Problems and Solutions in GROUPS & RINGSwilliamdemeo/SolutionsInGroupsAndRings.pdf · Problems and...

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, it’s 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 Nakayama’s 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 we’re 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 θ : G → Sym(G/H), defined for each g ∈ G byθ(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 θ 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, θ must have a non-trivialkernel, ker θ = {g ∈ G | θ(g) = idG/H} = {g ∈ G | gxH = xH for all x ∈ G}. Thus, (e) 6= ker θ 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. ut

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 gHg−1 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 g−1Pg (∀g ∈ G). Thus, if np = 1, then P = g−1Pg (∀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 ψ : G/(H ∩K) → G/H × G/K givenby ψ(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/G′

is 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.

ut

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 G×X → 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) = y−1x ∈ 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) ∈ G×G : ax = bx} = {(a, b) : b−1ax = x} = {(a, b) : b−1a ∈ 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 {aba−1b−1 : 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 algebra〈G/H;G〉. The map ϕ : 〈Gx;G〉 → 〈G/H;G〉 given by ϕ(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 ϕ(a(bx)) = ϕ((ab)x) = (ab)H = a(bH) = aϕ(bx). Thus, ϕ 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 : g−1agx = x}= {gbg−1 ∈ G : bx = x} = g{b ∈ G : bx = x}g−1 = gHg−1.

ut

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 {hkh−1k−1 | 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 )g−1 = N(P ). Then, since P 6 N(P ), it follows that gPg−1 6 N(P ). Also, P and gPg−1 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, gPg−1 = 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 Gn−1 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 ,

xhx−1h−1 ∈ [Gk, G] = Gk+1 6 H.

so xhx−1h−1 ∈ H . Therefore, xhx−1 ∈ 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 ), andso (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 and

e1, . . . , 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)|pei

i , by Lagrange’s theorem. Since p1, . . . , pr are distinct primes, o(a) - pej

j , for j 6= i.Therefore, again by Lagrange’s theorem, a /∈ P1 · · ·Pi−1Pi+1 · · ·Pr. This proves that, for each 1 6 i 6 r,Pi ∩ (P1 · · ·Pi−1Pi+1 · · ·Pr) = (e). Finally, P1 · · ·Pr < G and |P1 · · ·Pr| = pe11 · · · per

r = |G| togetherimply that G = P1 · · ·Pr. Therefore, by Corollary D.6. of the appendix, G ∼= P1 × · · · × Pr. ut

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 =⋃g∈G 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 = gKg−1. 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) ⇔ x−1y ∈ N(H) ⇔ x−1yH = Hx−1y ⇔ yHy−1 = xHx−1.

This proves that the map xN(H) 7→ xHx−1, 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)|. ut

b. We must show: G =⋃g∈G gHg

−1 iff H = G.

For every g ∈ G, |H| = |gHg−1|, so, for any x, y ∈ G, |(xHx−1) ∪ (yHy−1)| 6 (|H| − 1) + (|H| − 1) + 1 (atworst they intersect at the identity). More generally,∣∣∣∣∣∣

⋃g∈G

gHg−1

∣∣∣∣∣∣ 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, ∣∣∣∣∣∣

⋃g∈G

gHg−1

∣∣∣∣∣∣ 6 |G||N(H)|

(|H| − 1) + 1

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

(|H| − 1) + 1 6 |G|

with equality if and only if H = G.

8


Here’s 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. ut

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 that⋂ri=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 : Kg

i = Ki}

where Kgi = gKig

−1 = τg(Ki) are various common notations for the action (on E) of conjugation by G. (Thesubgroup N(Ki) is sometimes called the stabilizer of Ki under this action.) Let τ : G → Sym(E) be the mapwhich sends each g ∈ G to the action τg ∈ Sym(E), the group of permutations on the set E. It is trivial to checkthat τ ∈ Hom(G,Sym(E)), with kernel

ker τ = {g ∈ G : Kgi = Ki for all Ki ∈ E} =

r⋂i=1

N(Ki).

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

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

G/⋂N(Ki) ∼= imτ 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. ut

c. To deduce that a group of order 48 is not simple, suppose |G| = 48 = 24·3, 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

that⋂N(Ki) is a nontrivial normal subgroup of G, so G is not simple. ut

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.

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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 q−1, 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/K∩L is isomorphic to a subgroup ofG/K×G/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) ⇔ x−1y ∈ K ∩ L⇔ x−1y ∈ K and x−1y ∈ 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 ϕ denote the mapping, and fix two cosets x(K∩L)and y(K ∩ L) in G/(K ∩ L). Then

ϕ(x(K ∩ L)y(K ∩ L)) = ϕ(xy(K ∩ L)) = (xyK, xyL) = (xKyK, xLyL)

= (xK, xL) (yK, yL) = ϕ(x(K ∩ L))ϕ(y(K ∩ L)).

By exhibiting a monomorphism ϕ : G/K∩L ↪→ G/K×G/L, we have proved thatG/K∩L 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]

.

ut5cf. 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 ρ from G into Sn, thegroup of all permutations of an n-element set. Show that the kernel of ρ is contained in H , and the image of ρis 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 α ∈ Aut(G) such that α(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 | g−1ng = 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. ut

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 λg be the map of the set of left cosets ofH onto themselvesdefined by λg(xH) = gxH . Show that the map g 7→ λg 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). ConsiderG∩An. This is easily seen to be a normal subgroup ofG.(Normality: if σ ∈ G ∩An and g ∈ G, then gσg−1 ∈ G and gσg−1 ∈ An, since An P Sn, so gσg−1 ∈ 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 G∩An = G, of course, G 6 An and we aredone.

Proof 1: Suppose G∩An = (e). Then G contains only e and odd permutations. Let ζ ∈ G be an odd permutation.Then ζ2 is an even permutation in G, so it must be e. Thus, every nonidentity element of G has order 2. Supposeη is another odd permutation in G. Then ζη is an even permutation in G, so ζη = e. Therefore, ζζ = e = ζη, soη = ζ. This shows G has only two elements e and ζ. 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 Lagrange’s 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. ut

Remark: Although the first proof above is completely elementary, the second is worth noting since it reveals usefulinformation even when we don’t 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 λ : G → Sym(S) – defined by λ(g) = λg , where λg(xH) = gxH – is a monomorphism of Ginto Sym(S). To prove this, we first show that λ(g) = λg is indeed a bijection of S, then we show that λ is ahomomorphism of G, and finally we show that kerλ is (e).

To see that λg is injective, observe

λg(xH) = λg(yH) ⇔ gxH = gyH ⇔ (gy)−1gx ∈ H ⇔ y−1x ∈ H ⇔ xH = yH.

To see that λg is surjective, fix xH ∈ S, and notice that λg(g−1xH) = gg−1xH = xH .

To see that λ is a homomorphism, note that, for any xH ∈ S,

λg1g2(xH) = g1g2xH = λg1(g2xH) = λg1 ◦ λg2(xH).

That is, λ(g1g2) = λ(g1)λ(g2), which proves that λ is a homomorphism.

Finally, note that kerλ is a normal subgroup of G. Therefore, as G is simple, either kerλ = (e), or kerλ = G. But

kerλ = {g ∈ G : ∀x ∈ G,λg(xH) = xH} = {g ∈ G : ∀x ∈ G, x−1gx ∈ H} 6 H

(since e ∈ G). Therefore, H 6= G, so kerλ 6= G, so kerλ = (e). This proves that λ is a monomorphism. ut

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 λ 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. ut

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. ut

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., λ : G ↪→ A7. Thus G is (isomorphic to) a subgroup of A7 and|A7| = 7!/2 = 7 · 5 · 32 · 23. By Lagrange’s 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. ut

15


3. a. Letα be an element of the symmetric group Sn and let (i1i2 . . . ik) be a cycle in Sn. Prove thatα−1(i1i2 . . . ik)α =(i1αi2α . . . ikα). [Note that it is assumed that permutations act on the right, so α maps i to iα.]

b. Show that A4 is not simple.

c. Show that any five-cycle σ ∈ S5 and any two-cycle τ ∈ 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 isnormal.

b. If A,B and C are normal subgroups of G and C ⊆ A, prove Dedekind’s 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 ϕ be a homomorphism from G onto K, where K is an abelian group. Let N be the kernel of ϕ and supposeN is contained in Z. Suppose that there is an abelian subgroup H of G such that ϕ(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 ϕ : G→ A there exists ϕ : G/G′ → A suchthat ϕ = ϕ ◦ π where π : 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 onegroup 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 Sylow’s 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 the“even” permutations, which are the permutations that can be written as a product of an even number of 2-cycles or“transpositions.” 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 = π−11 (Z(G/C1))

where π1 : 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 = π−1

n (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 ∼= Z100

Z22 ⊕ Z5 ⊕ Z5∼= Z20 ⊕ Z5

Z2 ⊕ 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 November1.10

a. Define “injective module.”

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

0 // P ′

ϕ

��

// P

Q

with projective P is embeddable in a commutative diagram

0 // P ′

ϕ

��

// 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 have

Ra ⊆ Jn ⊆⋃Ji = Ra, and similarly for all Ji with i > n. ut

4. Let R be a ring and M a (left) R-module. Suppose M = K ⊕ L and ϕ is an R-endomorphism of M with theproperty that ϕ(L) ⊆ K. Prove that M = K ⊕ (1M + ϕ)(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,11

J = (1M − ϕ)(L), for some ϕ ∈ 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 ϕ(l) ∈ K. Therefore,

m = k + l = k − ϕ(l) + l + ϕ(l) = k′ + (1M + ϕ)(l).

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

5. Let R be any ring, M,N,K submodules of some R-module such that N 6 M . 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 M∩KN∩K ). 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 6 M , 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).

ut11I write 1M − ϕ instead of 1M + ϕ 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 α, β ∈ F , andlet ϕ : R → F be defined by ϕ(f(X,Y )) = f(α, β) for all polynomials f(X,Y ) ∈ R. Show that ker(ϕ) =(x− α, y − β).

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 =⊕i∈I

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 and

aij ∈ Aij . Therefore,

R = R1 = R(ai1 + · · ·+ ain) ⊆n⊕j=1

Aij ⊆⊕i∈I

Ai = R,

which proves that R is the finite direct sum⊕n

j=1Aij . ut

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 . ut(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 Rx⊗R 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∼=R

I⊕ R

J.

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 Gauss’s Lemma.)c. Show, in outline, how to use Gauss’s 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, that’s 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 = g−1c. 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. ut

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 φ(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 − α and b = y − β for some α, β ∈ 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 not

belong 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] are±ix, and so p(y) factors inC[x, y] as x2+y2 = (y+ix)(y−ix). 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 Q⊗Z Q.

b. Generalize the result of part a. to Q ⊗Z A, where A is a divisible abelian group. Note: the group A (writtenadditively) 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 nonunitsforms an ideal.

Solution: a. Let J be the set of all proper ideals in R and consider the partially ordered set (J ,⊆). Let{Aα : α ∈ A } be any chain in J – i.e. A is a totally ordered index set and for all α, β ∈ A , if α 6 β, thenAα ⊆ Aβ . Define J =

⋃α∈A

Aα. 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 Aα ∈J suchthat j ∈ Aα, so rj ∈ Aα ⊆ J . Therefore, J is an ideal. It is a proper ideal because otherwise 1R ∈ J , whichwould imply that 1R ∈ Aα for some α ∈ A . But then Aα = R which contradicts the definition of J .

28


Thus, we have J ∈ J , and J is an upper bound for {Aα : α ∈ A }. By Zorn’s lemma, J has a maximalelement. ut

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 Zorn’s 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. ut

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 don’t 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 A⊗R B.

b. Show that A⊗R 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.)

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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

I⊗R

R

J∼=

R

I + J.

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

Solution:

Lemma 2.1 Every element of RI ⊗RRJ 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

n∑i=1

(si + I)⊗ (ri + J) =n∑i=1

(1 + I)⊗ (siri + J) = (1 + I)⊗(

(∑

siri) + J).

This proves the lemma.

Now define the map θ : RI ×RJ →

RI+J by θ(s+ I, r+ J) = sr+ (I + J). This is clearly a well-defined mapping

from 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 − s′r′ = sr − sr′ + sr′ − s′r′ = s(r − r′) + (s− s′)r′ ∈ I + J.

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

A routine calculation shows that the mapping θ is bilinear, and therefore (by the universal property of tensorproducts) induces a unique homomorphism θ : RI ⊗R

RJ →

RI+J , which is clearly surjective. To show that θ is also

injective, suppose x ∈ ker θ. By lemma 2.1, x = (1 + I)⊗ (r + J) for some r ∈ R, and

0 = θ(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 θ = 0, so θ is an isomorphism. ut

2. a. Let A1 and A2 be modules over some ring, K1 ⊆ A1 and K2 ⊆ A2 be submodules, and γ1 : A1 → A2 andγ2 : A2 → A1 and be homomorphisms with the following properties.

i. γ1(K1) ⊆ K2 and γ2(K2) ⊆ K1, andii. (1− γ2γ1)(A1) ⊆ K1 and (1− γ1γ2)(A2) ⊆ K2.

Show that the maps

f : A1 ⊕K2 → A2 ⊕K1 : f((x1, y2)) = (γ1(x1)− y2,−x1 + γ2γ1(x1)− γ2(y2))

andg : A2 ⊕K1 → A1 ⊕K2 : g((x2, y1)) = (γ2(x2)− y1,−x2 + γ1γ2(x2)− γ1(y1))

are well-defined homomorphisms and that g = f−1.

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b. Let P1 and P2 be projective modules and M a module over some ring R. Suppose that β1 : P1 → M andβ2 : P2 →M are epimorphisms. Show that

P1 ⊕ ker(β2) ∼= P2 ⊕ ker(β1).

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 Lam’s 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 everysubmodule 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 ϕ is an endomorphism of M that is not identically zero.We must show that ϕ is invertible. It suffices to show that ϕ is bijective. M is simple, so {0} and M are theonly submodules. Of course, kerϕ and imϕ are both submodules of M (as is easily verified). Since ϕ is not thezero map, it must be the case that imϕ 6= {0} and kerϕ 6= M . Thus, ϕ is bijective. ut

b. By the standard Zorn’s 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. ut

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 . ut

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, and

b. 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)(a−

√−3) = a2 + 3. If we let a = 3, then

(3 +√−3)(3−

√−3) = 12 = 22 · 3,

giving two distinct factorizations of 12 in Z[√−1]. Therefore, Z[

√−1] is not a UFD. ut

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 ϕ : R→ R be defined by ϕ(r) = er. Then

ϕ(r1r2) = er1r2 = r1er2 = r1ϕ(r2)

andϕ(r1 + r2) = e(r1 + r2) = er1 + er2 = ϕ(r1) + ϕ(r2).

That is, ϕ is an R-module endomorphism (viewing R as a module over itself). Also, ϕ is itself an idempotentelement in the ring of endomorphisms EndRR. That is, ϕ ◦ ϕ = ϕ, as is easily verified.

Let T = imϕ = {ϕ(r) : r ∈ R} and let S = kerϕ = {r ∈ R : ϕ(r) = 0}. Then, for any r ∈ R we have

r = r − ϕ(r) + ϕ(r)

with ϕ(r) ∈ T and, since ϕ is idempotent, r− ϕ(r) ∈ S. Thus R = S + T . Now suppose r ∈ S ∩ T . Then r ∈ Timplies r = ϕ(r′) for some r′ ∈ R. On the other hand, r ∈ S implies 0 = ϕ(r) = ϕ(ϕ(r′)) = ϕ(r′) = r, thepenultimate equality by idempotence of ϕ. 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 ϕ is an R-module homomorphism, so ϕ(−s1) = ϕ(−1Rs1) − 1Rϕ(s1), where 1R is the unitof R. It then follows that

ϕ(s1 − s2) = ϕ(s1)− ϕ(s2) = 0− 0 = 0

so s1− s2 ∈ S, which proves that S is a subgroup of the abelian group reduct 〈R,+,−, 0〉 of R. Finally ϕ(rs1) =rϕ(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 . ut

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 ϕ : R→ Rx by ϕ(r) = rx. Then ϕ is clearly surjective. Also, ϕ(r1r2) = r1r2x = r1ϕ(r2), so ϕ is anR-module epimorphism. Finally,

kerϕ = {r ∈ R : rx = 0} = ann(x).

Therefore, by the first isomorphism theorem R/ann(x) ∼= imϕ = Rx. ut

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 α1, . . . , αr, r 6 n. Show that α1+· · ·+αr ∈ Z.

3. List all the ideals of the quotient ring R[x]/I , where I is the ideal generated by (x−5)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 ϕ : B →M , there exists an R-module homomorphism ψ : M → B such that ϕ ◦ ψ is theidentity on M .

c. Given a homomorphism ϕ : M → B and a surjection π : A → B, there exists a homomorphism ψ : M → Asuch that π ◦ ψ = ϕ.

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

sk

k where the p1, . . . , pk are (not necessarily distinct)primes inR and the s1, . . . , sk are positive integer. The rank ofE and the ideals (ps11 ), . . . , (psk

k ) are uniquelydetermined 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/(psk

k ).

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. ut

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

a. An injective Z[i, π]-module. (You may assume that π 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 Zorn’s 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 – it’s 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. ut

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. ut

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

⋃{Jα : α ∈ Λ}.

22This simply means that, for any α, β ∈ Λ, either Jα ⊆ Jβ or Jβ ⊆ Jα. Consequently, we can totally order the index set Λ by declaringα 6 β if and only if Jα ⊆ Jβ .

38


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

⋃α Jα. For, if 1R ∈M , then 1R ∈ Jγ for some γ ∈ Λ. But each

Jγ is a proper ideal, so this can’t 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 Zorn’s lemma,23 I has amaximal element. That is, J is contained in a maximal ideal. ut

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 . ut(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 A⊗R 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 A⊗R B. That is, we define A⊗R B = F/K.

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

ϕ(x1 + x2, y) = ϕ(x1, y) + ϕ(x2, y),

ϕ(x, y1 + y2) = ϕ(x, y1) + ϕ(x, y2),

ϕ(xr, y) = ϕ(x, ry).

23In proofs involving Zorn’s lemma, it is often better to use a symbol that signifies an arbitrary index set, like Λ (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. 207–209, 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 ϕ : Zm⊗Zn → Zd suchthat ϕι = ϕ where ι is the canonical middle linear map, (a, b) 7→ a⊗ b. In other words, we have the followingcommutative diagram:

Zm × Zn

ϕ$$IIIIIIIIIι // Zm ⊗ Zn

ϕzzuuuuuuuuu

Zd

The proof is complete if ϕ is actually an isomorphism. Equivalently, there should exist a ψ ∈ HomR(Zd,Zm ⊗Zn) such that ϕψ = 1Zd

and ψϕ = 1Zm⊗Zn. Indeed, define ψ on Zd to be the mapping x 7→ x⊗ 1. Then ψ is

a group homomorphism. (Proof: ψ(x+ y) = (x+ y)⊗ 1 = x⊗ 1 + y ⊗ 1 = ψ(x) + ψ(y) for all x, y ∈ Zd.)Also, for all x ∈ Zd,

ϕψ(x) = ϕ(x⊗ 1) = ϕι(x, 1) = ϕ(x, 1) = x1 mod d = x.

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

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

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

ψϕ(n∑i=1

xi ⊗ yi) =n∑i=1

xi ⊗ yi.

Whence, ψϕ = 1Zm⊗Zn , as desired. ut

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/12Z×M2(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 the

klth 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/12Z×M2(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 A2

are 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 R×R′.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. ut

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 A⊗R 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 ϕ : A→ B is an R-module monomorphism, then, under conditions (a) or (b), the natural map ϕ⊗ 1C :A⊗R C → B ⊗R C in injective. (To answer the question above, take ϕ to be the inclusion map.)

We begin with the simplest case, where C = R.

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

Proof 1: Define αA : A ⊗R R ∼= A by αA(a ⊗ r) = ar, and αB : B ⊗R R ∼= B by αB(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:

A⊗R RαA−−−−→ A

ϕ⊗1R

y yϕB ⊗R R

αB−−−−→ B

Fix a ∈ A and r ∈ R. By definition, (ϕ ⊗ 1R)(a ⊗ r) = ϕ(a) ⊗ 1R(r). Therefore, (αB (ϕ ⊗ 1R))(a ⊗ r) =αB(ϕ(a)⊗ r) = ϕ(a)r. Following the diagram in the other direction, (ϕαA)(a⊗ r) = ϕ(ar) = ϕ(a)r, since ϕ isan R-module homomorphism. Therefore, (αB (ϕ⊗ 1R))(a⊗ r) = (ϕαA)(a⊗ r). Since a ∈ A and r ∈ R werechosen arbitrarily, αB (ϕ⊗ 1R) and ϕαA agree on generators, so the diagram is commutative.

Now αA is an isomorphism, so αA and ϕ are both injective. Therefore, by lemma 2.7 (below), ϕαA is injective.Then, by commutativity of the diagram, αB (ϕ ⊗ 1R) is injective. Finally, lemma 2.9 implies that ϕ ⊗ 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 ∼=∑i∈I R, for some index set I . For ease of notation, in the sequel,

∑denotes

∑i∈I .

Claim 2: If ϕ : A→ B is anR-module monomorphism, then the natural map ϕ⊗1PR : A⊗R

∑R→ B⊗R

∑R

is injective.

Proof 2: Define Φ :∑A →

∑B by Φ({ai}) = {ϕ(ai)} for each {ai} ∈

∑A. Then Φ is an R-module

monomorphism. This is seen as follows: (∀{ai}, {a′i} ∈∑A) (∀r ∈ R)

Φ({ai}+{a′i}) = Φ({ai+a′i}) = {ϕ(ai+a′i)} = {ϕ(ai)+ϕ(a′i)} = {ϕ(ai)}+{ϕ(a′i)} = Φ({ai})+Φ({a′i}),

and Φ(r{ai}) = Φ({rai}) = {ϕ(rai)} = {rϕ(ai)} = r{ϕ(ai)} = rΦ({ai}).

If Φ({ai}) = {0} ∈∑B, then {ϕ(ai)} = 0, which implies ϕ(ai) = 0 for all i ∈ I , and therefore, {ai} = 0 ∈∑

A. Thus, Φ is a monomorphism.

By (the proof of) lemma 2.5, the map σ :∑A→

∑(A⊗R R) given by σ({ai}) = {ai ⊗ 1R} is an isomorphism.

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

(A⊗R R)→ A⊗R∑R, defined on generators by

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

τ(a1 ⊗ r1, a2 ⊗ r2, . . . ) = τ((a1 ⊗ r1, 0, . . . ) + (0, a2 ⊗ r2, 0, . . . ) + · · · )= τ(a1 ⊗ r1, 0, . . . ) + τ(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

σ−−−−→∑

(A⊗R R) τ−−−−→ A⊗R∑R

Φ

y yϕ⊗1PR∑

Bσ−−−−→

∑(B ⊗R R) τ−−−−→ B ⊗R

∑R

(5)

where σ and τ are the same maps as their bar-less counterparts, but defined on∑B and

∑(B⊗RR), respectively.

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

τ σ({ai}) = τ({ai ⊗ 1R}) = τ((a1 ⊗ 1R, 0, . . . ) + (0, a2 ⊗ 1R, 0, . . . ) + · · · )= a1 ⊗ (1R, 0, . . . ) + a2 ⊗ (0, 1R, 0, . . . ) + · · · .

∴ (ϕ⊗ 1PR) τ σ({ai}) = (ϕ⊗ 1P

R)(a1 ⊗ (1R, 0, . . . ) + a2 ⊗ (0, 1R, 0, . . . ) + · · · )= (ϕ⊗ 1P

R)(a1 ⊗ (1R, 0, . . . )) + (ϕ⊗ 1PR)(a2 ⊗ (0, 1R, 0, . . . )) + · · ·

= ϕ(a1)⊗ (1R, 0, . . . ) + ϕ(a2)⊗ (0, 1R, 0, . . . ) + · · · .

In the other direction, σΦ({ai}) = σ({ϕ(ai)}) = {ϕ(ai)⊗ 1R}, so

τ σΦ({ai}) = τ({ϕ(ai)⊗ 1R}) = τ((ϕ(a1)⊗ 1R, 0, . . . ) + (0, ϕ(a2)⊗ 1R, 0, . . . ) + · · · )= ϕ(a1)⊗ (1R, 0, . . . ) + ϕ(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 τ σΦ and (ϕ⊗ 1PR) τ σ agree on generators. Therefore, diagram (5) is commutative.

Now, σ and τ are R-module isomorphisms, and Φ is an R-module monomorphism. Therefore, by lemma 2.7below, τ σΦ is an R-module monomorphism, so commutativity of (5) implies that ϕ ⊗ 1P

R τ σ is also an R-module monomorphism. Finally, since σ and τ are isomorphisms, so is τ σ. Whence, ϕ ⊗ 1P

R 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 ϕ : A → B is an R-module monomorphism, then the natural mapϕ⊗ 1C : A⊗R C → B ⊗R C is injective.

Proof 3: SinceC is projective, there exists a freeR-module F and a (projective)R-moduleD such that F = C⊕D(lemma 2.4). Therefore,

A⊗R F = A⊗R (C ⊕D) ∼= (A⊗R C)⊕ (A⊗RD) and B⊗R F = B⊗R (C ⊕D) ∼= (B⊗R C)⊕ (B⊗RD),

where the isomorphisms are given by lemma 2.6.

By claim 2 above, the natural map ϕ⊗ 1F : A⊗R F → B ⊗R F is injective. Consider the diagram

A⊗R Cι1−−−−→ (A⊗R C)⊕ (A⊗R D) τ−−−−→ A⊗R (C ⊕D)

ϕ⊗1C

y yϕ⊗1F

B ⊗R Cι1−−−−→ (B ⊗R C)⊕ (B ⊗R D) τ−−−−→ B ⊗R (C ⊕D)

(6)

By (the proof of) lemma 2.6 the map τ : (A⊗R C)⊕ (A⊗R D)→ A⊗R (C ⊕D) given by τ(a1 ⊗ c, a2 ⊗ d) =a1 ⊗ (c, 0) + a2 ⊗ (0, d) is an R-module isomorphism. Given a⊗ c ∈ A⊗R C, then,

(ϕ⊗ 1F ) τ ι1(a⊗ c) = (ϕ⊗ 1F ) τ(a⊗ c, 0) = (ϕ⊗ 1F )(a⊗ (c, 0)) = ϕ(a)⊗ (c, 0).

Similarly, there exists τ : (B⊗RC)⊕(B⊗RD) ∼= B⊗R (C⊕D), with τ(b1⊗c, b2⊗d) = b1⊗(c, 0)+b2⊗(0, d),and

τ ι1 (ϕ⊗ 1C)(a⊗ c) = τ ι1 (ϕ(a)⊗ c) = τ(ϕ(a)⊗ c, 0) = ϕ(a)⊗ (c, 0).

Therefore, (ϕ⊗ 1F ) τ ι1 and τ ι1 (ϕ⊗ 1C) agree on generators of A⊗R C, so diagram (6) is commutative.

By lemma 2.7, (ϕ ⊗ 1F ) τ ι1 is injective, so, by commutativity, τ ι1 (ϕ ⊗ 1C) is injective. It now follows fromlemma 2.9 that ϕ⊗ 1C is injective, which completes the proof of claim 3 and part (b) of the problem. ut

The following six lemmas are used in the answer to problem 3 given above. The first is a standard theorem aboutprojective modules, the proof of which is not hard, and can be found, e.g., in Hungerford. The next two lemmas(2.5 and 2.6) are also standard, but I haven’t seen them proved in detail elsewhere and, as the solution given abovemakes repeated use of the maps defined in proving these lemmas, I include detailed proofs below. The last threelemmas are trivial verifications.

Lemma 2.4 Let R be a ring. The following conditions on an R-module P are equivalent:

(i) P is projective;

(ii) every short exact sequence of R-modules 0→ Af−→ B

g−→ P → 0 is split exact (hence B ∼= A⊕ P );

(iii) there is a free R-module F and an R-module N such that F ∼= N ⊕ P .

Lemma 2.5 If R is a commutative ring with 1R and A is a unitary R-module, then A⊗R R ∼= A.

44


Proof: Define α : A × R → A by α(a, r) = ar. Since α is clearly bilinear, there exists a unique R-module homomorphismα : A⊗R R→ A such that αι = α, where ι : A×R→ A⊗R R is the canonical bilinear map. Define β : A→ A⊗R R byβ(a) = a⊗ 1R. Then it is easy to verify that β is an R-module homomorphism and that

βα(a⊗ r) = β(ar) = ar ⊗ 1R = a⊗ r (∀a ∈ A)(∀r ∈ R).

Therefore, βα is the identity on generators of A ⊗R R. Also, αβ(a) = α(a ⊗ 1R) = a1R = a, so αβ = 1A. Therefore,α : A⊗R R ∼= A. ut

Lemma 2.6 Let R be a commutative ring with 1R and let A,B,C be unitary R-modules. Then A⊗R (B ⊕C) ∼=(A⊗R B)⊕ (A⊗R C).

Proof: Recall that if ϕ1 : A1 → D and ϕ2 : A2 → D are (group) homomorphisms, then there is a unique homomorphismΦ : A1 ⊕ A2 → D such that Φιi = ϕi (i = 1, 2), where ιi : Ai → A1 ⊕ A2 are the canonical injections. In other words,∃! Φ ∈ Hom(A1 ⊕A2, D) such that the following diagram is commutative:

A1

ϕ1$$IIIIIIIII

ι1 // A1 ⊕A2

Φ

��

A2

ι2oo

ϕ2zzuuuuuuuuu

D

We can apply this theorem to the Abelian group (A ⊗R B) ⊕ (A ⊗R C). Let ϕ1 : A × B → A ⊗R (B ⊕ C) be given byϕ1(a, b) = a⊗ (b, 0), and let ϕ2 : A×C → A⊗R (B⊕C) be given by ϕ2(a, c) = a⊗ (0, c). It is easily verified that ϕi arebilinear and thus induce R-module homomorphisms ϕ1 : A ⊗R B → A ⊗R (B ⊕ C) and ϕ2 : A ⊗R C → A ⊗R (B ⊕ C)such that ϕiκi = ϕi, where κ1 : A× B → A⊗R B and κ2 : A× C → A⊗R C are the canonical bilinear maps. Therefore,the two universal properties combine to give a unique homomorphism Φ : (A⊗RB)⊕ (A⊗R C)→ A⊗R (B⊕C) such thatthe following diagram is commutative:

A⊗R B

ϕ1 ((QQQQQQQQQQQQQι1 // (A⊗R B)⊕ (A⊗R C)

Φ

��

A⊗R C

ϕ2vvmmmmmmmmmmmmmι2oo

A×B

κ1

OO

ϕ1// A⊗R (B ⊕ C) A×B

κ2

OO

ϕ2oo

Now, for all a ∈ A and b ∈ B, ι1(a ⊗ b) = (a ⊗ b, 0) and ϕ1(a ⊗ b) = ϕ1κ1(a, b) = ϕ1(a, b) = a ⊗ (b, 0), so,Φ(a⊗ b, 0) = a⊗ (b, 0). Similarly, Φ(0, a⊗ c) = a⊗ (0, c), for all a ∈ A and c ∈ C. Therefore,

Φ(a⊗ b, a′ ⊗ c) = Φ(a⊗ b, 0) + Φ(0, a′ ⊗ c) = a⊗ (b, 0) + a′ ⊗ (0, c) (∀a, a′ ∈ A, b ∈ B, c ∈ C).

Next, define ψ : A× (B ⊕ C)→ (A⊗R B)⊕ (A⊗R C) by ψ(a, (b, c)) = (a⊗ b, a⊗ c). Again, ψ is bilinear, so there is aunique Ψ : A⊗R (B⊕C)→ (B⊕C)→ (A⊗RB)⊕(A⊗RC) such that Ψκ = ψ where κ : A×(B⊕C)→ A⊗R (B⊕C)is canonical. To complete the proof, we check that ΨΦ and ΦΨ given the appropriate identity maps:

ΨΦ(a⊗ b, 0) = Ψ(a⊗ (b, 0)) = (a⊗ b, a⊗ 0) = (a⊗ b, 0), and

ΨΦ(0, a⊗ c) = Ψ(a⊗ (0, c)) = (a⊗ 0, a⊗ c) = (0, a⊗ c).Thus, ΨΦ is the identity on generators of (A⊗R B)⊕ (A⊗R C), so ΨΦ = 1(A⊗RB)⊕(A⊗RC). Finally,

ΦΨ(a⊗ (b, c)) = Φ(a⊗ b, a⊗ c) = Φ((a⊗ b, 0) + (0, a⊗ c)) = a⊗ (b, 0) + a⊗ (0, c) = a⊗ (b, c).

Thus ΦΨ is the identity on generators ofA⊗R (B⊕C), so ΦΨ = 1A⊗R(B⊕C). This proves that Φ : (A⊗RB)⊕(A⊗RC) ∼=A⊗R (B ⊕ C). ut

Lemma 2.7 If h ∈ HomR(A,B), g ∈ HomR(B,C), and f ∈ HomR(C,D) are injective, then fgh ∈ HomR(A,D)is injective.

45


Proof: fgh(a) = 0⇒ gh(a) = 0 (since f is injective)⇒ h(a) = 0 (since g is injective)⇒ a = 0 (since h is injective). ut

Lemma 2.8 If g ∈ HomR(A,B), f ∈ HomR(B,C), if g is surjective, and if fg is injective, then f is injective.

Proof: Suppose f(b) = 0. Since g is surjective, there is an a ∈ A such that g(a) = b. Then fg(a) = f(b) = 0, which impliesa = 0, since fg is injective. Therefore, b = g(a) = g(0) = 0 since g is a homomorphism. ut

Lemma 2.9 If g ∈ HomR(A,B), f ∈ HomR(B,C), and if fg is injective, then g is injective.

Proof: ker g = {a ∈ A : g(a) = 0} ⊆ {a ∈ A : fg(a) = 0} = ker fg = {0}. The set containment holds since f is ahomomorphism, so f(0) = 0. ut

46

A BASIC DEFINITIONS AND THEOREMS

APPENDIX

A Basic Definitions and Theorems

A.1 AlgebrasAn algebra27 (or universal algebra) A is an ordered pair A = 〈A,F 〉 where A is a nonempty set and F is a family offinitary operations on A. The set A is called the universe of A, and the elements fA ∈ F are called the fundamentaloperations of A. (In practice we prefer to write f for fA when this doesn’t cause ambiguity.28) The arity of anoperation is the number of operands upon which it acts, and we say that f ∈ F is an n-ary operation on A if f mapsAn into A. An operation f ∈ F is called a nullary operation (or constant) if its arity is zero. Unary, binary, andternary operations have arity 1, 2, and 3, respectively. An algebra A is called unary if all of its operations are unary.An algebra A is finite if |A| is finite and trivial if |A| = 1. Given two algebras A and B, we say that B is a reduct ofA if both algebras have the same universe and A can be obtained from B by simply adding more operations.

Examples

groupoid A = 〈A, ·〉An algebra with a single binary operation is called a groupoid. This operation is usually denoted by + or ·, andwe write a+ b or a · b (or just ab) for the image of 〈a, b〉 under this operation, and call it the sum or product ofa and b, respectively.

semigroup A = 〈A, ·〉A groupoid for which the binary operation is associative is called a semigroup. That is, a semigroup is a groupoidwith binary operation satisfying (a · b) · c = a · (b · c), for all a, b, c ∈ A.

monoid A = 〈A, ·, e〉A monoid is a semigroup along with a multiplicative identity e. That is, 〈A, ·〉 is a semigroup and e is a constant(nullary operation) satisfying e · a = a · e = a, for all a ∈ A.

group A = 〈A, ·,−1 , e〉A group is a monoid along with a unary operation −1 called multiplicative inverse. That is, the reduct 〈A, ·, e〉 isa monoid and −1 satisfies a ·a−1 = a−1 ·a = e, for all a ∈ A. An Abelian group is a group with a commutativebinary operation, which we usually denote by + instead of ·. In this case, we write 0 instead of e to denotethe additive identity, and − instead of −1 to denote the additive inverse. Thus, an Abelian group is a groupA = 〈A,+,−, 0〉 such that a+ b = b+ a for all a, b ∈ A.

ring A = 〈A,+, ·,−, 0〉A ring is an algebra A = 〈A,+, ·,−, 0〉 such that

R1. 〈A,+,−, 0〉 is an Abelian group,

R2. 〈A, ·〉 is a semigroup, and

R3. for all a, b, c ∈ A, a · (b+ c) = a · b+ a · c and (a+ b) · c = a · c+ b · c.

A ring with unity (or unital ring) is an algebra A = 〈A,+, ·,−, 0, 1〉, where the reduct 〈A,+, ·,−, 0〉 is a ring,and where 1 is a multiplicative identity; i.e. a · 1 = 1 · a = a, for all a ∈ A.

27N.B. In this first paragraph, not all of the definitions are entirely precise. Rather, my goal here is to state them in a way that seems intuitive andheuristically useful.

28This convention creates an ambiguity when discussing, for example, homomorphisms from one algebra, A, to another, B; in such cases wewill adhere to the more precise notation fA and fB, for operations on A and B, respectively.

47

A.2 Congruence Relations and Homomorphisms A BASIC DEFINITIONS AND THEOREMS

field If A = 〈A,+, ·,−, 0, 1〉 is a ring with unity, an element r ∈ A is called a unit if it has a multiplicative inverse.That is, r ∈ A is a unit provided there exists r−1 ∈ A with r · r−1 = r−1 · r = 1. A division ring is a ring inwhich every non-zero element is a unit, and a field is a division ring in which multiplication is commutative.

module Let R = 〈R,+, ·,−, 0, 1〉 be a ring with unit. An R-module (sometimes called a left unitary R-module) isan algebra M = 〈M,+,−, 0, fr〉r∈R with an Abelian group reduct 〈M,+,−, 0〉, and with unary operations(fr)r∈R which satisfy the following four conditions for all r, s ∈ R and x, y ∈M :

M1. fr(x+ y) = fr(x) + fr(y)

M2. fr+s(x) = fr(x) + fs(x)

M3. fr(fs(x)) = frs(x)

M4. f1(x) = x.

If the ring R happens to be a field, an R-module is typically called a vector space over R.

Note that condition M1 says that each fr is an endomorphism of the Abelian group 〈M,+,−, 0〉. ConditionsM2–M4 say: (1) the collection of endomorphisms (fr)r∈R is itself a ring with unit, where the function compo-sition described in (M3) is the binary multiplication operation, and (2) the map r 7→ fr is a ring epimorphismfrom R onto (fr)r∈R.

Part of the importance of modules lies in the fact that every ring is, up to isomorphism, a ring of endomorphismsof some Abelian group. This fact is analogous to the more familiar theorem of Cayley stating that every groupis isomorphic to a group of permutations of some set.

bilinear algebra Let F = 〈F,+, ·,−, 0, 1〉 be a field. An algebra A = 〈A,+, ·,−, 0, fr〉r∈F is a bilinear algebra over Fprovided 〈A,+,−, 0, fr〉r∈F is a vector space over F and for all a, b, c ∈ A and all r ∈ F ,

(a+ b) · c = (a · c) + (b · c)c · (a+ b) = (c · a) + (c · b)a · fr(b) = fr(a · b) = fr(a) · b

If, in addition, (a · b) · c = a · (b · c) for all a, b, c ∈ A, then A is called an associative algebra over F. Thus anassociative algebra over a field has both a vector space reduct and a ring reduct. An example of an associativealgebra is the space of linear transformations (endomorphisms) of any vector space into itself.

A.2 Congruence Relations and HomomorphismsLet A be a set. A binary relation θ on A is a subset of A2 = A × A. If 〈a, b〉 ∈ θ we sometimes write a θ b. Thediagonal relation on A is the set ∆A = {〈a, a〉 : a ∈ A} and the all relation is the set ∇A = A2. (We write ∆ and ∇when the underlying set is apparent.)

equivalence A binary relation θ on a set A is an equivalence relation on A if, for any a, b, c ∈ A, it satisfies:

E1. 〈a, a〉 ∈ θ,

E2. 〈a, b〉 ∈ θ implies 〈b, a〉 ∈ θ, and

E3. 〈a, b〉 ∈ θ and 〈b, c〉 ∈ θ imply 〈a, c〉 ∈ θ.

We denote the set of all equivalence relations on A by Eq(A).

If θ ∈ Eq(A) is an equivalence relation on A and 〈x, y〉 ∈ θ, we say that x and y are equivalent modulo θ. Theset of all y ∈ A that are equivalent to x modulo θ is denoted by x/θ = {y ∈ A : 〈x, y〉 ∈ θ} and we call x/θ

48

A.2 Congruence Relations and Homomorphisms A BASIC DEFINITIONS AND THEOREMS

the equivalence class (or coset) of x modulo θ. The set {x/θ : x ∈ A} of all equivalence classes of A moduloθ is denote by A/θ. Clearly equivalence classes form a partion of A, which simply means that A =

⋃x∈A x/θ

and x/θ ∩ y/θ = ∅ if x/θ 6= y/θ.

To be continued...

49

A.3 Group Theory Basics A BASIC DEFINITIONS AND THEOREMS

A.3 Group Theory BasicsWe begin by collecting some facts that are very useful for solving elementary problems in group theory. (Notation:H 6 G means ‘H is a subgroup of G.’)

1. Aut(G) denotes the group of automorphisms of the group G. The set of inner automorphisms, denoted Inn(G),is the (normal) subgroup of Aut(G) defined as follows: ϕ ∈ Inn(G) iff (∃g ∈ G) (∀x ∈ G) (ϕ(x) = gxg−1).

2. Let H 6 G and let A ⊆ Aut(G) be a non-empty set of automorphisms of G. We say that H is an A-invariantsubgroup of G if hα ∈ H for all h ∈ H and α ∈ A. For instance, if A = 1 then, trivially, every subgroup of Gis A-invariant. In two important special cases we use special terms. If H is Aut(G)-invariant, H is said to becharacteristic in G. If H is Inn(G)-invariant, H is said to be normal in G.

The concept of normal subgroup dominates the whole of group theory, and a special notation is used. We writeH P G to mean ‘H is a normal subgroup of G’, and H R G to mean ‘H is not a normal subgroup of G’.

3. The normalizer of a subset S ⊆ G is the set

NG(S) = {g ∈ G : g−1Sg = S}.

In case S = {s} is a singleton, we write NG({s}) = NG(s). Also, we occasionally denote NG(S) by N(S)when the context makes clear which overgroup is intended. It is easy to verify that the normalizer is a subgroupof G. In fact, when S is a subgroup of G, NG(S) is the largest subgroup of G in which S is normal. Thus,S P NG(S) 6 G, and S P G iff NG(S) = G.

4. The centralizer of a subset S ⊆ G is the set

CG(S) = {g ∈ G : g−1sg = s for all s ∈ S}.

Here too, we write CG({s}) = CG(s) when S = {s} is a singleton, in which case we also have CG(s) =NG(s). Thus, it is clear from the definition that

CG(S) =⋂s∈S

CG(s) =⋂s∈S

NG(s).

Next note that, since S P NG(S), the group NG(S) acts on S by conjugation. The map τ : NG(S)→ Aut(S),defined by τ(g)(s) = gsg−1 (∀s ∈ S; ∀g ∈ NG(S)) is a group homomorphism with kernel

ker τ = {g ∈ NG(S) : g−1sg = s for all s ∈ S}.

But notice that if g ∈ G satisfies g−1sg = s for all s ∈ S, then certainly g ∈ NG(S). Therefore,

ker τ = {g ∈ NG(S) : g−1sg = s for all s ∈ S} = {g ∈ G : g−1sg = s for all s ∈ S} = CG(S).

Since kernels of homomorphisms are normal subgroups, we have prove CG(S) P NG(S).

5. A group G is called simple provided G has exactly two normal subgroups, namely G and (e) (the identity in G).Clearly if G is simple, then any homomorphism on G is a monomorphism (since the kernel of a homomorphismis a normal subgroup). A group G is called almost simple if G has a normal subgroup S P G which isnonabelian, simple, and has trivial centralizer CG(S) = (e).

50

A.3 Group Theory Basics A BASIC DEFINITIONS AND THEOREMS

Next we review a number of very useful results, as presented in John Dixon’s wonderful little book [1].29

1.T.1. If H 6 K 6 G and if [G : H] is finite, then [G : K] is finite and [G : H] = [G : K][K : H].

1.T.2. If A,B 6 G and [G : B] is finite, then A∩B is a subgroup of finite index in A. In fact, [A : A∩B] 6 [G : B],with equality iff G = AB.

In particular, if [G : A] is also finite, then [G : A ∩B] 6 [G : A][G : B] with equality iff G = AB.

1.T.3. If S is a subset of a group G, then the number of conjugates, x−1Sx (x ∈ G), of S in G is equal to [G : NG(S)]In the case that S consists of a single element s, then NG(S) = CG(s), and s has [G : C(s;G)] conjugates inG.

29As Dixon notes, the proofs of 1.T.1–3 can be found in Hall (Sec. 1.5–6, 2.1–3); Kurosh (Ch. 3); Rotman (Ch. 2); Schenkman (Sec. 1.1–3); andScott (Sec. 1.6–7, 2.2–3, 3.1–3).

51

A.4 Direct Products A BASIC DEFINITIONS AND THEOREMS

A.4 Direct ProductsD.1. For an arbitrary (possibly infinite) family of groups {Gi | i ∈ I}, define a binary operation on the Cartesian

product (of sets)∏i∈I

Gi as follows: If f, g ∈∏i∈I

Gi (that is, f, g : I →⋃i∈I

Gi and f(i), g(i) ∈ Gi for each i),

then fg : I →⋃i∈I

Gi is the function given by i 7→ f(i)g(i). Since each Gi is a group, f(i)g(i) ∈ Gi for every

i, whence fg ∈∏i∈I

Gi. It is easy to verify that∏i∈I

Gi together with this binary operation is a group, called the

direct product (or complete direct sum) of the family of groups {Gi | i ∈ I}.

D.2. The (external) weak direct product of a family of groups {Gi | i ∈ I}, denoted∏i∈I

wGi, is the set of all

f ∈∏i∈I

Gi such that f(i) = ei, the identity in Gi, for all but finitely many i ∈ I . If all Gi are (additive) abelian,

then∏i∈I

wGi is usually called the (external) direct sum and is denoted

∑i∈I Gi.

If I is finite, the weak direct product coincides with the direct product.

D.3. Theorem. If {Gi | i ∈ I} is a family of groups, then∏i∈I

wGi is a normal subgroup of

∏i∈I

Gi and, for each

k ∈ I ,

(i) the map πk :∏i∈I

Gi → Gk given by πk(f) = f(k) is an epimorphism of groups;

(ii) the map ιk : Gk →∏i∈I

wGi given by ιk(a) = (e1, . . . , ek−1, a, ek, . . . ) is a monomorphism of groups;

(iii) ιk(Gk) is a normal subgroup of∏i∈I

Gi.

As a memory aid, the following diagram might help:

0 −−−−→ Gkιk−−−−→

∏i∈I

wGi

P−−−−→∏i∈I

Giπk−−−−→ Gk −−−−→ 0

D.4. Theorem. Let {Ni | i ∈ I} be a family of normal subgroups of a group G such that

(i) G = 〈⋃iNi〉.

(ii) For each k ∈ I , Nk ∩ 〈⋃i 6=kNi〉 = (e).

Then G =∏wi∈I Ni.

D.5. The following special case arises frequently: if Ni P G for each 1 6 i 6 r, then 〈N1 ∪ N2 ∪ · · · ∪ Nr〉 =N1N2 · · ·Nr = {n1n2 · · ·nr | ni ∈ Ni}. In additive notation N1N2 · · ·Nr is written N1 +N2 + · · ·+Nr.

D.6. Corollary. If Ni P G for each 1 6 i 6 r, if G = N1N2 · · ·Nr, and if Nk ∩ (N1 · · ·Nk−1Nk+1 · · ·Nr) = (e)for each 1 6 k 6 r, then G ∼= N1 ×N2 × · · · ×Nr.

52

A.5 Ring Theory Basics A BASIC DEFINITIONS AND THEOREMS

A.5 Ring Theory Basicsmultiplicative A subset A of a ring (or groupoid) is called multiplicative iff it is closed under multiplication; i.e. x, y ∈ A

implies xy ∈ A.

prime ideal An ideal P in a ring R is prime iff P 6= R and, for any ideals A,B of R,

AB ⊆ P ⇒ A ⊆ P or B ⊆ P. (7)

Theorem A.1 If R is a commutative ring, then a proper ideal P ( R is prime if and only if, for all elements a, b ∈ R,

ab ∈ P ⇒ a ∈ P or b ∈ P. (8)

If R is a general (non-commutative) ring, then condition (8) implies P is prime but not conversely.

Proof: See Hungerford [2], Theorem III.2.15.

Consider the forward implication of Theorem A.1– i.e. P prime implies condition (8). To see that this does not hold ingeneral for non-commutative rings, consider this simple counter-example: Let M2(Z2) be the ring of 2 × 2 matricesover Z2 = Z/2Z (or over any division ring, for that matter). Then the ideal P = (0), that is, the ideal generated by

the zero matrix(

0 00 0

), is the only ideal of M2(Z2) (this easily follows from Proposition A.1 below.) As the only

ideal, P trivially satisfies condition (7), so P is prime. However, if a =(

1 00 0

)and b =

(0 00 1

), then ab ∈ P , but

a /∈ P and b /∈ P .

Corollary A.1 If P is a prime ideal of a commutative ring R, then

a1a2 · · · an ∈ P ⇒ ai ∈ P for some i ∈ {1, . . . , n}. (9)

Theorem A.1 can also be stated this way:

Corollary A.2 In a commutative ring R, an ideal P ( R is prime if and only if the set R \ P is multiplicative.

Theorem A.2 If S is a multiplicative subset of a ring R which is disjoint from an ideal I of R, then there exists anideal P which is maximal in the set of all ideals R disjoint from S and containing I . Furthermore, any such ideal P isprime.

Note: this theorem is frequently used in the case I = (0). (See, e.g., the April 2008 exam 2.10, Problem 2, part d.) Toprove the theorem, let I be the set of all ideals ofR which are disjoint from the multiplicative subset S. Of course, Iis not empty since I ∈ I . If {Iλ : λ ∈ Λ} is a chain of ideas in I , it is not hard to show that J =

⋃{Iλ : λ ∈ Λ} is

an ideal, and that J must be disjoint from S. Therefore J ∈ I and Zorn’s lemma implies that I contains a maximalideal. The proof is completed by the following result, which is a useful general principal, so we state it separately as

Lemma A.1 Suppose R is a commutative ring, S ⊆ R is a multiplicative subset, and P is an ideal of R which ismaximal with respect to the property P ∩ S = ∅. Then P is a prime ideal.

Proof: Suppose a, b ∈ R \ P . We show that ab /∈ P , which will prove that P is prime. Since P is maximal among all ideals withthe property P ∩ S = ∅, we have

(P +Ra) ∩ S 6= ∅ 6= (P +Rb).

Therefore, p+ ra = s for some p ∈ P , r ∈ R, and s ∈ S, and p′ + r′b = s′ for some p′ ∈ P , r′ ∈ R, and s′ ∈ S. Now compute

ss′ = (p+ ra)(p′ + r′b) = pp′ + pr′b+ rap′ + rr′ab.

The first three terms on the right are in P since P is an ideal, while the left-hand-side ss′ ∈ S, as S is multiplicative. Therefore,rr′ab cannot be in P , so ab /∈ P . ut

53

A.6 Factorization in Commutative Rings A BASIC DEFINITIONS AND THEOREMS

A.6 Factorization in Commutative RingsWe assume throughout this section that R is a commutative ring with identity e. First recall some basic definitions.

unit An element of R that has an inverse is called a unit. That is r ∈ R is a unit iff there is an a ∈ R withar = ra = e.

irreducible

prime A nonzero, nonunit element r in a commutative ring R is called

(i) irreducible provided that r = ab implies a or b is a unit

(ii) prime provided that r|ab implies r|a or r|b.

A.7 Rings of PolynomialsWe review some basic facts about a the polynomial rings R[x] and R[x1, . . . , xn]. Most of the material in this sectionhas been shamelessly copied from Hungerford ([2], section III.6) – sometimes word-for-word – since, in my opinion,Hungerford’s presentation cannot be improved upon.

Let R be a ring. The degree of a nonzero monomial axk11 xk22 · · ·xkn

n ∈ R[x1, . . . , xn] is the nonnegative integerk1 + k2 + · · · + kn. If f is a nonzero polynomial in R[x1, . . . , xn], then f =

∑mi=0 aix

ki11 · · ·xkin

n , and the (total)degree of the polynomial f is the maximum of the degrees of the monomials aixki1

1 · · ·xkinn such that ai 6= 0

(i = 1, 2, . . . ,m). The total degree of f is denoted deg f . Clearly a nonzero polynomial f has degree zero iff f is aconstant polynomial and f = a0 = a0x

01 · · ·x0

n. Recall that, for each 1 6 k 6 n, R[x1, . . . , xk−1, xk+1, . . . , xn] is asubring of R[x1, . . . , xn]. The degree of f in xk is the degree of f considered as a polynomial in one indeterminatexk over the ring R[x1, . . . , xk−1, xk+1, . . . , xn].

Theorem A.3 (i) deg(f + g) 6 max{deg f, deg g}.

(ii) deg(fg) 6 deg f + deg g.

(iii) If R has no zero divisors, deg(fg) = deg f + deg g.

(iv) If n = 1 and the leading coefficient of f or g is not a zero divisor in R (in particular, if it is a unit), thendeg(fg) = deg f + deg g.

This theorem is also true if deg f is taken to mean “degree of f in xk.”

It is easy to verify that R[x] is commutative (resp. an integral domain) whenever the same is true of R.If F is a field, then F [x] is a Euclidean domain, hence a PID and a UFD, and the set of units of F [x] is precisely

the set of nonzero constant polynomials, which can be identified with the set F× = F \ {0}. ([2] Theorem 6.14)IfD is a UFD, we have the following important facts about the units and irreducible elements ofD[x] ([2], p. 162):

(i) The units in D[x] are precisely the constant polynomials that are units in D.

(ii) If c is an irreducible element of D, then the constant polynomial c is irreducible in D[x] [use Thm A.3 and (i)].

(iii) Every first degree polynomial whose leading coefficient is a unit in D is irreducible in D[x]. In particular, everyfirst degree polynomial over a field is irreducible.

(iv) Suppose D is a subring of an integral domain E and f ∈ D[x] ⊆ E[x]. Then f may be irreducible in E[x] butnot in D[x] and vice versa, as is seen in the following examples.

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A.8 Nakayama’s Lemma A BASIC DEFINITIONS AND THEOREMS

Examples. 2x+ 2 is irreducible in Q[x] by (iii) above. However, 2x+ 2 = 2(x+ 1) and neither 2 nor x+ 1 is a unitin Z[x] by (i), whence 2x+ 2 is reducible in Z[x]. On the other hand x2 + 1 is irreducible in R[x] but factors in C[x]as (x+ i)(x− i). By (i), neither x+ i nor x− i is a unit in C[x], so x2 + 1 is reducible in C[x].

If D is a UFD, then so is D[x1, . . . , xn]. ([2] Theorem 6.14) Since a field F is trivially a UFD, F [x1, . . . , xn] is aUFD.

A.8 Nakayama’s LemmaTheorem A.4 (Nakayama) If J is an ideal in a commutative ring R with identity, 1R, then the following conditionsare equivalent.

(i) J is contained in every maximal ideal of R.

(ii) 1R − j is a unit for every j ∈ J .

(iii) If A is a finitely generated R-module such that JA = A, then A = 0.

(iv) If B is a submodule of a finitely generated R-module A such that JA+B = A, then B = A.

Remark. As we will see in the next section, the lemma is true even when R is noncommutative, provided that (i) isreplaced with the condition that J is contained in the Jacobson radical of R.

Proof: ((i)⇒ (ii)) Suppose (i) holds and suppose 1R− j is not a unit. Then (1R− j) is a proper ideal, so (1R− j) ⊆M for somemaximal ideal M ( R. Thus, 1R − j ∈ M and J ⊆ M , and we have j ∈ M so 1R − j + j = 1R ∈ M . Of course, this impliesthe contradiction M = R. Therefore, 1R − j must be a unit. [(ii) ⇒ (iii)] Suppose (ii) holds and let A be a finitely generatedR-module with JA = A. There is a minimal generating set for A, say, {a1, . . . , an}. If A 6= 0, then a1 is not zero. Recall,

JA = {mXi=1

jixi | ji ∈ J, xi ∈ A,m ∈ N},

where each xi ∈ A is of the form xi = ri1a1 + · · ·+ rinan, and upon rearranging terms, we see that jixi = eachj

A.9 Miscellaneous Results and ExamplesHere is a basic fact that is sometimes useful (e.g., in the counter-example after Theorem A.1 above). It is elementaryand probably appears as an exercise in some of the standard references, though I have not seen it elsewhere.30

Proposition A.1 Let R be a commutative ring with 1R 6= 0. Then the ideals of Mn(R) are precisely the subsetsMn(J) ⊆Mn(R), where J is an ideal of R.

Proof: We prove it for the case n = 2. The proof in the general case should be almost identical. Suppose J is an ideal andA1, A2 ∈ M2(J). Then A1 − A2 has all elements in J , since J is, in particular, a subgroup. Therefore M2(J) is a subgroup of〈M2(R); +,0〉. If M = (mij) ∈ M2(R) and A = (aij) ∈ M2(J), then MA = (

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 ofM2(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 the

klth 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. ut

30I have since found a nice, short proof of this result in Lam’s book [5].

55

A.9 Miscellaneous Results and Examples A BASIC DEFINITIONS AND THEOREMS

Proposition A.2 Let R and R′ rings. Then the ideals of R×R′ have the form I × J , where I is an ideal of R and Jis an ideal of R′.

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 A2 are idealsof 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 × A2.31 Fix a ∈ 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 someb ∈ 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, so A1 is asubgroup of 〈R,+, 0〉. Fix r ∈ R. Then (ra, a′) = (ra, 1a′) = (r, 1) · (a, a′) ∈ A, since A is an ideal of R × R′. Therefore,ra ∈ A1, which proves that A1 is an ideal of R. The same argument, mutatis mutandis, proves that A2 is an ideal of R′. Thiscompletes the proof of (ii) and establishes the proposition. ut

Example:32 The ideals of Z/12Z×M2(R) are all sets of the form I × J , where I ∈ {(0), (1), (2), (3), (4), (6)} andJ ∈ {M2(0),M2(R)}.

The following is a nice proof of a standard problem that I learned from the book by Lam [5].

Proposition A.3 Let R be a ring with identity 1. If a and b are nonzero elements in R, then 1− ab is invertible if andonly if 1− ba is invertible.

Proof: It is clear by symmetry that we need only prove one direction. Suppose 1− ab is invertible. The ideal R(1− ba) containsRa(1− ba) = R(1− ab)a = Ra and, therefore, contains 1− ba+ ba = 1. Thus, R(1− ba) = R, which implies r(1− ba) = 1for some r ∈ R.

A related linear algebra problem is to show that if A and B are square matrices, then AB and BA have the sameeigenvalues. This can be solved exactly as above, with one modification: replace 1 with λI , where I is the identitymatrix of the same dimensions as A and B.

31Although 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.)

32This example appears in the first problem of the ring theory section of the November 2008 comprehensive exams.

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REFERENCES REFERENCES

References[1] John Dixon. Problems in Group Theory. Dover, New York, 1974.

[2] Thomas W. Hungerford. Algebra. Springer-Verlag, New York, 1974.

[3] Nathan Jacobson. Basic Algebra I. W. H. Freeman and Co., New York, 1985.

[4] Friedrich Kasch and Adolf Mader. Rings, Modules, and the Total. Birkhauser, Basel, 2004.

[5] T. Y. Lam. A First Course in Noncommutative Rings. Springer-Verlag, New York, second edition, 2001.

[6] Harvey E. Rose. A Course on Finite Groups. Springer-Verlag, London, 2009.

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