.e., (r) is not properlycontained in any other principal ideal.

b. Prove that a Noetherian integral domain R is a unique factorization domain if and only if all principal idealsgenerated by irreducible elements are prime.

2. Let R be a unique factorization domain. A polynomial f R[X] is called primitive if f / aR[X] for any properideal a of R.

a. Prove that a polynomial f is primitive if and only if f / pR[X] for any prime ideal p of R.b. Prove that the product of two primitive polynomials is primitive. (This is essentially Gausss Lemma.)c. Show, in outline, how to use Gausss Lemma to prove that the polynomial ring R[X] is a unique factorization

domain.

3. Prove that an integral domain R has the property that every submodule of a free R-module is free if and only if Ris a principal ideal domain.

4. Let I and J be ideals in a commutative ring R such that I + J = R.

a. Prove that IJ = I J .b. Prove that R/IJ = R/I R/J .c. Prove that I J = R IJ .

5. a. Prove that a ring R is left Noetherian if and only if all submodules of finitely generated left R-modules arefinitely generated.

b. Prove that over any ring, a direct summand of a finitely generated module is finitely generated. (I.e., if M Nis finitely generated, then M is finitely generated.)

c. Prove that over any ring R, a module is finitely generated and projective if and only if it is isomorphic to a directsummand of a free R-module with finite rank (i.e., of the form Rn for some finite n).

6. Let M be a module over a commutative ring R. Let p be a prime ideal in R and let Mp be the localization of M atp.

a. Prove that for m M , 0 6= m1 Mp if and only if ann(m) p, where ann(m) = {r R | rm = 0}.b. Prove that an R-module M is non-trivial if and only if Mm 6= 0 for some maximal ideal m of R.

19Actually, thats not quite the right definition we should also assume r is a nonzero nonunit element of R.

26


2.4 1996 April1. Identify the following rings (integral domain, unique factorization domain, principle ideal domain, Dedekind do-

main, Euclidean domain, etc), citing theorems to justify your answers and giving examples to show why it is not ofa more specialized class.

a. F2[x, y]b. F2[x, y]/(y2 + y + 1)c. R[x, y]/(x2 + y2)d. C[x, y]/(x2 + y2)e. Z[

5]

Solution:Lemma:20 If D is an integral domain containing an irreducible element c D, then D[x] is not a PID.Proof: Let c be an irreducible element in D (i.e. c is a nonzero nonunit and c = ab only if a or b is a unit). SupposeD[x] is a PID, and consider the ideal (c, x) of D[x]. Then (d) = (c, x) for some d D[x]. This implies c (d),so c = gd, for some g D[x]. By irreducibility of c, either g is a unit or d is a unit. If d is a unit, then (d) wouldbe all of D[x]. But (d) = (c, x) 6= D[x] (for one thing, 1 / (c, x)), so d is not a unit. Therefore, g must be a unit,so we can write d = g1c. This implies d (c), so (c, x) = (d) = (c). It follows that x (c), and thus x = fcfor some f D[x]. Finally, x is irreducible,21 so f must be a unit, but this puts both f and c, hence x, in D, whichis impossible. uunionsq

Corollary 2.1 If F is a field and n > 2, then F[x1, . . . , xn] is not a PID.

This follows from the lemma above since x1 is irreducible in F[x1, . . . , xn]. (If x1 = ab for some a, b F[x1, . . . , xn], then, arguing by degree, either a F or b F.)

Theorem 2.1 If D is a UFD, then so is the polynomial ring D[x1, . . . , xn].

By corollary 2.1 above, however, D[x1, . . . , xn] need not be a PID when n > 2. (Note that a field is trivially aUFD.)

a. F2[x, y]Since F2 is a field, it is a UFD, so theorem 2.1 implies that F2[x, y] is a UFD. However, as in the corollaryabove, F2[x, y] is not a PID. In particular, (x, y) is a (maximal) ideal in F2[x, y] which is not principal, as thereader may easily verify.

b. F2[x, y]/(y2 + y + 1)Letting D = F2[x], we have F2[x, y]/(y2 + y + 1) = D[y]/(y2 + y + 1). It is easily verified that D is aEuclidean domain with norm (f) = deg(f). Now, if the polynomial g(y) = y2 + y+ 1 is irreducible in D[y],then (g(y)) is a maximal ideal and so D[y]/(g(y)) is a field. To show that g(y) is, indeed, irreducible, supposeg(y) = ab, for some a, b D[y]. Then degx(a) = degx(b) = 0 so a, b F2[y]. If they are to be nonunits, aand b must have degy(a) = degy(b) = 1. The only choices are a = y and b = y for some , F2.That is, g(y) = y2 + y + 1 must have roots in F2, which it does not (g(0) = 1 and g(1) = 3 1). Therefore,g(y) is irreducible in D[y] = F2[x, y]. We have thus shown that E = F2[x, y]/(y2 + y + 1) is a field.

20This is Exercise III.6.1 of Hungerford [2], page 165.21To see that x is irreducible in D[x], argue by degree.

27


c. R[x, y]/(x2 + y2)First note that (x2 +y2) is not a maximal ideal of R[x, y]. For, (x2 +y2) ( (x, y) ( R[x, y]. LettingD = R[x],and p(y) = x2 + y2, we have R[x, y]/(x2 + y2) = D[y]/(p(y)). The roots of p(y) are 1x which do notbelong to D. Thus p(y) is irreducible in D[y]. What else can we say about R[x, y]/(x2 + y2) = D[y]/(p(y))?Finish this one!

d. C[x, y]/(x2 + y2)Let D = C[x] and consider C[x, y]/(x2 + y2) = D[y]/(p(y)), where p(y) = x2 + y2. The roots of p(y) inD[y] = C[x, y] areix, and so p(y) factors inC[x, y] as x2+y2 = (y+ix)(yix). Therefore,C[x, y]/(x2+y2)has zero divisors: let a = y + ix + (x2 + y2) and b = y ix + (x2 + y2). Then ab = (x2 + y2) = 0 C[x, y]/(x2 + y2). Therefore, C[x, y]/(x2 + y2) is not a domain.

e. Z[5]

This is a domain (as a subring of C) which is not a UFD. For example, 9 has a non-unique factorization as aproduct of irreducibles:

9 = 3 3 = (2 +

5) (2

5).

Some useful references for problem 1 are Hungerford [2] (in particular, p. 139 and cor. 6.4, p. 159), and Jacobson [3](p. 141).

2. a. Describe the ring QZ Q.b. Generalize the result of part a. to Q Z A, where A is a divisible abelian group. Note: the group A (written

additively) is called divisible if for any integer n > 2 and any element a A, there exists an element b Asuch that nb = a.

3. a. Show that any commutative ring has a maximal ideal.b. Give an example of a commutative ring with only one maximal ideal.c. A ring as in b. is called a local ring. Show that a commutative ring is local if and only if the set of nonunits

forms an ideal.

Solution: a. Let J be the set of all proper ideals in R and consider the partially ordered set (J ,). Let{A : A } be any chain inJ i.e. A is a totally ordered index set and for all , A , if 6 , thenA A . Define J =

A

A. We verify that J is a proper ideal of R. Of course, as a union of subgroups of

the abelian group reduct R; +,, 0, J is a subgroup. Fix r R and j J . Then there exists A J suchthat j A, so rj A J . Therefore, J is an ideal. It is a proper ideal because otherwise 1R J , whichwould imply that 1R A for some A . But then A = R which contradicts the definition ofJ .

28


Thus, we have J J , and J is an upper bound for {A : A }. By Zorns lemma, J has a maximalelement. uunionsq

b. For an example of a commutative ring with only one maximal ideal, take the set Z4 = {0, 1, 2, 3} and define+ and on Z4 to be addition mod 4 and multiplication mod 4, respectively. Then the ideal generated by 2 is(2) = {0, 2}, and this is the only maximal ideal. For 1 and 3 are units, so (1) = (3) = Z4. (You shouldgeneralize this argument and convince yourself that, if p is prime and n > 1, then Zpn is a ring with uniquemaximal ideal (p).)

c. A ring is called a local ring iff it has a unique maximal ideal. Let R be a commutative ring with 1R. We mustshow that R is local iff the set of nonunits forms an ideal. To do so, we prove the equivalence of the followingthree statements:

(i) R is a local ring.(ii) All nonunits of R are contained in some ideal M 6= R.

(iii) The set of all nonunits of R forms an ideal.

Throughout, we let S denote the set of all nonunits in R. We will prove (ii) (iii) (i) (ii).Suppose (ii) holds so that S M where M is a proper ideal of R. Of course, an ideal is a proper ideal iff itcontains only nonunits. Therefore, S M S, so S = M , which proves (iii).Suppose (iii) holds, and let S J R be a chain of ideals. Then, as above, J = R unless J S, so S containsall proper ideals of R. By (iii), S itself is and ideal, so it is the unique maximal ideal, which proves (i).If (i), then there is a unique maximal ideal M ( R. Fix a S. Now, there is always a maximal ideal of Rwhich contains the ideal (a) (by Zorns lemma). But M is the unique maximal ideal of R. Therefore, (a) M .Since a S was arbitrary, this proves that S M and (ii) holds. uunionsq

4. a. Let R be a commutative ring and M an R-module. Show that HomR(R,M) and M are isomorphic as leftR-modules.

b. It is clear that HomR(R,M) HomZ(R,M). Give an explicit example of a Z-module homomorphism whichis not an R-module homomorphism.

5. Let M be a module over an arbitrary ring R.

a. Define what it means for M to be projective.b. Prove that M is projective if and only if it is a direct summand of a free R-module.

29

2.5 1999 March 2 RING THEORY

2.5 1999 MarchInstructions: Do as many problems as you can. You are not expected to do all of the problems; you should try to havea collection of solutions balanced among the sections. You may use earlier parts of a problem to solve later parts, evenif you cannot solve the earlier part. Most importantly, give careful solutions.

Ph.D. candidates should not do the starred problems.

1. * Suppose D is a commutative ring (not necessarily with 1) having no zero divisors. Give a definite constructionof a field in which D can be embedded along with the embedding. You dont need to prove anything beyond thatthe operations are well defined and that the set of your construction is closed under the operations.

2. a. Every Euclidean domain is a principal ideal domain.b. Every principal ideal domain satisfies the ascending chain condition for ideals.c. Give an example of a unique factorization domain that is not a principal ideal domain.

3. Any finitely generated module over a principal ideal domain is the direct sum of its torsion submodule and a freemodule.

4. Suppose R is a ring.

a. Suppose A is a right R-module and B is a left R-module. Define the Abelian group AR B.b. Show that AR R and A are isomorphic Abelian groups.

5. Let I be an infinite set. Suppose for each i I that Fi is a field.

(Fi : i I) denotes the direct product of thefamily of fields, and

(Fi : i I) denotes the direct sum, and the latter set consists of those sequences belonging

to the direct product that have 0 in all but finitely many places. Show that the ring

(Fi : i I)/

(Fi : i I)has a homomorphic image which is a field. (Any assertion you make about the existence of an ideal should beproved.)

30


2.6 2003 April1. Let R be a ring, I a right ideal and J a left ideal of R. The expression I+J stands for the Abelian group generated

by I J . Prove thatR

IR R

J= RI + J

.

Define your maps carefully with particular attention to well-definedness.

Solution:

Lemma 2.1 Every element of RI R RJ can be written as a simple tensor (1 + I) (r + J), for some r R.

Note that every element can certainly be written as a finite sum of generators, say, (s1 + I) (r1 + J) + +(sn + I) (rn + J). Also, (si + I) (ri + J) = (1 + I) (siri + J), so

ni=1

(si + I) (ri + J) =ni=1

(1 + I) (siri + J) = (1 + I)(

(

siri) + J).

This proves the lemma.

Now define the map : RI RJ RI+J by (s+ I, r+ J) = sr+ (I + J). This is clearly a well-defined mappingfrom one abelian group to another. For, if (s+ I, r + J) = (s + I, r + J), then s s I and r r J , so

sr sr = sr sr + sr sr = s(r r) + (s s)r I + J.Therefore, (s+ I, r + J) = (s + I, r + J) implies (s+ I, r + J) = (s + I, r + J).

A routine calculation shows that the mapping is bilinear, and therefore (by the universal property of tensorproducts) induces a unique homomorphism : RI R RJ RI+J , which is clearly surjective. To show that is alsoinjective, suppose x ker . By lemma 2.1, x = (1 + I) (r + J) for some r R, and

0 = (x) = r + (I + J) r I + J r = i+ j,for some i I and j J . Therefore,

x = (1 + I) (i+ j + J) = (1 + I) (i+ J) = (i+ I) (1 + J) = I (1 + J) = 0.

This shows that ker = 0, so is an isomorphism. uunionsq

2. a. Let A1 and A2 be modules over some ring, K1 A1 and K2 A2 be submodules, and 1 : A1 A2 and2 : A2 A1 and be homomorphisms with the following properties.

i. 1(K1) K2 and 2(K2) K1, andii. (1 21)(A1) K1 and (1 12)(A2) K2.

Show that the maps

f : A1 K2 A2 K1 : f((x1, y2)) = (1(x1) y2,x1 + 21(x1) 2(y2))and

g : A2 K1 A1 K2 : g((x2, y1)) = (2(x2) y1,x2 + 12(x2) 1(y1))are well-defined homomorphisms and that g = f1.

31


b. Let P1 and P2 be projective modules and M a module over some ring R. Suppose that 1 : P1 M and2 : P2 M are epimorphisms. Show that

P1 ker(2) = P2 ker(1).

3. Let Mn(R) be the ring of matrices with coefficients in a ring R (with 1), where n is a natural number > 1. For a(two-sided) ideal I of R, let

Mn(I) = {[aij ] Mn(R) : aij I}.a. Prove that Mn(I) is an ideal of Mn(R).b. Prove that every ideal of Mn(R) is of the form Mn(I) for a suitable ideal I of R.

Solution: The first part is very easy. The second part is proved for n = 2 below (see proposition A.1), and for thegeneral case in Lams book [5].

4. Recall that a module is simple if it is nontrivial and has no proper submodules.

a. Prove that the endomorphism ring of a simple module is a division ring (= skew field).b. Prove that a ring R with identity 1 contains a left ideal I such that R/I is a simple left R-module.c. Let R be a ring with identity 1 and suppose that M = RM is a left R-module with the property that every

submodule is a direct summand (everybody splits). Show that every non-zero submodule K of M contains asimple submodule.

Solution: a. Let M be a simple R-module, and suppose is an endomorphism of M that is not identically zero.We must show that is invertible. It suffices to show that is bijective. M is simple, so {0} and M are theonly submodules. Of course, ker and im are both submodules of M (as is easily verified). Since is not thezero map, it must be the case that im 6= {0} and ker 6= M . Thus, is bijective. uunionsq

b. By the standard Zorns lemma argument, we can find a maximal left ideal I in R. It is easily verified that R/Iis a left R-module. Suppose M is a left R-submodule of R/I . Then, in particular, M is a left ideal of R/I .Recall that the left ideals of R/I are in one-to-one correspondence with the left ideals of R which contain I .Moreover, each left ideal in R/I has the form J/I for some left ideal J in R containing I (see, e.g., Hungerford[2], p. 126). Thus, we have a chain of left ideals I J R with M = J/I . Since I is maximal, either J = I ,in which case M is the zero submodule of R/I , or J = R, in which case M = R/I . Since M was an arbitrarysubmodule of R/I , this proves that R/I is a simple module. uunionsq

c. Let k be a nonzero element of K. Clearly Rk K is also a submodule of M . Therefore, M = Rk N forsome submodule N of M . uunionsq

32


2.7 2003 November1. Let R be a commutative ring, and let N be the set of all nilpotent elements of R. Prove that

a. N is an ideal of R, andb. R/N has no nonzero nilpotent elements.

2. a. Show that Z[1] is a unique factorization domain (UFD).

b. Show that Z[3] is not a UFD.

Solution: a. Z[1] can be embedded in C which is a UFD, so Z[1] is a UFD.

b. Consider (a+3)(a3) = a2 + 3. If we let a = 3, then

(3 +3)(33) = 12 = 22 3,

giving two distinct factorizations of 12 in Z[1]. Therefore, Z[1] is not a UFD. uunionsq

3. Let R be a commutative ring with unit. Show that if R contains an idempotent element e, then there exist ideals S,T of R such that R = S T .

Solution: Let : R R be defined by (r) = er. Then

(r1r2) = er1r2 = r1er2 = r1(r2)

and(r1 + r2) = e(r1 + r2) = er1 + er2 = (r1) + (r2).

That is, is an R-module endomorphism (viewing R as a module over itself). Also, is itself an idempotentelement in the ring of endomorphisms EndRR. That is, = , as is easily verified.Let T = im = {(r) : r R} and let S = ker = {r R : (r) = 0}. Then, for any r R we have

r = r (r) + (r)

with (r) T and, since is idempotent, r (r) S. Thus R = S + T . Now suppose r S T . Then r Timplies r = (r) for some r R. On the other hand, r S implies 0 = (r) = ((r)) = (r) = r, thepenultimate equality by idempotence of . This proves that S T = 0, so R = S T .We prove that S and T are ideals by simply checking the required conditions. Fix s1, s2 S, and r R. First notethat s1 S since is an R-module homomorphism, so (s1) = (1Rs1) 1R(s1), where 1R is the unitof R. It then follows that

(s1 s2) = (s1) (s2) = 0 0 = 0so s1 s2 S, which proves that S is a subgroup of the abelian group reduct R,+,, 0 of R. Finally (rs1) =r(s1) = r 0 = 0, so rs1 S, which proves that S is an ideal.The proof that T is an ideal is similar; we verify that T is a subgroup of the abelian group reduct R,+,, 0, andthen check that rt T whenever r R and t T . uunionsq

33


4. Prove that if R is a commutative ring and I, J are ideals of R, then there is an R-module isomorphism

R/I R/J = R/(I + J).

Solution: This is the special case (in which R is commutative) of problem 1 of April 2003.

5. Let M be a left R-module and x M . Let ann(x) = {r R : rx = 0}.a. Show that ann(x) is a left ideal of R.b. Prove that there is an R-module isomorphism Rx = R/ann(x).

Solution: a. Fix r1, r2 ann(x) and r R. Then (r1 r2)x = r1x r2x = 0 so r1 r2 ann(x) so ann(x)is a subgroup of the abelian group reduct R,+,, 0. Also, rr1x = r0 = 0 so rr1 ann(x) which provesthat ann(x) is a left ideal of R.

b. Define : R Rx by (r) = rx. Then is clearly surjective. Also, (r1r2) = r1r2x = r1(r2), so is anR-module epimorphism. Finally,

ker = {r R : rx = 0} = ann(x).

Therefore, by the first isomorphism theorem R/ann(x) = im = Rx. uunionsq

34


2.8 2004 NovemberNotation: Z denotes the ring of integers and Q denotes the field of rational numbers.

1. Let R be a commutative ring.

a. Show that if R is an integral domain, then the only units in the polynomial ring R[x] are the units of R.b. Give counterexamples when R is not an integral domain,c. Show that R[x] is a principal ideal domain if and only if R is a field.

2. Let f Z[x] be a monic polynomial of degree nwith distinct roots 1, . . . , r, r 6 n. Show that 1+ +r Z.

3. List all the ideals of the quotient ring R[x]/I , where I is the ideal generated by (x5)2(x2 + 1). Identify which ofthe ideals are prime and which are maximal. Does your answer change if the field R is replaced by C? (Explain.)

4. Let M be a left module over a ring R.

a. Suppose that M is finitely generated and that R is commutative and Noetherian. Sketch a proof that M isNoetherian; i.e., M satisfies the ascending chain condition on submodules.

b. Suppose that M is Noetherian and that f : M M is a surjective homomorphism. Show that f is anisomorphism.

5. Let R be a ring with 1 and let M be an R-module. Show that the following are equivalent:

a. There exists a module N such that M N is free.b. Given any surjection : B M , there exists an R-module homomorphism : M B such that is the

identity on M .

c. Given a homomorphism : M B and a surjection pi : A B, there exists a homomorphism : M Asuch that pi = .

6. Let R = Z[i] be the ring of Gaussian integers.

a. Show that any nontrivial ideal must contain some positive integer.b. Find all the units in R.c. If a+ bi is not a unit, show that a2 + b2 > 1.

35

2.9 2008 January 2 RING THEORY

2.9 2008 January1. State a structure theorem for finitely generated modules over a PID, including uniqueness conditions on the direct

summands.

Solution: If A is a finitely generated module over a PID R, then

(i) A is a direct sum of a free submodule F of finite rank and finitely many cyclic torsion modules. The cyclictorsion summands (if any) are of orders r1, . . . , rt where the r1, . . . , rt are (not necessarily distinct) nonzerononunits in R satisfying r1|r2| |rt. The rank of F and the ideals (r1), . . . , (rt) are uniquely determined byA.

(ii) A is a direct sum of a free submodule E of finite rank and a finite number of cyclic torsion modules. Thecyclic torsion summands (if any) are of orders ps11 , . . . , p

skk where the p1, . . . , pk are (not necessarily distinct)

primes inR and the s1, . . . , sk are positive integer. The rank ofE and the ideals (ps11 ), . . . , (pskk ) are uniquely

determined by A (except for the order in which the pi appear).

Symbolically, the situation in part (i) of this theorem is as follows:

A = F Ra1 Rat = F R/(r1) R/(rt),

where (ri) = ann(ai) = {r R : rai = 0}. Note that r1|r2| |rt holds if and only if ann(at) ann(a1),in which case rta = 0 for all a Tor(A). The situation in part (ii) of the theorem can be written symbolically as

A = E R/(ps11 ) R/(pskk ).

2. Suppose that f : M N and g : A B are homomorphisms of right and left R-modules, respectively. Provethat there is a group homomorphism h : M R A N B with h(m a) = f(m) g(a) for all a A andm M .

3. Let A be an (R,S)-bimodule and B be an (R, T )-bimodule and let M = HomR(A,B).

a. Give the actions of S and T on M making it an S-module and a T -module (left or right, as appropriate). Noproofs required.

b. Assuming your module actions from the first part, prove that M is in fact an (S, T ) or (T, S) bimodule(whichever is appropriate).

Solution: a. An action of S on M making M into a left S-module is given as follows: for all s S, f 7 sf ,where (sf)(a) = f(as), for all a A. An action of T on M making M into a right T -module is given asfollows: for all t T , f 7 ft where (ft)(a) = f(a)t, for all a A.

b. That M is an (S, T )-bimodule can be seen by simply checking that it satisfies the definition; i.e., by checkingthat M is both a left S-module and a right T -module. Indeed, for all a A, s1, s2, s S, f1, f2, f M , wehave

((s1 + s2)f)(a) = f(a(s1 + s2)) = f(as1 + as2) (since A is a right S-module)= f(as1) + f(as2) = (s1f)(a) + (s2f)(a) = (s1f + s2f)(a),

(s(f1 + f2))(a) = (f1 + f2)(as) = f1(as) + f2(as) = sf1(a) + sf2(a) = (sf1 + sf2)(a),

36

2.9 2008 January 2 RING THEORY

and (s1(s2f))(a) = (s2f)(as1) = f((as1)s2) = f(a(s1s2)) = ((s1s2)f)(a). The penultimate equality holds,again, because A is a right S-module. This proves that M is a left S-module. The proof that M is a rightT -module is similar, so we leave it to the reader to check that, since B is a right T -module, the following holds:For all a A, t1, t2, t T, f1, f2, f M ,

(f(t1 + t2))(a) = (ft1 + ft2)(a),

((f1 + f2)t)(a) = (f1t)(a) + (f2t)(a),

(f(t1t2))(a) = ((ft1)t2)(a).

It follows that M is an (S, T )-bimodule. uunionsq

4. Give an example (no proof required) or prove that none exists:

a. An injective Z[i, pi]-module. (You may assume that pi is transcendental over Q.)b. A ring which is a UFD but which does not have the ascending chain condition on ideals.c. Two nonzero Q-modules M and N with M Q N = 0.d. A Euclidean domain which is not a PID.

37


2.10 2008 April1. Let R be a commutative ring.

a. Define what it means for R to be Noetherian.b. Prove that if R is Noetherian, the polynomial ring R[x] is Noetherian.

Solution: a. A commutative ring R is Noetherian provided R satisfies the ascending chain condition (ACC) onideals. That is, for any chain I1 I2 of ideals ofR, there is anN N such that In = In+1 for all n > N .(By Zorns lemma, a condition which is equivalent to the ACC is that every nonempty set of ideals contains amaximal element.) More generally, an R-module is Noetherian provided it satisfies the ACC on submodules.

b. This is the Hilbert Basis Theorem, which can be found in many standard references, for example, TheoremVIII.4.9 of Hungerford [2]. The proof in Hungerford uses a fundamental theorem ([2], VIII.1.9) which is worthnoting: a module M is Noetherian if and only if every submodule of M is finitely generated. In particular, acommutative ring R is Noetherian if and only if every ideal of R is finitely generated. Therefore, to show thatR[x] is Noetherian, it suffices to show that every ideal J in R[x] is finitely generated.

2. Let R be a commutative ring with identity. Recall that an ideal P R is prime if P 6= R and ab P a P orb P . Let J R be an ideal.a. Show that J is prime if and only if R/J is an integral domain.b. Show that J is maximal if and only if R/J is a field.c. Prove that if J 6= R, then J is contained in a maximal ideal.d. Prove that an element of R is nilpotent if and only if it belongs to every prime ideal of R.

Solution: a. First, note thatR/J is a commutative ring with multiplication (a+J)(b+J) = ab+J , for a, b R,additive identity 0 = J , and multiplicative identity 1R+J . (That the ring axioms are satisfied is easily verified.)() Suppose J is prime and (a+ J)(b+ J) = ab+ J = 0 = J . Then ab J so either a J and a+ J = 0or b J and b+ J = 0. That is, R/J has no zero divisors.() Suppose R/J is an integral domain. Then, since an integral domain has a (non-zero) multiplicative identity in this case, 1R + J its clear that J 6= R. Now, suppose ab J . Then (a + J)(b + J) = ab + J = 0. Itfollows that either a+J = 0 or b+J = 0, since R/J is a domain. That is, either a J or b J , which provesthat J is prime. uunionsq

b. () Suppose J is maximal and a+J 6= 0 inR/J . Then a / J . Consider the ideal generated by a and J , denoted(a, J). Since J ( (a, J) R and J is maximal, it must be the case that (a, J) = R. Whence, 1R (a, J),so 1R = ab + rj for some j J , r, b R. Therefore, 1R ab J so (a + J)(b + J) = ab + J = 1R + J ,which shows that a+ J is a unit in R/J . Since a+ J was an arbitrary non-zero element, this proves that R/Jis a field.() Suppose R/J is a field. Let J I R be a chain of ideals. Suppose there exists an element a I \ J .Then a + J is non-zero, so there is a b + J R/J such that ab + J = 1R + J . Now, since a I , an ideal,clearly ab I as well. Also, ab 1R J I . Therefore, 1R I , which implies I = R. This proves that J ismaximal. uunionsq

c. LetI be the set of all proper ideals ofR which contain J . Then I , is a partially ordered set. Let {J : } be a chain of subsets of I indexed by (totally ordered by set inclusion).22 Define M = {J : }.

22This simply means that, for any , , either J J or J J. Consequently, we can totally order the index set by declaring 6 if and only if J J .

38


We verify that (1) M is an ideal and (2) M belongs to I : First, a, b M implies a J and b J for some, . Without loss of generality, suppose J J . Then both a and b are in J , so a b J M ,which proves that M is a subgroup of R,+, 0R. If r R, then ra J M , so (1) is proved. To see that Mis proper, note that 1R cannot belong to M =

J. For, if 1R M , then 1R J for some . But each

J is a proper ideal, so this cant happen. (Any ideal which contains a unit is all of R.) We have thus shownthat every chain in the poset I , has an upper bound in the set I . Therefore, by Zorns lemma,23 I has amaximal element. That is, J is contained in a maximal ideal. uunionsq

d. () Let a R be nilpotent and let P R be a prime ideal. Then P is, in particular, a subgroup of R,+, 0R,so 0R P . Let n N be such that an = 0R. Then an P and, recall the following fact (Corollary A.1) forprime ideals in a commutative ring: for b1, b2, , bk R,

if b1b2 bk P then bi P for some i {1, . . . , k}.

Applied in the present case, an P implies a P .() Suppose a is not nilpotent; i.e. for all n N, an 6= 0. Consider the set A = {an : n N}. This isa multiplicative set which is disjoint from the zero ideal (0). By Theorem A.2, then, there is a prime ideal Pwhich is disjoint from A, whence a / P . uunionsq(See also November 92 (6).)

3. Let R be a ring with identity.

a. Suppose that A is a right R-module and B is a left R-module. Define the abelian group AR B.b. Show that Zm Zn = Zd, where d = gcd(m,n).

Solution: a.24 Let F be the free abelian group on the set A B and let K be the subgroup of F generated by allelements of the form

(i) (a1 + a2, b) (a1, b) (a2, b)(ii) (a, b1 + b2) (a, b1) (a, b2)

(iii) (ar, b) (a, rb)where a, a1, a2 A, b, b1, b2 B, and r R. The quotient group F/K is an abelian group called the tensorproduct of A and B, denoted by AR B. That is, we define AR B = F/K.

b. Define the map : Zm Zn Zd by (x, y) = xy (mod d), for x Zm, y Zn. Then it is easy to checkthat is a middle linear map. That is, satisfies, for all x, x1, x2 Zm, y, y1, y2 Zn, and r Z,

(x1 + x2, y) = (x1, y) + (x2, y),

(x, y1 + y2) = (x, y1) + (x, y2),

(xr, y) = (x, ry).

23In proofs involving Zorns lemma, it is often better to use a symbol that signifies an arbitrary index set, like (as opposed to, say, N), as wetypically want the argument to work for uncountable collections of sets. This may seem like a minor point, but examiners are likely to notice thissort of thing.

24In my opinion, Hungerford [2] section IV.5, pp. 207209, gives the clearest, most succinct exposition of the tensor product and its basicproperties, and the solution given here reflects this bias.

39


Therefore, by the universal property, there is a unique (abelian group) homomorphism : ZmZn Zd suchthat = where is the canonical middle linear map, (a, b) 7 a b. In other words, we have the followingcommutative diagram:

Zm Zn

$$III

IIII

II // Zm Zn

zzuuuuuuuuu

Zd

The proof is complete if is actually an isomorphism. Equivalently, there should exist a HomR(Zd,Zm Zn) such that = 1Zd and = 1ZmZn . Indeed, define on Zd to be the mapping x 7 x 1. Then isa group homomorphism. (Proof: (x+ y) = (x+ y) 1 = x 1 + y 1 = (x) + (y) for all x, y Zd.)Also, for all x Zd,

(x) = (x 1) = (x, 1) = (x, 1) = x1 mod d = x.

That is, = 1Zd . Next consider the action of on generators of Zm Zn: for x Zm, y Zn,

(x y) = (x, y) = (xy mod d) = xy mod d 1 = x mod d y mod d = x y. (4)

Since is a homomorphism, the identity (4) extends to finite sums of generators:

(ni=1

xi yi) =ni=1

xi yi.

Whence, = 1ZmZn , as desired. uunionsq

40


2.11 2008 NovemberAll rings are unitary, i.e., have an identity.

1. List the ideals of the rings:

a. Z/12Zb. M2(R) (the ring of 2 2 real matrices)c. Z/12ZM2(R).

Solution: a. An ideal is, in particular, a subgroup of the additive group. Since Z12; +, 0 is cyclic, all subgroupsare cyclic. Thus, the subgroups of Z12; +, 0 are as follows: (0) = {0} (1) = {0, 1, . . . , 11} (2) = {0, 2, 4, 6, 8, 10} (3) = {0, 3, 6, 9} (4) = {0, 4, 8} (6) = {0, 6}

Of course, 5, 7, and 11 are relatively prime to 12, so (1) = (5) = (7) = (11) = Z12. The remaining generators,8, 4, and 9, are also redundant: (8) = (4), (9) = (3), and (10) = (2). It is easy to verify that each of the 6distinct subgroups listed above call them {Gi : 1 6 i 6 6} have the property that, if r Z12 and a Gi,then ra Gi. Thus, each Gi, 1 6 i 6 6, is an ideal of Z12.(Remark: This also shows Z12 is a PID, hence the proper prime ideals are the maximal ideals, (2) and (3).)

b. By the following lemma, since R is a field (and thus has no non-trivial proper ideals), the only ideals of M2(R)are M2(0) and M2(R).

Lemma 2.2 Let R be a commutative ring with 1R 6= 0. Then the ideals of M2(R) are precisely the subsetsM2(J) M2(R), where J is an ideal of R.

Proof: Suppose J is an ideal and A1, A2 M2(J). Then A1 A2 has all elements in J , since J is, in particular, asubgroup. Therefore M2(J) is a subgroup of M2(R); +,0. If M = (mij) M2(R) and A = (aij) M2(J), thenMA = (

P2k=1 mikakj). Clearly, all the elements of MA are in J, since J is an ideal and aij J . Similarly, all the

elements of AM are in J . This proves that M2(J) is an ideal of M2(R) whenever J is an ideal of R.Now, let I M2(R) be an ideal of M2(R), and let J be the set of all a R such that a is an element of some A I .Clearly 0 J, since A I implies 0 = A A I . Let a, b J . Suppose a is the ijth entry of A I and b is theklth entry of B I . Let Mij be a matrix with 1 in the ijth position and 0 elsewhere. Then M1iAMj1 =

a 00 0

I

and M1kBMl1 =b 00 0

I . Therefore,

a b 0

0 0

I , so a b J , whence, J is a subgroup. Fix r R. Then

r 00 0

a 00 0

=

ra 00 0

I , so ra J , and J is an ideal.

c. By the following lemma, and parts a. and b., the ideals of Z/12ZM2(R) are all sets of the form I J , whereI {(0), (1), (2), (3), (4), (6)} and J {M2(0),M2(R)}.

Lemma 2.3 Let R and R rings. Then the ideals of R R have the form I J , where I is an ideal of R andJ is an ideal of R.

41


Proof: Let A R R be an ideal. I will show (i) A = A1 A2, for some A1 R and A2 R, and (ii) A1 and A2are ideals of R and R, respectively.(i) Define

A1 = {a R : (a, a) A for some a R} and A2 = {a R : (a, a) A for some a R}.That is, A1 (resp. A2) is the set of all first (resp. second) coordinates of elements in A. I claim that A = A1 A2.25 Fixa A1 and b A2. We show (a, b) A. Since a A1, there is some a A2 such that (a, a) A. Similarly,(b, b) A, for some b A1. SinceA is an ideal, (1, 0) (a, a) = (a, 0) A, and (0, 1) (b, b) = (0, b) A. Therefore,(a, 0) + (0, b) = (a, b) A, as claimed. This proves A1 A2 A. The reverse inclusion is obvious.(ii) Fix a, b A1. Then, there exist a, b A2 such that (a, a) A and (b, b) A. Now, since A is, in particular, asubgroup of the additive group of R R, we have (a, a) (b, b) = (a b, a b) A. Therefore, a b A1, soA1 is a subgroup of R,+, 0. Fix r R. Then (ra, a) = (ra, 1a) = (r, 1) (a, a) A, since A is an ideal of RR.Therefore, ra A1, which proves that A1 is an ideal of R. The same argument, mutatis mutandis, proves that A2 is anideal of R. This completes the proof of (ii) and establishes the lemma. uunionsq

3. Let R be a commutative ring and let A,B, and C be R-modules with A a submodule of B. Which (if any) of thefollowing conditions guarantee that the natural map AR C B R C is injective?(a) C is free,

(b) C is projective.

Solution: I will prove a slightly more general result (as the increased generality comes at no additional cost).That is, if : A B is an R-module monomorphism, then, under conditions (a) or (b), the natural map 1C :AR C B R C in injective. (To answer the question above, take to be the inclusion map.)We begin with the simplest case, where C = R.

Claim 1: If : A B is an R-module monomorphism, then the natural map 1R : A R R B R R isinjective.

Proof 1: Define A : A R R = A by A(a r) = ar, and B : B R R = B by B(b r) = br. (Seelemma 2.5 below for proof that these are, indeed, isomorphisms.) I claim that the following diagram of R-modulehomomorphisms is commutative:

AR R A A1R

y yB R R B B

Fix a A and r R. By definition, ( 1R)(a r) = (a) 1R(r). Therefore, (B ( 1R))(a r) =B((a) r) = (a)r. Following the diagram in the other direction, (A)(a r) = (ar) = (a)r, since isan R-module homomorphism. Therefore, (B ( 1R))(a r) = (A)(a r). Since a A and r R werechosen arbitrarily, B ( 1R) and A agree on generators, so the diagram is commutative.Now A is an isomorphism, so A and are both injective. Therefore, by lemma 2.7 (below), A is injective.Then, by commutativity of the diagram, B ( 1R) is injective. Finally, lemma 2.9 implies that 1R isinjective, which proves claim 1.

25Although this may seem trivial, and to some extent it is trivial, there is something to verify here, since we could have (a, a) A, (b, b) A,so that (a, b) A1 A2, and yet (a, b) / A. That is, in general, A need not be equal to the product of all first and second coordinates ofelements of A. (Consider, for example, the diagonal {(0, 0), (1, 1), (2, 2)}, which cannot be written as A1 A2.)

42


Next, let C be a free R-module. Then C = iI R, for some index set I . For ease of notation, in the sequel, denotes

iI .

Claim 2: If : A B is anR-module monomorphism, then the natural map 1PR : ARR BRRis injective.

Proof 2: Define :A B by ({ai}) = {(ai)} for each {ai} A. Then is an R-module

monomorphism. This is seen as follows: ({ai}, {ai} A) (r R)

({ai}+{ai}) = ({ai+ai}) = {(ai+ai)} = {(ai)+(ai)} = {(ai)}+{(ai)} = ({ai})+({ai}),

and (r{ai}) = ({rai}) = {(rai)} = {r(ai)} = r{(ai)} = r({ai}).If ({ai}) = {0}

B, then {(ai)} = 0, which implies (ai) = 0 for all i I , and therefore, {ai} = 0

A. Thus, is a monomorphism.

By (the proof of) lemma 2.5, the map :A(AR R) given by ({ai}) = {ai 1R} is an isomorphism.

By (the proof of) lemma 2.6, there is an isomorphism :

(AR R) ARR, defined on generators by

(a r, 0, . . . ) = a (r, 0, . . . ), (0, a r, 0, . . . ) = a (0, r, 0, . . . ), . . . , and

(a1 r1, a2 r2, . . . ) = ((a1 r1, 0, . . . ) + (0, a2 r2, 0, . . . ) + )= (a1 r1, 0, . . . ) + (0, a2 r2, 0, . . . ) + = a1 (r1, 0, . . . ) + a2 (0, r2, 0, . . . ) + .

(The sum has finitely many non-zero terms, since {ai} is non-zero for finitely many indices i I .)Consider the following diagram of R-module homomorphisms:

A (AR R) ARR

y y1PRB

(B R R) B RR(5)

where and are the same maps as their bar-less counterparts, but defined onB and

(BRR), respectively.

I claim that diagram (5) is commutative. Indeed, for any {ai} A, we have26

({ai}) = ({ai 1R}) = ((a1 1R, 0, . . . ) + (0, a2 1R, 0, . . . ) + )= a1 (1R, 0, . . . ) + a2 (0, 1R, 0, . . . ) + .

( 1PR) ({ai}) = ( 1PR)(a1 (1R, 0, . . . ) + a2 (0, 1R, 0, . . . ) + )= ( 1PR)(a1 (1R, 0, . . . )) + ( 1PR)(a2 (0, 1R, 0, . . . )) + = (a1) (1R, 0, . . . ) + (a2) (0, 1R, 0, . . . ) + .

In the other direction, ({ai}) = ({(ai)}) = {(ai) 1R}, so

({ai}) = ({(ai) 1R}) = (((a1) 1R, 0, . . . ) + (0, (a2) 1R, 0, . . . ) + )= (a1) (1R, 0, . . . ) + (a2) (0, 1R, 0, . . . ) + ,

26Again, the sums have finitely many non-zero terms, since {ai} is non-zero for finitely many indices i I . The same comment applies to thesums in the following two sets of equations.

43


which proves that and ( 1PR) agree on generators. Therefore, diagram (5) is commutative.Now, and are R-module isomorphisms, and is an R-module monomorphism. Therefore, by lemma 2.7below, is an R-module monomorphism, so commutativity of (5) implies that 1PR is also an R-module monomorphism. Finally, since and are isomorphisms, so is . Whence, 1PR is injective, bylemma 2.8. This which proves claim 2 and completes part (a) of the problem.

Claim 3: If C is a projective R-module and : A B is an R-module monomorphism, th

solutions in groups and rings

Documents

g g gxh

g g bygxh

g symgh

homomorphism of g

group of order g

ifh p g

ker p g sincethe kernel

group theory1 group