UCLA Algebra Qualifying Exam Solutions

Ian Coley

August 25, 2014

1 Spring 2014

Problem 1.Let L : C → D be a functor, left adjoint to R : D → C. Show that if the counit L ◦R→ idDis a natural isomorphism, then R is fully faithful.

Solution.R is fully faithful if for every objects X, Y in D, we have MorD(X, Y ) ∼= MorC(R(X), R(Y )).Recall further that L and R being left and right adjoints means that, for every object A inC and B in D, we have

MorC(A,R(B)) ∼= MorD(L(A), B).

As such, we see, again with X, Y in D,

MorC(R(X), R(Y )) = MorD(L ◦R(X), Y ) = MorD(X, Y ),

as required.

Problem 2.Let A be a central division algebra (of finite dimension) over a field k. Let [A,A] be thek-subspace of A spanned by the elements ab− ba with a, b ∈ A. Show that [A,A] 6= A.

Solution.Let K be the algebraic closure of k, and consider B = A⊗k K. Then B ∼= Mn(K) for somen ∈ N, and thus we can understand [B,B] ∼= [A,A]⊗k K. In this case, [B,B] contains onlymatrices of trace 0, since it is clear that tr(XY − Y X) = 0 for every X, Y ∈ B. Therefore[B,B] 6= B. As such, we could not have had [A,A] = A.

Problem 3.Given ϕ : A→ B a surjective morphism of rings, show that the image by ϕ of the Jacobsonradical of A is contained in the Jacobson radical of B.

Solution.We use the following characterisation of the Jacobson radical: J(R) = {x ∈ R : xy − 1 ∈

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R× for all y ∈ R}. As such, let x ∈ J(A). Then consider ϕ(x)b − 1 for any b ∈ B. Since ϕis surjective, we have b = ϕ(y) for some y ∈ A. Hence we have

ϕ(x)b− 1 = ϕ(x)ϕ(y)− 1 = ϕ(xy − 1).

Since xy−1 is invertible by assumption, ϕ(xy−1) is also invertible, with inverse ϕ((xy−1)−1).Therefore ϕ(x) ∈ J(B), and we are done.

Problem 4.Let G be a group and H a normal subgroup of G. Let k be a field and let V be an irreduciblerepresentation of G over k. Show that the restriction of V to H is semisimple.

Solution.This is known as Clifford’s theorem, if we assume that V is a finite-dimensional representation(which we do here). The following proof is (crudely) taken from Wikipedia, but a bettersource likely exists.

Let VH be the restriction of V to H, and let U ⊂ VH be an irreducible k[H]-module. Foreach g ∈ G, g · U is an irreducible k[H]-module of VH . Further,

∑g∈G g · U is a nontrivial

k[G]-submodule of V , so must be V itself by irreducibility. Therefore∑

g∈G g · U = VH , soVH is a sum of irreducible submodules. This is one of the equivalent characteristics of asemisimple module, so we are done.

Problem 5.Let G be a finite group acting transitively on a finite set X. Let x ∈ X and let P be a Sylowp-subgroup of the stabiliser of x in G. Show that NG(P ) acts transitively on XP .

Solution.Recall that any transitive G-action is G-isomorphic to the action of G by left multiplicationon G/H for some subgroup H. As such, we can solve the problem in these terms. First, weneed to get a handle on what Gx, the stabiliser of the coset xH ∈ G/H, is. We have

Gx = {g ∈ G : gxH = xH} = {g ∈ G : x−1gxH = H}= {g ∈ G : x−1gx ∈ H} = {g ∈ G : g ∈ xHx−1} = xHx−1.

Hence P is a Sylow p-subgroup of xHx−1. Further, XP is given by

XP = {yH ∈ G/H : pyH = yH for all p ∈ P} = {xH ∈ G/H : y−1pyH = H for all p}= {yH ∈ G/H : y−1py ∈ H for all p} = {yH ∈ G/H : p ∈ yHy−1 for all p}= {yH ∈ G/H : P ⊂ yHy−1}.

In particular, we see that xH ∈ XP . We will show that, for any coset yH ∈ XP ,there exists p ∈ P so that pxH = yH, which will prove transitivity. Indeed, since we haveP ⊂ xHx−1 and P ⊂ yHy−1, we have x−1Px, y−1Py ⊂ H. These are two Sylow p-subgroupsof H, so they are conjugate by some h ∈ H. Hence

hx−1Pxh−1 = y−1Py =⇒ yhx−1Pxh−1y−1 = P =⇒ yhx−1 ∈ NG(P ).

This is the required element, as

yhx−1 · xH = yh ·H = yH.

Hence the action of NG(H) is transitive.

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Problem 6.Let A be a ring and M a noetherian A-module. Show that any surjective morphism ofA-modules M →M is an isomorphism.

Solution.This question has been asked in various forms time and time again. See, for example, Fall2013 #9, whose argument boils down to this statement.

Problem 7.Let G be a finite group and let s, t ∈ G be two distinct elements of order 2. Show that thesubgroup of G generated by s and t is a dihedral group. (Recall that the dihedral groupsare the groups D2m = 〈g, h : g2 = h2 = (gh)m = 1〉 for some m ≥ 2.)

Solution.The definition given here is by no means standard, and makes the problem fairly trivial.First, let H denote the subgroup in question. We know that |st| = n < ∞ for some n ∈ Nsince G is finite. Moreover, because |s| = 2, s−1 = s, so in particular s−1 6= t so st 6= 1.Therefore n ≥ 2, which gives a surjection f : H → D2n, where f(s) = g, f(t) = h. We needonly show it is injective. Note that |ts| = |st| since

t = t(st)n = (ts)nt =⇒ 1 = (ts)n

and if |ts| < n we would have a contradiction by the same reasoning. Thus without loss ofgenerality assume (st)ks ∈ ker f for k < n. Then we have

f((st)ks) · f(s) = f((st)k) = g =⇒ f((st)2k) = 1.

Additionally,f((st)ks) · f(t) = f((st)k+1) = h =⇒ f((st)2k+2) = 1.

This implies that f((st)2) = 1. Therefore since (st)ks ∈ ker f as well, this implies that eithersts or s is in ker f , depending on the parity of k. The second case is impossible. For thefirst case, this implies that

ghgh = f(stst) = f(t) = h,

which is also a contradiction. Therefore f is injective as well, and we are done.

Problem 8.Let F be a finite field. Without using any of the theorems on finite fields, show that F hasa field extension of degree 2.

Solution.It suffices to construct an irreducible quadratic polynomial. If f is such a polynomial, thenF [X]/(f) is a priori a 2-dimensional algebra over F . Since F [X] is a PID, any irreducibleelement is prime, and any (nonzero) prime ideal is maximal. Hence F [X]/(f) is a field ofdegree 2 over F .

Now, let S = {X2 + aX + b : a, b ∈ F}, the set of all monic quadratic polynomials inF [X]. Further, let T = {(X − a)(X − b) : a, b ∈ F}, the set of reducible monic quadratic

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polynomials. We use the fact that a quadratic polynomial is reducible over F if and onlyif it has a root in F . Clearly T ⊂ S. By direct calculation, if |F | = q, then |S| = q2 and|T | =

(q2

)+ q, which correspond to when a 6= b and a = b, respectively. Hence |S| > |T | for

any q > 1, so there are more monic quadratic polynomials than reducible ones, whence onemust be irreducible. This completes the proof.

Problem 9.Let G be a finite group. Show that there exist fields F ⊂ E such that E/F is Galois withgroup G.

Solution.There is a usual way of performing this construction. Let |G| = n. Let E = C(X1, . . . , Xn),the field of rational functions in n variables. Let k = C(s1, . . . , sn), the subfield generatedby the symmetric polynomials, which are defined by

sk =∑

1≤j1<···<jk≤n

Xj1 · · ·Xjk .

E/F is separable since char k = 0. We need only show that it is a (finite) normal extension.But this is clear, since E is the splitting field of

g(Y ) =n∏i=1

(Y −Xi) ∈ k[Y ],

whose coefficients are easily seen to lie in k because of the inherent symmetry of g(Y ).Therefore E/k is Galois, and by construction has Gal(E/k) = Sn. We have a canonicalembedding of G into Sn by the group action of G on itself by left multiplication. Let H ⊂ Snbe the image of that embedding. By Galois correspondence, there exists an intermediatefield F so that Gal(E/F ) = H. Since H ∼= G, we are done.

Problem 10.Let F be a field. Show that the polynomial ring F [X] has infinitely many prime ideals. Alsoprove that algebraically closed fields are infinite.

Solution.Since F [X] is a PID, we need to show that there are infinitely many irreducible polynomialsin the polynomial ring of a field. If F itself were infinite, then the set of polynomials{X −α : α ∈ F} is infinite, and all of these are irreducible since they are monic of degree 1.

If F is finite, then let |F | = q = pn. Then we may consider the field extensions Fqk/F foreach k ∈ N. These are finite separable extensions, so they are simple, so let Fqk = F (αk).Then the (irreducible) minimal polynomial of αk over F is of degree k. Hence there areinfinitely many irreducible polynomials over a finite field as well.

Let K be an algebraically closed field. Let F ⊂ K be its prime field. In particular, Kcontains (an embedded copy of) the algebraic closure of F , so K contains (an embeddedcopy of) the roots of every irreducible polynomial in F [X]. By the above, there are infinitelymany such polynomials, so there are infinitely many distinct roots. Hence K is infinite.

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2 Fall 2013

Problem 1.How many groups are there up to isomorphism of order pq where p < q are prime integers?

Solution.Let nq denote the number of Sylow q-subgroups (and respectively np). Then by the Sylowtheorems, nq | p and nq ≡ 1 mod q. since q + 1 > p, we must have nq = 1, so every groupG with |G| = pq has a normal Sylow q-subgroup, which we denote Q / G. Now for thep-subgroups of G, by the Sylow theorems, np | q and np ≡ 1 mod p. Since np | q, we havenp = 1 or q. We always have 1 ≡ 1 mod p, but we do not always have q ≡ 1 mod p.

Let P be a Sylow p-subgroup of G. Since P and Q have coprime orders, they have trivialintersection. Hence

|PQ| = |P | · |Q||P ∩Q|

= pq.

which implies that G = Q oϕ P , for some map ϕ : P → AutQ. Since P ∼= Z/pZ, we haveAutQ ∼= Z/(q − 1)Z. Further, since P has no proper subgroups, ϕ is injective or trivial.Suppose that p - q− 1. Then ϕ must be trivial, since |ϕ(P )| must divide the order of AutQ.Therefore G = Q × P is a proper direct product, so G ∼= Z/pqZ is the only group of orderpq.

Otherwise, if p | q − 1, then there is a unique subgroup of AutQ of order p by thefundamental theorem of cyclic groups. We may map the generator of P to any elementof this subgroup, for a total of p different homomorphisms ϕ, one of which is the trivialhomomorphism. This yields p groups of order pq of the form G = Q oϕ P for each ϕ.However, for any two nontrivial ϕ, ψ : P → AutQ, we can construct an isomorphism ofQ oϕ P and Q oψ P via an automorphism of P , since P is cyclic. Hence there are twononisomorphic groups of order pq, which correspond to a trivial ϕ and a nontrivial ϕ.

Problem 2.Show that there are up to isomorphism exactly two nonabelian groups G of order 8. Provethat each of them has an irreducible complex representation of dimension 2.

Solution.First, suppose that no element of G has order 4. Then every nonidentity element has order2, which implies that G is abelian. Therefore G contains an element of order 4, say a ∈ G.Suppose there exists b ∈ G of order 4 so that ba 6= ab. Then let A,B ⊂ G be the subgroupsgenerated by a, b respectively. Then if A ∩ B = {1}, then |AB| = |A| · |B| = 16, which is acontradiction. Therefore there is an element of order 2 in A∩B, which we denote −1, whichsatisfies a2 = b2 = −1. Then we can write all the elements of G : {1,−1, a,−a, b,−b, ab, ba},and in fact ba = −ab. We see that G ∼= Q, the quaternion group.

Suppose now that there is no element of order 4 that does not commute with a. Thenthere must exist an element of order 2 that does not commute with a, which we also callb. Then since A ∩ B = {1}, |AB| = |A| · |B| = 8, so we can write all the elements of Gagain: {1, a, a2, a3, b, ab, a2b, a3b}, which we see is isomorphic to D8, the dihedral group of thesquare. These are the only two nonabelian groups of order 8, and they are not isomorphicsince Q contains more elements of order 4 than does D8.

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Recall that there is an irreducible complex representation of a group G for each of itsconjugacy classes, and further the sum of the squares of the dimensions of the representationsmust equal the order of the group. The only way to sum squares to 8 is 1+1+1+1+1+1+1+1,1+1+1+1+4, or 4+4. The first case corresponds to 8 conjugacy classes of G, which couldimply G is abelian, a contradiction. Therefore G must contain an irreducible representationof dimension 2, so that 22 = 4 is part of this sum.

Problem 3.For a positive integer n, let Φn(X) be the nth cyclotomic polynomial. If a is an integer andp a prime not dividing n, such that p divides Φn(a), show that the order of a mod p is n.Using this prove that there are infinitely many primes p such that p is 1 modulo n.

Solution.If p | Φn(a), then Φn(a) = 0 in Fp. As such a is an nth root of unity modulo p, so it is a rootof Xn− 1 = 0. We need to show that it is a primitive nth root of unity, so that the order ofa modulo p is 1. We know that

Xn − 1 =∏d|n

Φd(X).

Suppose that a were a dth root of unity modulo p. Then Φd(a) = Φn(a) = 0, which wouldmake a a multiple root of Xn − 1. However since p - n, the polynomial Xn − 1 is separable,so it cannot have a multiple root. Therefore a is a primitive nth root of unity.

Since an = 1 mod p, we know that in F×p we have an = 1, so n | p − 1, which is to saythat p ≡ 1 mod n. Therefore we need to show that there are infinitely many primes thatdivide the set of values Φn(a) for a ∈ Z. First, since Φn(a) → ∞ as a → ∞, there is someprime p - n dividing some value of Φn(a). But this is (or at least should be) clear, so we aredone.

Problem 4.Given a field K of characteristic p, when is an α that is algebraic over K said to be separable?Show that if α is algebraic over K, then α is separable if and only if K(α) = K(αp

n) for all

positive integers n.

Solution.An algebraic element α is separable over K if its minimal polynomial mα ∈ K[X] is separable,that is, its formal polynomial derivative is nonzero.

First, consider the extension K(αpn) ⊂ K(α). Then since α is a root of the polynomial

Xpn−αpn , we have mα | Xpn−αpn over K(αpn)[X]. But since we are working in characteristic

p, Xpn −αpn = (X −α)pn

in K(α), mα is not a separable polynomial. In fact, this extensionis purely inseparable.

Suppose that α is separable. Then the subextension K(αpn) ⊂ K(α) is also separable,

so since it is also purely inseparable, it is trivial. Hence K(α) = K(αpn). Conversely, if α

is not separable, then the minimal polynomial of α is of the form f(Xpk) where f(X) is aseparable polynomial and k ≥ 1. But then αp is a root of the polynomial f(Xpk−1

), so theextension K(αp) is strictly of lower degree than K(α). Therefore K(α) ) K(αp).

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Now suppose that we are in this case and K(α) = K(αpn). Then since K(αp) ⊃ K(αp

n),

this is a contradiction that K(α) is strictly larger. This completes the proof.

Problem 5.Let G be a finite group which has the property that for any element g ∈ G of order n, andan integer r prime to n, the elements g and gr lie in the same conjugacy class. Then showthat the character of every representation of G takes values in the rational numbers Q (infact even the integers Z). (Hint: Use Galois theory.)

Solution.Let ρ : G→ GL(V ) be a complex representation. Consider an element ρ(g). Then since thisis a complex representation, we may represent ρ(g) by a diagonal matrix diag(λ1, . . . , λm).Since ρ(g)n = ρ(gn) = Im, we must have λi is an nth root of unity for each i, so we writeρ(g) = diag(ζa1n , . . . , ζ

amn ).

If χ is a character of ρ, then χ(ρ(g)) =∑m

i=1 ζain ∈ Q(ζn) ⊂ C. Consider Gal(Q(ζn)/Q).

This group consists precisely of the automorphisms σ(ζn) = ζrn for some (r, n) = 1. We claimthat χ(ρ(g)) is fixed by every σ ∈ Gal(Q(ζn)/Q). If this is so, then χ(ρ(g)) lies in the fixedfield of the Galois group, which is Q, so we would be done.

To prove the claim, we have

σ(χ(ρ(g))) = σ

(m∑i=1

ζain

)=

m∑i=1

ζai·rn = tr (diag(ζa1n , . . . , ζamn ))r = χ(ρ(gr)).

But since χ is a class function and g, gr lie in the same conjugacy class, we have σ(χ(ρ(g))) =χ(ρ(g)). This proves the claim.

Problem 6.Let I be an ideal of a commutative ring and a ∈ R. Suppose the ideals I + Ra and(I : a) := {x ∈ R : ax ∈ I} are finitely generated. Prove that I is also finitely generated.

Solution.We have I + Ra/Ra = I/I ∩ Ra = I/a(I : a) by their definitions. Therefore if I + Ra isfinitely generated, its quotient is also finitely generated. We have an exact sequence

0→ a(I : a)→ I → I/a(I : a)→ 0.

Since (I : a) is finitely generated, a(I : a) is also finitely generated. We further havesurjections

0 −−−→ Rm −−−→ Rm+n −−−→ Rn −−−→ 0y y y0 −−−→ a(I : a) −−−→ I −−−→ I/a(I : a) −−−→ 0y y y

0 0 0where the middle surjection follows from the Five Lemma. Hence I is finitely generated aswell.

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Problem 7.Give an example of a 10 × 10 matrix over R with minimal polynomial (X + 1)2(X4 + 1)which is not similar to a matrix with rational coefficients.

Solution.In R, we may factor (X4 + 1) = (X2 −

√2X + 1)(X2 +

√2X + 1). Therefore choose any

matrix in M10(R) with invariant factors {X2 +√

2X + 1, X2 +√

2X + 1, (X + 1)2(X4 + 1)},which we know exists by appealing to any canonical form. Suppose that there was a matrixα with rational coefficients with these invariant factors. Then in particular, the determinantof the matrix X · I10−α would be the product of the invariant factors, a polynomial in R[X]but not in Q[X]. But since α has only rational entries, and the determinant is a function onthe entries of the matrix, we must have det(X · I10−α) ∈ Q[X], and this is preserved undersimilarity. Therefore α cannot have these invariant factors, so we are done.

Problem 8.Suppose that E/F is an algebraic extension of fields such that every nonconstant polynomialin F [X] has at least one root in E. Show that E is algebraically closed.

Solution.We need to show that every nonconstant polynomial in E[X] has at least one root in E. Ifthis is true, then let α be a root of f(X) ∈ E[X]. Then f(X) = (X − α)g(X). Then g hasa root in E, and by induction on the degree of the polynomial, f splits over E. Thereforeevery polynomial over E[X] splits in E, so E is algebraically closed.

Let α be an algebraic element over E. Since E/F is algebraic, α is algebraic over F , solet mα be its minimal polynomial over F . Let mα split in some K/F . If mα is separable,then K = F (β) for some element β. Since mα(γ) = 0 for some γ ∈ E, we can embed K intoE by β 7→ γ, so that mα also splits in E. In particular, E is perfect.

However, suppose F is not perfect. Let charF = p. Now take the perfect closure of Fin E, i.e. Fperf = {a ∈ E : ap

n ∈ F for some n > 0}. By construction, Fperf is a perfect field.We claim that every irreducible (hence separable) polynomial over Fperf has a root in E,

which would complete the proof. But every f ∈ Fperf[X] has fpk ∈ F [X] for a large enough

k > 0, and f and fpk

have the same roots. Therefore f has a root in E, which completesthe proof.

Problem 9.Prove that if ab = 1 in a semisimple ring, then ba = 1.

Solution.A semisimple ring is artinian and noetherian. Let `a : R → R be left multiplication by a.This map is surjective because abc = c for any c. We claim that it is also injective. Let Rdenote our ring and let Ni = ker `ia be a chain of ideals of R. We have N1 ⊂ N2 ⊂ . . ., sinceax = 0 =⇒ a2x = 0. Therefore this is an ascending chain, and since R is Noetherian, itstabilises at some term n, so that ker `na = ker `n+ma for all m ∈ N. Suppose that x ∈ ker `a.Since `a is surjective, there exists y ∈ R so that `na(y) = any = x. Therefore y ∈ Nn+1.But since Nn+1 = Nn, we have any = 0, hence x = 0. Therefore ker `a = 0, so the map isinjective.

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Take `a(ba− 1). We have aba = a so a(ba− 1) = aba− a = a− a = 0, so by injectivitythis implies that ba− 1 = 0, so ba = 1 as required.

Problem 10.Let A be the functor from the category of groups to the category of (unital) rings taking agroup G to the group ring Z[G] of all finite formal sums

∑g∈G agg with ag ∈ Z. Prove that

A has a right adjoint functor.

Solution.A right adjoint functor to A would be a functor B from Ring to Group such that, for agroup G and ring R,

HomRing(A(G), R) ∼= HomGroup(G,B(R)).

For a ring homomorphism ϕ : Z[G]→ R, every coefficient ag must be mapped to 1R + . . .+ 1R︸ ︷︷ ︸ag times

.

Therefore this ring homomorphism is completely determined by ϕ(g) for each g ∈ G. How-ever, since we we have ϕ(g)ϕ(g−1) = 1R, we must have ϕ(g) ∈ R×. This is only restriction,however, so we claim that the functor B is B(R) = R×. We have shown how every map inthe left set yields a map in the right set. Conversely, given a map ψ : G→ R×, this yields amap ψ : Z[G]→ R by

ψ

(∑g∈G

ag g

)=∑g∈g

1R + . . .+ 1R︸ ︷︷ ︸ag times

· ψ(g).

Therefore B is indeed a right adjoint to A.

3 Spring 2013

Problem 1.Let G be a free abelian group of rank r, so G ∼= Zr as groups. Show that G has only finitelymany subgroups of a given finite index n.

Solution.Let H ⊂ G be a subgroup of index n. Then G acts on G/H by left multiplication, whichinduces a group homomorphism ϕ : G→ Sn. We know that H is the stabiliser of the cosetH ∈ G/H. Labelling this coset 1 (where we view the symmetric group as the permutationsof {1, . . . , n}), the stabiliser is all those elements whose presentation does not contain (1).This is a subgroup H ′ ⊂ Sn which satisfies ϕ−1(H ′) = H.

The above shows that that any subgroup of index n determines a unique morphismϕ : G→ Sn, as the original subgroup is recovered by the action. SinceG is free abelian of rankr, we know that any morphism G → Sn comes (uniquely) from a set map {1, . . . , r} → Sn,of which there are only r ·n!. Therefore there are only finitely many maps G→ Sn, so thereare only finitely many subgroups of index n.

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Problem 2.Assume that L is a Galois extension of the field of rational numbers Q and that K ⊂ L isthe subfield generated by all roots of unity in L. Suppose that L = Q[a], where an ∈ Q forsome positive integer n. Show that the Galois group Gal(L/K) is cyclic.

Solution.We may assume that n is the minimal such n. Since L normal, it is a splitting field ofXn − an over Q since one root of this polynomial lies in L. We see that ζn ∈ L for all nthroots of unity, so µn ⊂ K. An element σ ∈ Gal(L/K) can only send a 7→ ζrna, since theseare the only roots of Xn− an over K. Since L = K[a] and σ(ζn) = ζn since ζn ∈ K, see thatGal(L/K) ↪→ Z/nZ, so it is cyclic.

Problem 3.Let K ⊂ L be an algebraic extension of fields. An element a ∈ L is called abelian if K[a] is aGalois extension of K with abelian Galois group Gal(K[a]/K). Show that the set of abelianelements L is a subfield of L containing K.

Solution.Let L denote the set of abelian elements of L. Since for any a ∈ K, Gal(K[a]/K) = 0, thetrivial abelian group, K ⊂ L. Suppose that a, b ∈ L are arbitrary elements. Then sinceK[−a] = K[a−1], we know that these elements are also abelian. Therefore we need onlyshow that a+ b, ab are abelian elements.

Note that any subfield of an abelian extension is an abelian extension. If K[a]/K isabelian and K ⊂ E ⊂ K[a], then E corresponds to a subgroup H ⊂ Gal(K[a]/K). Since His a normal subgroup of Gal(K[a]/K), E/K is a Galois extension with Gal(E/K) = G/H,which is still abelian.

Now suppose a, b ∈ L. Then by a theorem of Galois theory (e.g. Lang VI, Theorem1.14), Gal(K[a]K[b]/K) ↪→ Gal(K[a]/K)×Gal(K[b]/K). Since the direct product of abeliangroups if an abelian group, K[a]K[b] is an abelian extension of K. Since a+b, ab ∈ K[a]K[b],they generate abelian subextensions of K[a]K[b], and by the above reasoning a + b, ab ∈ L.This completes the proof.

Problem 4.Let F2 be the field with 2 elements and let R = F2[X]. List, up to isomorphism, all R-moduleswith 8 elements.

Solution.We use the classification theorem of modules over a PID. Since R is a finite module, it is inparticular an F2 vector space. We can write

M ∼= R/n1R⊕R/n2R⊕ · · · ⊕R/nrR

for polynomials n1 | n2 | · · · | nr. In our case, we have∑r

i=1 deg ni = 3, so we have threeoptions: r = 1, deg n1 = 3, r = 2, deg n2 = 2, and r = 3, deg n3 = 1. The first case yields 8options. For the second case, we need n2 to be reducible, so we have X(X+1), X2, (X+1)2 aschoices. The first choice yields 2 decompositions, and the latter choices yield 1 decomposition

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each, for a total of 4. For the linear case, we need the same linear term repeated thrice,which is 2 choices. Therefore there are 14 in all, listed by invariant factors below:

{X3 + (0/1)X2 + (0/1)X + (0/1)}{X2, X}, {X2 +X,X + (0/1)}, {X2 + 1, X + 1}{X,X,X}, {X + 1, X + 1, X + 1}.

Problem 5.Let R be a commutative local ring, so R has a unique maximal ideal m.

(a) Show that if x ∈ m then 1− x is invertible.

(b) Show that if R is Noetherian and if a is an ideal such that a2 = a then a = 0.

Solution.(a) If 1 − x is not invertible, then it is contained in some maximal ideal m′ ⊂ R. But R

is a local ring, so we must have 1 − x ∈ m. But then 1 − x + x ∈ m =⇒ m = R, acontradiction. Hence 1−x is not contained in any maximal ideal, so 1−x is invertible.

(b) Viewing a as an R-module, since R is Noetherian, a is finitely generated. Since R islocal, m is its Jacobson radical. Applying Nakayama’s lemma, since a · a = a, there issome element r ∈ R so that r = 1 mod a such that ra = 0. Since r = 1 mod a, wehave 1− r ∈ a ⊂ m. As such 1− (1− r) = r ∈ R×. Then M = r−1rM = 0.

Problem 6.Let D be a division ring of characteristic 0. Assume that D has dimension 2 as a Q-vectorspace. Show that D is commutative.

Solution.We know that 0 6= 1 ∈ D, and since D has characteristic 0, Z ⊂ D. Indeed, Q ⊂ D sinceD is a division ring. Therefore we realise D as a division algebra of dimension 2 over Q. Assuch, Q ⊂ Z(D), and in particular qX = Xq for any q ∈ Q. Therefore take a basis {1, X}for D over Q. Then every element in D is of the form a+ bX. As such, we have

(a+bX)(c+dX) = ac+bXc+adX+bXdX = ca+cbX+dXa+dXbX = (c+dX)(a+bX).

Therefore D is commutative.

Problem 7.Let F = F2 be the field with 2 elements. Show that there is a ring homomorphismF [GL2(F )] → M2(F ) that sends the element g in the group ring to the matrix g ∈ M2(F ).Show that this homomorphism is surjective. Let K be the kernel; since it is a left ideal, it isa (left) GL2(F )-module. Is this module indecomposable? (Reminder: a module is indecom-posable if it is not the direct sum of two proper submodules.) Describe the simple modulesin its composition series.

11

Solution.We can see that 1 · GL2(F ) injects into GL2(F ) ⊂ M2(F ). It is also easy to verify that themap is surjective by direct computation.

Calculating out the rest of this is nasty, but it is easily done. I may update this sectionat a later time.

Problem 8.Let C and D be additive categories, and let Φ : C → D be a functor. Show that if Φ has aright adjoint, then Φ commutes with direct sums and for any two objects x and y in C, themap Φx,y : HomC(x, y)→ HomD(Φ(x),Φ(y)) is a homomorphism.

Solution.Let Ψ : D → C be the right adjoint, so that for a ∈ C and b ∈ D, we have

HomD(Φ(a), b) ∼= HomC(a,Ψ(b)).

Suppose that x =∐

i∈I xi is a direct product in C. Then for any y ∈ D,

HomD(Φ(x), y) ∼= HomC

(∐xi,Ψ(y)

)∼=∏

HomC(xi,Ψ(y))

∼=∏

HomD(Φ(xi), y) ∼= HomD

(∐Φ(xi), y

).

Therefore in D both∐

Φ(xi) and Φ(∐xi) satisfy the universal property of the coproduct,

so they must be equal. Hence Φ commutes with direct sums.We write + for the abelian group action on Hom sets. Then for f, g ∈ HomC(x, y), we

see that

x⊕ xx x

y

incl incl

f g

where the middle arrow induced by the direct sum is f + g. Therefore applying Φ to thisdiagram, we see that Φ(f + g) must be Φ(f) + Φ(g) since Φ commutes with the directsum.

Problem 9.Let D be an associative ring without zero divisors, and assume the centre of D is a field overD which is a finite-dimensional vector space. Prove that D is a division algebra.

Solution.Let 0 6= α ∈ D. Then we claim the vector space endomorphism `α : D → D where β 7→ αβis injective. Indeed, since D has no zero divisors, αβ = 0 implies β = 0. Therefore since Dis a finite dimensional vector space, `α is surjective as well by counting dimensions. Hencethere exists a unique γ ∈ D so that αγ = 1, so α ∈ D×. Therefore D× = D \ {0}, so D is adivision algebra.

12

Problem 10.Let G be a finite group of order n and ρ : G→ GL(V ) be a complex representation of G ofdimension n. Show that ρ cannot be irreducible.

Solution.By general representation theory, we have the sum of the squares of the dimensions of theirreducible representations that comprise ρ must equal the order of the group. Suppose thatρ is irreducible. Then (dim ρ)2 = n2 6= n unless G is the trivial group. Hence ρ cannot beirreducible.

4 Fall 2012

Problem 1.Let p be a prime integer and let G be a (finite) p-group. Write C for the subgroup of centralelements x ∈ G satisfying xp = 1. Let N be a normal subgroup of G such that N ∩C = {1}.Prove that N = {1}.

Solution.Assume that N is nontrivial. First, we claim that any nontrivial normal subgroup N ofa p-group intersects its centre nontrivially. To see this, consider the action of G on N byconjugation. Since G is a p-group, we have |N | ≡ |NG| mod p, where NG = {n ∈ N :g · n = n for all g ∈ G}, which is a consequence of the orbit-stabiliser theorem. Since|N | ≡ 0 mod p, then p | |NG|. Since 1 ∈ NG, |NG| ≥ 1, hence NG is a nontrivial subgroupof N which intersects Z(G) nontrivially.

Since NG is a subgroup of a p-group, it too is a p-group, so it contains an element oforder p. Hence that element x ∈ N ∩ C, a contradiction. Therefore N must be a trivialnormal subgroup.

Problem 2.Let A be an n× n matrix over F having only one invariant factor. Prove that every n× nmatrix over F that commutes with A is a polynomial in A with coefficients in F .

Solution.Consider A acting on V = F n. Since A has only one invariant factor, there exists a vectorv ∈ V so that {Av, . . . , Anv} is a basis for V . Then we can write B(Av) =

∑ni=1 aiA

iv usingthe given basis. But we also have B(Av) = A(Bv), i.e.

n∑i=1

aiAiv = A(Bv) =⇒ Bv =

n∑i=i

aiAi−1v.

Let C =∑n

i=i aiAi−1. We claim that B = C on all of V . Since any vector in w ∈ V may be

written as p(A)v for some p ∈ F [X] of degree n− 1 (using the basis {v,A, . . . , An−1v} now),we have

Bw = Bp(A)v = p(A)Bv = p(A)Cv = Cp(A)v = Cw.

Therefore B is a polynomial in A.

13

Problem 3.Let F be a field and let n be a positive integer such that F has no nontrivial field extensionsof degree less than n. Let L = F (x) be a field extension with xn ∈ F . Prove that everyelement in L is a product of elements of the form ax+ b with a, b ∈ F .

Solution.Since x satisfies Xn − xn = 0, we know [L : F ] ≤ n. If [L : F ] < n, then L = F byassumption, so any element b ∈ F is expressible in the form 0 · x + b, and we are done.Therefore suppose [L : F ] = n.

Consider y ∈ L. Then y =∑m

i=0 aixi for some ai ∈ F , where m < n. Consider the

corresponding polynomial f(X) =∑m

i=0 aiXi. Then we claim f splits over F . Suppose not.

Then let α be a root of f . then [F (α) : F ] ≤ deg f = m < n, so F (α) = F , i.e. α ∈ F .Hence we can write f(X) = (X −α)g(X). By induction, this shows that f splits. Thereforewe obtain a factorisation of f(X) into linear factors:

f(X) =m∏i=i

(ciX + di).

This corresponds with a factorisation

β =m∏i=1

(cix+ di)

as claimed.

Problem 4.Let F be the functor from the category of rings to the category of sets taking a ring R tothe set {x2 : x ∈ R}. Determine whether F is representable.

Solution.Suppose it were representable. Then let F = Hom(R,−). Then in particular, Hom(R,R)must be in bijection with R2. Let x ∈ R2 be the element corresponding to the identity1R ∈ Hom(R,R), and let x = y2. Then for a map g : R → S, we have the commutativediagram

Hom(R,R)g◦−−−−→ Hom(R, S)

∼yαR ∼

yαS

F (R)F (g)−−−→ F (S)

1R has the unique property that g ◦ 1R = g, so we see what this means for the bottom row.x ∈ F (R) has the unique property that for every b2 ∈ F (R), there is a unique g : R→ S sothat F (g)(x) = b2, i.e. g(x) = b2.

But this is evidently not the case. Consider S = Z[X]. Then there should be a uniquegroup homomorphism g : R → S such that g(x) = X2, so evidently g(y) = ±X. But sincethere is an automorphism ϕ of S interchanging X and −X, we see that (ϕ ◦ g)(x) = g(x) =X2, so this homomorphism is not unique. Therefore F is not representable.

14

Problem 5.Let G and H be finite groups and let V and W be irreducible (over C) G- and H-modulesrespectively. Prove that the G×H-module V ⊗W is also irreducible.

Solution.We use the theory of characters. If the character of V ⊗W is 1, then V ⊗W is irreducible.Since tr(V ⊗W ) = trV · trW , if ρG, ρH are characters on G and H, then the character θ onG×H is given by θ(g, h) = ρG(g)ρH(h). Then

|θ|2 = 〈θ, θ〉 =1

|G×H|∑

(g,h)∈G×H

θ(g, h)θ(g, h).

Rearranging this to suit our case,

|θ|2 =1

|G|∑g∈G

(ρG(g)ρG(g)

(1

|H|∑h∈H

ρH(h)ρH(h)

))

=1

|G|∑g∈G

(ρG(g)ρG(g)|ρH |2

)= |ρH |2

(1

|G|∑g∈G

ρG(g)ρG(g)

)= |ρG|2|ρH |2 = 1

since ρG and ρH were irreducible characters on G and H. Therefore |θ|2 = 1, so θ is anirreducible character, and V ⊗W is irreducible as desired.

Problem 6.Let Dn be a dihedral group of order 2n > 4; so, it contains a cyclic subgroup C of order n onwhich σ ∈ Dn outside C acts as σcσ−1 = c−1 for all c ∈ C. When is the cyclic subgroup Cwith the above property unique? Determine all n for which Dn has a unique cyclic subgroupC and justify your answer.

Solution.Let r denote the rotation and s the reflection in Dn. We claim that all elements of order nare of the form rk. Indeed, we have (srk)2 = (srk)(r−ks) = 1 for any k ∈ {0, . . . , n − 1}.Therefore the only elements of order n are the usual rk with (k, n) = 1. Therefore in everycase, the cyclic subgroup C is 〈r〉.

Problem 7.Let D be a central simple division algebra of dimension 4 over a field F . If a quadraticextension K/F can be isomorphically embedded into D as F -algebras, prove that D ⊗F Kis isomorphic to M2(K) as K-algebras.

Solution.Assume charF 6= 2. Let K = F (

√α). Recall that for quaternion algebras, the cases

of division algebra and matrix algebra and mutually exclusive and collectively exhaustive.Since D is central and simple, D ⊗F K is a central simple algebra over K of dimension 4,

15

so it suffices to prove that it is not division. Indeed, since K may be embedded into D, wehave K ⊗F K ⊂ D ⊗F K as a subalgebra. We claim that

√α⊗ 1− 1⊗

√α

is a zero divisor. Indeed,

(√α⊗ 1 + 1⊗

√α)(√α⊗ 1− 1⊗

√α) = α⊗ 1−

√α⊗√α +√α⊗√α− 1⊗ α

= α⊗ 1− 1⊗ α = (α− α)⊗ 1 = 0

since in the last step, we may move α over the tensor product. Hence D⊗FK is not division,so it is isomorphic to M2(K).

If the characteristic of F is 2 and the extension is separable, the above argument holdswith (roughly) β ⊗ 1 + 1⊗ γ, where β, γ are the roots of the quadratic polynomial of whichK is the splitting field.

Problem 8.How many monic irreducible polynomials over Fp of prime degree l are there? Justify youranswer.

Solution.We claim that the product of all irreducible polynomials of degree d | n equals f(X) =Xpn − X. Given any α ∈ Fpn , f(α) = 0. Hence the irreducible polynomial of α dividesf , and the degree of α divides the degree of the extension n. Conversely, for any monicirreducible polynomial g of degree d | n, g has roots Fpd . Since Fpd ⊂ Fpn , all the roots of αlie in Fpn , so g | f . Hence

Xpn −X =∏

deg g=d|n

g(X).

Now, since l is prime, the only divisors of l are 1 and l. There are evidently p irreduciblepolynomials of degree 1, and so the degree of the product of the polynomials of degree l is

pl − p. Since each of those polynomials has degree l itself, their number is pl−pl

.

Problem 9.Consider a covariant functor F : R 7→ R× from the category of commutative rings withidentity to the category of sets. LetG = Autfunctors(F ) be made up of natural transformationsI : F → F having an inverse J : F → F ∈ G such that I ◦ J = J ◦ I is the identity naturaltransformation. Prove that F is representable, that G is a finite group, and find the orderof the group G with justification.

Solution.First, F is representable by Z[X,X−1], the polynomial ring. Let ϕ ∈ Hom(Z[X,X−1], R) bea homomorphism. Then ϕ(1) = 1R, and by extension ϕ(n) = n · 1R for any n ∈ Z. Further,X may be sent to any invertible element in R, and this determines the image of X−1 aswell. Further, these are the only homomorphisms Z[X,X−1] → R. Hence F is representedby Z[X,X−1].

16

We claim that Autfunctors F ∼= AutRings Z[X,X−1]. Assuming this, the only nontrivialring automorphism of Z[X,X−1] sends X 7→ X−1, and we have Autrings Z[X,X−1] ∼= Z/2Z.Indeed, by Yoneda’s lemma we have

Homfunctors(F ) ∼= Homfunctors(hZ[X,X−1], F ) ∼= F (Z[X,X−1]) = {±1, X±1}.

However, the homomorphisms corresponding to X 7→ {±1} are not invertible homomor-phisms, so we are left with the automorphisms X 7→ {X±1}. This completes the proof.

Problem 10.For a finite field of order q, consider the polynomial ring R = F[X], and let L be a freeR-module of rank 2. Give the number of R-submodules of M such that XL ( M ( L andjustify your answer.

Solution.First, we simplify the problem by quotienting out by XL, so obtain the chain

0 ( N ( L/XL ∼= F× F.

where N is a submodule. Therefore we need to find all dimension 1 subspaces of F2. Theseare entirely determined by nonzero vectors of F2, of which there are q2 − 1, but divided bythe nonzero spans of each vector space, which yields

q2 − 1

q − 1= q + 1.

5 Spring 2012

Problem 1.Let V be a finite dimensional space over Q and G ⊂ GL(V ) a finite subgroup. Prove thatthe Q-subalgebra of End(V ) generated by G is semisimple.

Solution.By Maschke’s theorem, the group algebra of a finite group by a field of characteristic zero issemisimple. In particular [1]:

Assume charF - |G|. Let V be a finite dimensional F[G]-module, and let W ⊂ V be asubmodule. Choose an idempotent element e ∈ EndF(V ) so that eV = W . Let

e :=1

|G|∑g∈G

geg−1,

where we view g ∈ EndF(V ). Then

he =1

|G|∑g∈g

hgeg−1 =1

|G|∑g∈G

(hg)e(hg)−1h = eh.

17

for all h ∈ G, and thus e ∈ EndF[G](V ). Since W ⊂ V is a submodule, e satisfies eV ⊂ Wand e|W = idW . Hence e is an idempotent satisfying eV = W , so we have V = W ⊕ (1− e)V .Repeating this process, we can reduce V into a direct sum of irreducible submodules, soevery F[G]-module is semisimple.

Problem 2.Let V be the vector space of all a ∈ Rn such that a1 + . . . + an = 0. The symmetric groupSn acts naturally on V . Prove that the Sn-module V is simple.

Solution.Note that the vectors B = {e1 − ei : i ∈ {2, . . . , n}} form a basis for V . V is an (n − 1)-dimensional vector space since it is precisely the orthogonal complement to e1 + . . . + enunder the usual inner product, and the n− 1 vectors B are linearly independent and in V .

Therefore let W be a submodule of V , and let v = (v1, . . . , vn) ∈ W be a nonzero vector.Then two entries vi 6= vj are both nonzero, since only one nonzero entry contradicts

∑vi = 0.

Therefore applying an appropriate permutation, we may assume v1 6= v2 are nonzero. Thenapplying the transposition τ = (1 2), we have

τv − v = (v1 − v2, v2 − v1, 0, . . . , 0) ∈ W.

Multiplying by (v1 − v2)−1, we have e1 − e2 ∈ W . Then applying the transpositions (2 i) fori ∈ {3, . . . , n}, we see that B ⊂ W . Therefore W = V , so V is simple.

Problem 3.Let R be a commutative local ring, and P a finitely generated projective R-module. Showthat P is a free module.

Solution.This is a consequence of Nakayama’s lemma. Let P = Rm1 + . . .+Rmk be an expression ofP with the minimal number of elements. Then we have an exact sequence

0→ N = kerϕ→ Rk ϕ→ P → 0,

and since P is projective, we have Rk ∼= P ⊕ N . Therefore we need to prove that N = 0.Let m be the unique maximal ideal in R. Applying the functor −⊗R/m, we have

(R/m)k ∼= P/mP ⊕N/mN.

We claim that dimR/m(R/m)k = dimR/m(P/mP ) = k. If dimR/m(P/mP ) < k, then byNakayama’s lemma P can be generated by fewer than k elements, which is a contradiction.Thus we have N/mN = 0. By Nakayama’s lemma, since J(R) = m, mN = N implies N = 0.Hence Rk ∼= P , so P is free.

Problem 4.Let R be the subring of M3(R) consisting of all matrices (aij) with a31 = a32 = 0. Determinethe Jacobson radical of R.

18

Solution.We use that x ∈ J(R) if and only if xy − 1 ∈ R× for all y ∈ R. We use the general formula a b c

d e f0 0 g

α β γδ ε η0 0 ζ

− 1 0 0

0 1 00 0 1

=

aα + bδ − 1 aβ + bε aγ + bη + cζdα + eδ dβ + eε− 1 dγ + eη + fζ

0 0 gζ − 1

.

If g 6= 0, then ζ = 1/g creates a zero row on the bottom, so the matrix cannot be invertible.Playing around with this more, you will see that J(R) = {α ∈ R : a = b = d = e = g =0}.

Problem 5.Let G be a finite group, K a normal subgroup, and P a Sylow p-subgroup of G. Prove thatP ∩K is a Sylow p-subgroup of K.

Solution.Let C = P∩K. Since C ⊂ P , C is a p-subgroup of K, so it sits inside some Sylow p-subgroupD of K. Likewise D sits inside some Sylow p-subgroup Q of G. Since Q is conjugate to P(as all Sylow p-subgroups are), write gQg−1 = P . We have |C| ≤ |D| = |Q ∩K|. Further,|C| = |P ∩K| = |Q ∩K|. To see this, suppose q ∈ P ∩K. Then gqg−1 ∈ g(P ∩K)g−1 =gPg−1 ∩K = Q ∩K, so the elements are in bijective correspondence. Hence |C| = |D|, soC is a Sylow p-subgroup.

Problem 6.Determine the Galois group of the polynomial X8 + 16 over Q.

Solution.Let f be this polynomial. Beginning with one root, we see ζ16

√2 is a root of f , where ζ16 is

a primitive 16th root of unity. Therefore we construct the Galois group in two steps:

Q ⊂ Q(ζ16) ⊂ K,Q ⊂ Q(√

2) ⊂ K

where K is the splitting field of X8 + 16. Evidently Q(ζ16) and Q(√

2) are both Galois overQ, but their intersection is not Q. In particular, we have

ζ216 = ζ8 = eiπ/4 = cos(π/4) + i sin(π/4) =

√2 + i

√2

2.

So that Q(√

2) ∩ Q(ζ8) = Q(√

2). Therefore we see that K = Q(ζ16), and it is easy toverify that for any σ ∈ Gal(K/Q), we have σ(

√2) =

√2. Hence the Galois group of this

polynomial is (Z/16Z)× ∼= Z/2Z× Z/4Z.

Problem 7.Let F be a finite field with pn elements. Find the number of elements of F that can bewritten in the form ap − a for some a ∈ F .

19

Solution.Let f = Xp −X ∈ Fp[X], which is a vector space endomorphism of F . Its kernel is simplyroots of the polynomial Xp − X, of which there are p. Therefore | im f | = |F |/| ker f |, wehave | im f | = pn−1. Therefore pn−1 elements of F can be written in this form.

Problem 8.Let R be a reduced (no nonzero nilpotent elements) commutative ring that has a uniqueproper prime ideal. Show that R is a field.

Solution.If the prime ideal in question is the zero ideal, then R is a field, since its unique proper primeideal is in particular maximal, so that R/0 ∼= R is a field.

Since R is reduced, it has a zero nilradical. But the nilradical is precisely the intersectionof all prime ideals, of which there is only one. Therefore the only prime must be zero, so weare done.

Problem 9.Let R be a flat commutative Z-algebra, and ModR the category of R-modules. Supposethat I is an injective R-module. Show that the underlying abelian group of I is divisible.

Solution.Suppose that I is indivisible as an abelian group. Therefore there exists n ∈ N and x ∈ Iso that x 6= n · y for any y ∈ I. Consider the map f : Z→ Z given by 1 7→ n, and the mapg : Z → I given by 1 7→ x. Then if I were injective, there exists a unique lifting h : Z → Iso that h ◦ f = g. In particular,

x = g(1) = h(f(1)) = h(n) = n · h(1),

which is a contradiction since x is not divisible by n. Therefore as a Z-module, I is notinjective. Suppose that I is injective as an R-module. Then tensoring over Z by R, exactsequences are preserved by flatness, so we have f : R → R, g : R → I, and a unique lifth : R → I so that g = h ◦ f . Then we have x = g(1) = n · h(1) as before. Since nothinghere depends on the R-module structure, this map descends back down to Z, which is acontradiction. Therefore I cannot be an injective R-module, and we are done.

Problem 10.Determine the automorphism group of the symmetric group S3 up to isomorphism.

Solution.Recall that AutS3 is comprised of inner automorphisms and outer automorphisms, and thatfor a general group G, the inner automorphisms of G is isomorphic to G/Z(G). In our case,Z(S3) ∼= 1, so InnS3

∼= S3.We claim that every automorphism of S3 is inner. By the above, we have a map S3 →

AutS3 by σ 7→ ϕσ, where ϕσ(ρ) = σρσ−1. For any automorphism ϕ, the set of transpositionsin S3 is preserved, which induces a group homomorphism from AutS3 to the symmetric groupof set of transpositions, i.e. AutS3 → S3. This map is injective since the set of transpositionsgenerates S3, so two different automorphisms cannot permute the transpositions in the sameway. Therefore |AutS3| ≤ |S3|, so our original injective map S3 → AutS3 must be surjectiveas well. Hence S3

∼= AutS3.

20

6 Fall 2011

Problem 1.For a finite field F, prove that the order of the group SL2(F) is divisible by 6.

Solution.Let F = q. Consider the determinant map det : GL2(F) → F×, which is a group homomor-phism since det(ab) = (det a)(det b). Then SL2(F) ∼= ker det. Additionally, det is clearly asurjective map. By the first isomorphism theorem,

|GL2(F)/SL2(F)| = |F×|.

Since all groups are finite, we may rewrite this

|SL2(F)| = |GL2(F)|/|F×|.

|F×| = q − 1, since only 0 ∈ F \ F×. Now we calculate |GL2(F)|. Recall that a nonsingularmatrix is equivalent to choosing two linearly independent vectors in F2. For the first vector,choose any v 6= (0, 0) ∈ F2. For the second, we cannot select any vector in the span of v, ofwhich there are q. Hence

|GL2(F)| = (q2 − 1)(q2 − q) = (q − 1)2q(q + 1) =⇒ |SL2(F)| = (q − 1)q(q + 1).

Given any three consecutive integers, one of them is divisible by 3, and one of them isdivisible by 2. Hence their product is divisible by 6.

Problem 2.Let G be a non-trivial finite group and p a prime. If every subgroup H 6= G has indexdivisible by p, prove that the centre of G has order divisible by p.

Solution.Let G act on itself by conjugation. Then the orbits of this map are precisely the conjugacyclasses of G. We have, letting x run over representatives of conjugacy classes,

G =⋃x∈G

Gx = Z(G) ∪⋃

x∈G,|Gx|>1

Gx.

By the orbit-stabiliser theorem,

|G| = |Z(G)|+∑

x∈G,|Gx|>1

|Gx| = |Z(G)|+∑

x∈G,|Gx|>1

|G : Gx|,

where Gx is the stabiliser of x in G under this action. Since each Gx is a subgroup, byassumption it has index divisible by p. Since |G : {e}| = |G| is also divisible by p, we musthave p | |Z(G)| as well. This completes the proof.

Problem 3.Let R be a local UFD of Krull dimension 2. Let π ∈ R be neither zero nor a unit. Provethat R[1/π] is a PID.

21

Solution.Since prime ideals in R[1/π] correspond precisely to prime ideals in R not containing π, wesee that R[1/π] has dimension at most 1, since the unique maximal ideal m ⊂ R is no longera prime ideal in the localisation. If R[1/π] has dimension 0, then it is a field, so a fortiori aPID.

Now R[1/π] is a dimension 1 UFD, so we claim that every prime ideal is principal. Letp be a prime ideal. Then we may find a ∈ p which is an irreducible element by repeatedlyreducing any element. Since we are in a UFD, (a) is a prime ideal, so we have a chain0 ( (a) ⊂ p. Since R[1/π] has dimension 1, we must have (a) = p, so every prime ideal isprincipal.

We claim that this is sufficient to prove that every ideal of R[1/π] is principal. Otherwise,consider the set S of nonprincipal ideals in R[1/π]. For some ascending (by inclusion) chain{ai} ∈ S, the ideal a =

⋃ai is an upper bound of this chain. If a /∈ S, then a is principal.

Let a = (x). Then x ∈ ai for some i, and hence a = (x) = ai ⊂ a. But this implies that aprincipal ideal is in S, which is not possible. Therefore a is not principal, so applying Zorn’slemma we let b be a maximal element of S.

Since b is not principal, in particular it is not prime. Therefore there exists a, b /∈ b sothat ab ∈ b. Let ba = b + (a) and bb = b + (b). Since b ( ba, bb, these ideals are not in S,so they are principal. Let ba = (α). Examine the quotient ideal

k = (b : ba) := {x ∈ R[1/π] : xba ⊂ b}.

We claim that b = bak. Clearly we have bak ⊂ b. Conversely, suppose that x ∈ b. Then wecan write x = yα. But then (x) = (yα) = yba ⊂ b, so y ∈ k, so x ∈ bak.

Now we finally note that k ⊂ bb. If x ∈ (a), y ∈ (b), then for z, z′ ∈ b we have

(z + x)(z′ + y) = zz′ + zx+ z′y + xy ∈ b.

Therefore k /∈ S, so k = (β) for some β. Hence b = (α)(β) = (αβ), a contradiction. ThereforeS must be empty, so R[1/π] is a PID.

Problem 4.Let p be a prime. Prove that the nilradical of the ring Fp[X] ⊗Fp[Xp] Fp[X] is a principalideal.

Solution.We claim that the nilradical is generated by X ⊗ 1− 1⊗X. Let n = (X ⊗ 1− 1⊗X). Thenfirst, we see

(X ⊗ 1− 1⊗X)p = Xp ⊗ 1− 1⊗Xp = 0

since the lower degree terms are zero in characteristic p and we may move Xp over the tensorproduct, so n is contained in the nilradical N .

Now we use the property that R/N has no nontrivial nilpotents. We examine R/n. Sincewe are modding out by the relation X ⊗ 1 = 1 ⊗ X, we may move X across the tensorproduct freely. That is,

R/n ∼= Fp[X]⊗Fp[X] Fp[X].

But this is isomorphic to Fp[X] alone, which is a domain, so has no nontrivial nilpotents.Hence n = N , so N is principal.

22

Problem 5.Let F be a finite field and let F be an algebraic closure of F. Let K be a subfield of Fgenerated by all roots of unity over F. Show that any simple K-algebra of finite dimensionover K is isomorphic to the matrix algebra Mn(K) for a positive integer n.

Solution.By the Artin-Wedderburn theorem, a simple K-algebra of finite dimension of isomorphic tosome matrix algebra Mn(D) for a division algebra D/K of finite dimension. Therefore itsuffices to show that D = K.

We claim that, in fact, K = F. Let F/L/F be an algebraic extension. Then fine F/Fp isa finite extension over its prime field, we have L/Fp is an algebraic extension, so L ∼= Fpnfor some n ∈ N. Then every element of L satisfies the equation Xpn − X = 0, and allnonzero elements satisfy Xpn−1 = 1. Therefore the elements of L are pn− 1th roots of unity,so L ⊂ K. Therefore K contains all algebraic extensions of Fp, so it contains all algebraicextensions of F, and therefore K = F. Therefore there does not exist a nontrivial divisionalgebra over K = F, so every simple K-algebra must be isomorphic to Mn(K) for somen > 0.

Problem 6.Let R be a commutative ring and let M be a finitely generated R-module. Let f : M →Mbe R-linear such that f ⊗ id : M ⊗R R[T ] → M ⊗R R[T ] is surjective. Prove that f is anisomorphism.

Solution.We have an exact sequence

0→ ker f →Mf→ im f → 0.

Applying the tensor −⊗R R[T ] is right exact, so we have an exact sequence

M ⊗R R[T ]f⊗id−→ im f ⊗R R[T ]→ 0

Therefore f is surjective. We need to prove that f is also injective. We claim that anysurjective map f : M →M for a finitely generated module is also injective.

Let A = EndRM , the endomorphism ring over which M is also a module, and considerthe left ideal a = (f) ⊂ A. Then aM = M since f is surjective. Then by Nakayama’slemma, there exists g ∈ A so that gM = 0 and g ≡ 1M mod a. Therefore

g(m) = 0 =⇒ m+ f(m) · h(m) = 0

for some h ∈ A. Suppose that f(m) = 0. Then we must have m = 0 as well, so f is injective.Therefore f is an isomorphism.

Problem 7.Let C be the category of semi-symplectic topological quantum paramonoids of Rice-Paddytype, satisfying the Mussolini-Rostropovich equations at infinity. Let X, Y be objects of Csuch that the functors MorC(X,−),MorC(Y,−) are isomorphic as covariant functors from Cto Sets. Show that X and Y are isomorphic in C.

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Solution.We know by Yoneda’s lemma that the natural transformations between F and hA, thecovariant functor represented by A, is isomorphic to F (A). In our case, we have a naturaltransformation α : hX

∼−→ hY , so there is a corresponding natural map in f ∈ hY (X) =MorC(Y,X). We claim that this map is the desired isomorphism.

Let α−1 : hY∼−→ hX be the inverse of α. This corresponds to a map g : X → Y , and we

claim f ◦ g = idX . But since α−1 ◦ α = idhX , and idhX must correspond to the identity map1X , we are done.

Problem 8.Let Γ be the Galois group of the polynomial X5 − 9X + 3 over Q. Determine Γ.

Solution.First, by Eisenstein’s criterion, this polynomial f(X) is irreducible. For a root α of f(X),we know [Q(α) : Q] = 5, so 5 | |Γ|. Therefore Γ contains an element of order 5, which mustbe a 5-cycle since Γ ⊂ S5.

Now we claim Γ contains a transposition. If so, by the well-known fact that Sp is generatedby a p-cycle and any transposition, Γ = S5. To prove the claim, first note that f(X) hasone real root. We claim that it has exactly three real roots. We see f ′(X) = 5X4− 9, whichhas real roots at ± 4

√9/5 only. Therefore the graph of f(X) crosses the X-axis at most

thrice, and indeed by the intermediate value theorem we can check that it crosses the X-axisprecisely three times. Therefore a valid permutation of the roots is complex conjugation,which is a transposition in Γ because there are only two complex roots. This completes theproof.

Problem 9.

(a) Is there a group G with C[G] isomorphic to C× C×M2(C)?

(b) Is there a group G with Q[G] isomorphic to Q×Q×M3(Q)?

Solution.We use throughout the Artin-Wedderburn theorem since group algebras of finite groups byfields of characteristic zero are semisimple. Further, we know that dimC[G] = |G|, and thenumber of factors is equal to the number of conjugacy classes.

(a) Since dimC[G] = 6, we need a group of order 6 with three conjugacy classes. Thegroup G ∼= D6 = S3 works, because the three conjugacy classes are the three cycletypes.

(b) Since dimQ[G] = 11, we must have G = Z/11Z. But this group has 11 conjugacyclasses, so we cannot have a decomposition as above. Therefore no such group exists.

Problem 10.Let K/k be an extension of finite fields. Show that the norm NK/k : K → k is surjective.

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Solution.Let charK = p. Recall that

NK/k(α) =∏

σ∈Gal(K/k)

σ(α).

Let [K : k] = n. Then in particular, Gal(K/k) ∼= Z/nZ, and it is generated by the Frobeniusmap ϕ(a) = aq, where q = |k|. Therefore in general,

NK/k(α) =∏

σ∈Gal(K/k)

σ(α) = α · ϕ(α) · · ·ϕn−1(α) = α1+q+···+qn−1

Let β be a generator of K×. Then we claim that NK/k(β) generates k×. We see that

NK/k(β)q−1 = (β1+q+···+qn−1

)q−1 = βqn−1 = 1.

In words, since β has order qn− 1, β1+q+···+qn−1has order qn− 1/(1 + q+ · · ·+ qn−1) = q− 1.

Therefore this element generates k×, so in particular the norm map is surjective.

7 Spring 2011

Problem 1.Let G be a group of order n. Show that there are two subgroups H1 and H2 of the symmetricgroup Sn, both isomorphic to G such that h1h2 = h2h1 for all h1 ∈ H1 and h2 ∈ H2.

Solution.Consider the action on G on itself by left multiplication. Since this action is faithful, wehave an injection of G into Sn, so we may identify G ∼= H1 ⊂ Sn. Similarly, the action ofG on itself by right multiplication, which must be given by g · x = xg−1, is also faithful andyields an embedding onto H2 ⊂ Sn. We claim that these subgroups of Sn commute.

Indeed, given x ∈ G, we have h1h2 · g = h2h1 · g, since (if we identity group elementsin G with group elements in Sn) we have h1(gh

−12 ) = (h1g)h−12 by associativity of group

multiplication. Therefore these elements commute in Sn, so we are done.

Problem 2.Let G be a finitely generated group. Show that G contains only finitely many subgroups ofany fixed finite index.

Solution.Let G = 〈g1, . . . , gn〉, and let m > 0 be a positive integer. Let H be a subgroup of indexm. Then G acts on the set of cosets G/H by left multiplication, which induces a grouphomomorphism ϕ : G→ Sm. We can realise H as the set {g ∈ G : ϕ(g) ·H = H}. Thereforeany choice of H produces a unique ϕ : G → Sm. Since G is generated by n elements, thereare at most n · |Sm| homomorphisms G → Sm. Therefore there are at most n · |Sm| < ∞subgroups of index m.

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Problem 3.Let R be a Noetherian domain. Show that every nonzero nonunit in R is a product ofirreducible elements and that R is a UFD if and only if every nonzero nonunit in R is aproduct of prime elements, where an element is prime if it generates a nonzero prime ideal.

Solution.Let S ⊂ R be the family of the principal ideals of elements without an irreducible factorisa-tion. If this family is nonempty, then since R is Noetherian it contains a maximal element(a). Since a does not have an irreducible factorisation, it itself is not irreducible, so we maywrite a = bc for some b, c ∈ R \ R×. As such, (a) ( (b). Therefore (b) /∈ S, so b has a fac-torisation into irreducible elements p1 · · · pn. Similarly, (a) ( (c), so c = q1 · · · qm. Thereforea = bc = p1 · · · pn · q1 · · · qm is a factorisation of a into irreducible elements, hence S is empty.

Suppose that R is a UFD, so that this factorisation is unique. Then we claim that, infact, every irreducible element is prime. Indeed, let x be an irreducible element, and letp = (x). Then suppose that y, z ∈ R/p, where y, z ∈ R, satisfy yz = 0. Then yz ∈ (x),which implies that yz = xa for some a ∈ R. Therefore the unique irreducible factorisation(up to association) of yz contains x, so the unique irreducible factorisation of either y or zcontains x, which implies that y or z is an element in (x). Hence y or z = 0, so R/p is adomain. Therefore (x) is prime, so x is a prime element. As such, since every element of Rhas a factorisation into irreducible elements, and each of these elements is prime, then everyelement of R has a factorisation into prime elements.

Conversely, let x = p1 · · · pn be a factorisation of x into prime elements, and let x =q1 · · · qm be a factorisation into irreducible elements. Then since p1 | q1 · · · qm, we have p1 | qifor some i, and with reordering we may assume i = 1. Therefore write q1 = p1a1. Since q1was irreducible, we must have a1 ∈ R×. Since we are in a domain, we may cancel the termp1 from both sides to obtain

p2 · · · pn = q2 · · · qm · a1.

Continuing this process, we obtain

1 = qm−n · · · qn · a1 · · · an.

Hence we must have m = n. Therefore the two factorisations above were the same up torearrangement and association, which implies that R is a UFD.

Problem 4.Prove that every prime ideal in Z[t] can be generated by two elements.

Solution.Let p ∈ Z[t] be a prime ideal. Then p∩Z is a prime ideal in Z, since (Z[t]/p)∩Z = Z/(p∩Z).Therefore p ∩ Z is a principal ideal, either 0 or pZ for p a prime element.

Suppose that p ∩ Z = 0. Then we claim that p is principal in Z[t]. Let f ∈ p be apolynomial of minimal degree in p, which by assumption has deg f > 0 since p contains noconstant terms. Write f = c(f)f ′, where f ′ is a primitive polynomial. Then since p is prime,either c(f) ∈ p or f ′ ∈ p, but c(f) /∈ p since c(f) ∈ Z. Hence f ′ is a primitive polynomialin p of minimal degree. We claim that p = (f ′). Let g ∈ p. Suppose that f ′ - g. Then

26

the greatest common divisor of f ′ and g would have degree less than that of f ′ (since f ′ isprimitive). By the Euclidean algorithm we may realise gcd(f ′, g) as a linear combination off ′ and g, so gcd(f ′, g) ∈ p, which contradicts the minimality of f ′. Hence f ′ | g, so (f ′) = p.

Now suppose that p∩Z = pZ. Consider the image of p in Z/pZ[X]. Then p is a principalideal, so write p = f · Z/pZ[X]. If we let f be a polynomial lying over f in p, then we havef = f + g(X), for g(X) ∈ pZ[X]. But since p ∈ p, g(X) ∈ p, so in fact the same f lies in p,which we rename f ′ to avoid confusion.

We claim that p = (p, f ′). Certainly (p, f ′) ⊂ p. Let g ∈ p, and examine g in Z/pZ[X].Since g ∈ p, we have g = h · f ′, where h is a polynomial in Z/pZ[X]. Then moving back topreimages, let h = h+ h′ and g = g + g′ for g′, h′ ∈ pZ[X]. Then we have

g = (h+ h′)f ′ + g′ ∈ (f ′) + (p) = (p, f ′)

which proves our claim.

Problem 5.Let F be a field having no nontrivial field extensions of odd degree and K/F a finite fieldextension. Show if K has no field extensions of degree two, then F is perfect and K isalgebraically closed.

Solution.Suppose that F is not perfect, and let E/F be an extension. Then in particular, let α ∈ Ebe an inseparable element so that F (α)/F is a purely inseparable extension. Then [F (α) :F ] = pn for some n ≥ 0. Suppose p is odd. Then we must have [F (α) : F ] = 1 since everyextension of odd degree is trivial. Therefore α ∈ F . Since this is true for any α, E/F is aseparable extension, so F is perfect.

If p = 2, then the case is somewhat different. We claim that K is a perfect field. LetL/K be a finite extension, and let α ∈ L be an inseparable element not in K. Then[K(α) : K] = 2n for some n > 0, where the minimal polynomial of α is X2n − α2n . Hencethe degree of α2n−1

is 2, so α2n−1 ∈ K, which is a contradiction. Hence L/K is separable, soK is perfect.

We claim that any finite subextension of a perfect field is perfect. We prove this ingeneral. Let K/F be a finite extension, where K is perfect and charK = p. Then let α ∈ Kbe an inseparable element. Choose n minimal so that αp

n ∈ F , so [F (α) : F ] = pn, andassume n > 1. Since K is perfect, Kp = K, so let β ∈ K so that βp = α. Hence βp

n+1 ∈ F ,so [F (β) : F ] = pn+1. Continuing this process, we can create arbitrarily large subextensionsof K over F , so K/F is infinite, a contradiction. Hence we must have α ∈ F , so K/F is aseparable extension. Suppose now that L/F is a finite field extension. Then if K/L/F , wehave L/F is separable since K/F is separable. If L/K/F , we have L/K is separable sinceK is perfect, so L/F must be separable since each subextension is separable. Therefore Fis perfect.

Therefore let E/K be a finite field extension and L be its normal closure over F . ThenL/F is Galois. Let G be the Galois group of L/F , and write |G| = 2nm, where m is odd. ByGalois correspondence, there exists a field of degree m over F , which by assumption must betrivial. Hence m = 1, so G is a 2-group. By the general theory of p-groups, the subgroups ofG are nested with orders consecutive powers of 2. In particular, let H ⊂ G be the subgroup

27

corresponding to K/F . Then there is a subgroup H ′ ⊃ H corresponding to a field M/K ofdegree 2. But K has no field extensions of degree 2, so no such H ′ can exist, which impliesthat G = H. Hence this collapses L = K, so in particular E = K. Hence K has no finiteextensions, therefore it has no algebraic extensions, so it is algebraically closed.

Problem 6.Prove that over a finite field there are irreducible polynomials of any positive degree.

Solution.Let Fq be a finite field, where q = pn. Then for any positive integer m there is a finite field oforder qm = pnm, and Fp ⊂ Fq ⊂ Fqm is a tower of field extensions. Therefore [Fqm : Fq] = m.Further, since (Fqm)× is cyclic with some generator α, we have Fq(α) = Fqm . Therefore wehave m = [Fqm : Fq] = degmα, the minimal (irreducible) polynomial of α. Therefore thereis an irreducible polynomial of degree m for any m > 0.

Problem 7.Let T : V → V be a linear operator on a finite dimensional vector space. Prove thatthe characteristic polynomial pT is irreducible if and only if T has no nontrivial invariantsubspaces.

Solution.Suppose that pT is irreducible. Then pT is also the minimal polynomial of T , so we have

V ∼= F [X]/(pT ).

Since (pT ) is a prime ideal in a PID, it is also a maximal ideal, so V is actually a field. Sincean invariant subspace of T corresponds to a ideal of V , no nontrivial example exists.

Conversely, suppose that T has no nontrivial invariant subspaces. Then there must beonly one invariant factor. Otherwise, the Jordan normal form would have multiple blocks,which correspond to invariant subspaces. In particular, the characteristic polynomial isthe minimal polynomial of T . Now, if the minimal polynomial of T were reducible, thenthere would be an invariant subspace corresponding to an irreducible factor of it, which byassumption does not exist. Hence pT is irreducible as well.

Problem 8.Let V be a finite dimensional vector space over a field F . Prove that every right ideal inEndF (V ) is of the form {T : imT ⊂ W} for a unique subspace W of V .

Solution.Let a ⊂ EndF (V ) be a right ideal. Let T ∈ a be an endomorphism of maximal rank. Thenwe claim that for all S ∈ a, imS ⊂ imT . Let V = kerT⊕U be an orthogonal decomposition.Then T : U → imT is an isomorphism. Therefore we can invert T on imT via Q ∈ EndF (V )so that Q(v) = T−1(v) if v ∈ imT and Q(v) = 0 if v ∈ kerT and linear extension to the restof V . Since a is a right ideal, T ◦Q ∈ a, and this is an isomorphism from imT to itself.

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Let S ∈ a and suppose that S(v) = w /∈ imT for some v ∈ V . Let R ∈ EndF (V ) so thatR(u) = v for some v ∈ kerT and R ≡ 0 outside of the span of u. Then S ◦ R ∈ a, so themap T ◦Q+ S ◦R ∈ a. Then

(T ◦Q+ S ◦R)(u) = 0 + S(v) = w.

Further, since S ◦ R ≡ 0 on imT and T ◦Q : imT∼−→ imT is an isomorphism, the rank of

T ◦ Q + S ◦ R is strictly larger than the rank of T . But this is a contradiction. ThereforeimS ⊂ imT .

Let W = imT . Then we claim that a = {S ∈ EndF (V ) : imS ⊂ W}. The inclusion⊂ is proven above. Now, let S ∈ EndF (V ) so that imS ⊂ W . Let TS ∈ EndF (V ) sothat TS(v) = T−1(S(v)), which is sensible since S(v) ∈ imT . Then T ◦ TS = S, and sinceT ◦ TS ∈ a, we have S ∈ a. This completes the proof.

Problem 9.Let G be a finite group of invertible linear operators on a finite dimensional vector spaceV over the field of complex numbers. Prove that if (dimV )2 > |G|, then there is a propernonzero subspace W ⊂ V such that g(w) ∈ W for every g ∈ G and every w ∈ W .

Solution.By the Artin-Wedderburn theorem, we know that C[G] =

⊕ri=1Mni

(C), where∑r

i=1 n2i =

|G|. We would like to show that every vector space V as above cannot be irreducible, thatis, cannot be a simple C[G]-module. Indeed, simple C[G]-modules correspond precisely tosimple modules over one of the matrix algebras in the direct sum decomposition of C[G],and these are only Cni . Therefore suppose dimV = n. If it is simple, it must be a simplemodule over the matrix algebra Mn(C). But then n2 > |G|, which contradicts the aboveequality from Artin-Wedderburn. Hence no such V is irreducible, so it has a subspace asabove, and we are done.

Problem 10.Show that there is a (covariant) functor from the category of groups to the category ofsets taking a group G to the set of all subgroups of G. Determine whether this functor isrepresentable.

Solution.Let F : Groups → Sets be that functor. We need to examine how F acts on grouphomomorphisms f : G → H. We let F (f) : F (G) → F (H) send the set of subgroups{G1, . . . , Gn} to {f(G1), . . . , f(Gn)}, where we delete any repeated elements. Since theimage of a subgroup is a subgroup, this set of images of subgroups of G is a subset of thesubgroups of H.

Let 1G be the identity map on G. Then {G1, . . . , Gn} is mapped to itself under 1G, soF (1G) = 1F (G). Further, if f : G→ H and g : H → K are two homomorphisms, we have

{G1, . . . , Gn}F (f)7→ {f(G1), . . . , f(Gn)} F (g)7→ {g(f(G1)), . . . , g(f(Gn)}F (g◦f)7→ {(g ◦ f)(G1), . . . , (g ◦ f)(Gn)}.

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Since these sets are the same (up to deleting repeats), we have F (g ◦ f) = F (g) ◦ F (f).Therefore F is a functor.

However F is not representable. Suppose that F = Hom(X,−) for a group X. Then inparticular,

2 = |F (Z/3Z)| = |Hom(X,Z/3Z)|.

Let ϕ ∈ Hom(X,Z/3Z) be the nontrivial homomorphism. Let ψ ∈ Aut(Z/3Z) be thenontrivial automorphism 1 7→ 2. Then ψ ◦ ϕ is another nontrivial homomorphism inHom(X,Z/3Z), so |Hom(X,Z/3Z)| > 2, which is a contradiction. Therefore F is notrepresentable.

8 Fall 2010

Problem 1.Let Group be the category of groups and Ab be the category of abelian groups. If F :Ab → Group is the inclusion of categories, then find a left adjoint to F and prove that itis a left adjoint.

Solution.We claim that the left adjoint to F is G : Group → Ab so that G(X) = X/[X,X], theabelianisation of the group X. G would need to satisfy, for all groups X and abelian groupsY ,

HomAb(G(X), Y ) = HomGroup(X,F (Y )).

Since both sets are group homomorphisms, we will suppress the subscript on Hom. First,we look at any homomorphism ϕ from an arbitrary group X into an abelian group Y . Then

ϕ(ghg−1h−1) = ϕ(g)ϕ(h)ϕ(g)−1ϕ(h)−1 = eY

for every element ghg−1h−1 ∈ [X,X]. Therefore ϕ factors through [X,X], so it descends to ahomomorphism ϕ : X/[X,X] → Y . Conversely, given a homomorphism ψ : X/[X,X[→ Y ,we can define a homomorphism ψ : X → Y by ψ(g) = ψ(g[X,X]). This is well defined since

ψ(g · hkh−1k−1) = ψ(g)ψ(h)ψ(k)ψ(h)−1ψ(k)−1 = ψ(g · 1),

so ψ does not depend on the choice of coset representative. Therefore these Hom-sets areequal, so G is the left adjoint to F .

Problem 2.Let F : C → Sets be a covariant functor on a category C. Assume that F is representable byan object CF ∈ Obj(C). Identity which of the following statements are necessarily correct,proving your answers in each case.

(a) If C ∈ Obj(C) and F (C) 6= ∅, then there is an element f ∈ MorC(CF , C);

(b) If G is a left adjoint of F , then G is representable;

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(c) If C,D ∈ Obj(C), and there is a map of sets f : F (C) → F (D), then there exists ag ∈ MorC(C,D);

(d) If C,D ∈ Obj(C), and there exists a map g ∈ MorC(C,D), then there exists a map ofsets f : F (C)→ F (D) (note that we are not guaranteed in advance that F (D) 6= ∅);

(e) If h ∈ MorC(D,CF ), then for any C ∈ Obj(C), there is a map of sets F (C) →MorC(D,C).

Solution.Throughout, we suppress C when it is obvious.

(a) We know that MorC(CF , C) = F (C), and since F (C) 6= ∅, there is an f correspondingto some element in F (C).

(b) If G is left adjoint of F , then G : Sets→ C. Therefore if C 6⊂ Sets, it makes no senseto talk about representing the functor G. Hence G is not necessarily representable.

(c) This is not necessarily true. Let C be the category of unital rings. Then considerthe rings R and Q. Suppose ϕ : R → Q is a ring homomorphism. Then we musthave ϕ(

√2)2 = 2, but there is no square root of 2 in Q. Therefore there are no ring

homomorphisms from R to Q. Now consider the functor F : Ring → Sets thatsends a ring to its set of units. This functor is representable by Z[X,X−1]. SinceF (R), F (Q) 6= ∅, there is a map F (R)→ F (Q). But there is no corresponding map inRing to this map.

(d) This is false if and only if we can construct this situation so that F (C) 6= ∅ andF (D) = ∅, since there is always a map into any nonempty set and from the emptyset. Therefore suppose that F (C) 6= ∅. Then there exists a map f ∈ Mor(CF , C).Then g ◦ f ∈ Mor(CF , D), and there exists an element in F (D), so the situation neveroccurs. Therefore there is always a map F (C)→ F (D).

(e) If we identify F (C) = Mor(CF , C) we are looking for a set map from Mor(CF , C) →Mor(D,C). If f : CF → C, then f ◦h : D → C. so we indeed have a set map via −◦h.We can construct the actual map F (C)→ Mor(D,C) via the natural isomorphism.

Problem 3.Prove that there is no simple group of order 120.

Solution.Let G be a simple group of order 120. Examine the Sylow 5-subgroups of G. We have n5 ≡ 1mod 5 and n5 | 24. By assumption, n5 6= 1 so we see n5 = 6. Let G act on its set of Sylow5-subgroups by conjugation. This induces a homomorphism ϕ : G→ S6. Since G is simple,we know that kerϕ is either {e} or G itself. Clearly kerϕ 6= G since the Sylow subgroupsare conjugate. Therefore ϕ : G ↪→ S6.

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Suppose that G ⊂ A6. Then there is an action of A6 on A6/G, which is a three elementset. Since A6 is also simple, the induced action ψ : A6 → S3 is injective, which is acontradiction since |A6| > |S3|. Therefore if G is not a subset of A6, then consider G ∩ A6.This intersection is nontrivial since the product of any two odd permutations is even, solands inside A6. Therefore since A6 / S6, A6 ∩G / G. But this implies that G is not simple.Therefore no such group exists.

Problem 4.(a) Show from first principles (i.e. without using the classification theorem) that a sub-

group of a finitely generated abelian group is finitely generated.

(b) Let M ⊂ Z3 be the subgroup generated by the elements (13, 9, 2), (29, 21, 5), and(2, 2, 2). Determine the isomorphism class of the quotient group Z3/M .

Solution.(a) See Fall 2009, Problem 3. I did that one before this one, and the result there is stronger

anyway.

(b) This question is equivalent to reducing the matrix 13 9 229 21 52 2 2

via row and column operations. Painstakingly, one can verify that we obtain 1 0 0

0 1 00 0 16

.

This corresponds to the quotient Z/Z× Z/Z× Z/16Z ∼= Z/16Z.

Problem 5.Prove that if a finite group G acts transitively on a set S having more than one element thenthere exists an element of G which fixes no element of S.

Solution.Consider the set X in G× S where (g, s) ∈ X if g(s) = s. Then we see

|X| =∑g∈G

|{s ∈ S : g(s) = s}| =∑s∈S

|{g ∈ G : g(s) = s}|.

The first sum is the quantity with which we are concerned. The second quantity is thestabiliser of s, denoted Gs, so we have

|X| =∑s∈S

|Gs| =∑s∈S

|G||Gs|

32

where Gs is the orbit of s by the orbit-stabiliser theorem. Since this action is transitive, wehave |Gs| = |S|. Hence ∑

g∈G

|{s ∈ S : g(s) = s}| =∑s∈S

|G||S|

= |G|.

Let the sets on the left be denoted Xg. We know that |Xe| ≥ 2 since e(s) = s for all s ∈ S.Therefore if |Xg| > 0 for all g ∈ G, we have∑

g∈G

|{s ∈ S : g(s) = s}| ≥ 1 + |G| > |G|,

a contradiction. Hence some g ∈ G has Xg = ∅, as required.

Problem 6.Let R be the 5-dimensional tautological representation of S5. Show that R is isomorphic tothe direct sum of the trivial representation and an irreducible 4-dimensional representation.

Solution.First, let R take e1, e2, e3, e4, e5 as a basis, where S5 acts on R by permuting these basiselements. We see that the subspace spanned by e1 + e2 + e3 + e4 + e5 is S5-invariant (sinceits generator is), so it is corresponds to a 1-dimensional representation of S5, which must beisomorphic to the trivial representation.

LetW denote this subrepresentation of R, generated by the vector w = e1+e2+e3+e4+e5.Then consider R = W ⊕ W⊥. We claim that W⊥ is also irreducible. To see this, letv = (vi) ∈ W⊥ be a nonzero vector. Then since 〈v, w〉 = 0, we have

∑5i=1 vi = 0. Since

vi 6= 0 for some i, we must have vi > 0 and vj < 0 for some i 6= 0. Then we have

(1 i)(2 j) · v ∈ W⊥ =⇒ (1 2)(1 i)(2 j) · v − (1 i)(2 j) · v = (vi − vj, vj − vi, 0, 0, 0) ∈ W⊥.

Normalising, we see that e1 − e2 ∈ W⊥. Applying to this the transpositions (2 j) for j =3, 4, 5, we see that there are four linearly independent vectors in W⊥, so that W⊥ is generatedby any nonzero element, so it is irreducible. This completes the proof.

Problem 7.Let k be a field and let f be an irreducible element of the polynomial ring k[X, Y ].

(a) Describe the localisation k[X, Y ]p, where p = (f). Prove that it is a subring of therational functions k(X, Y ).

(b) For any r ∈ k(X, Y ), prove that

r ∈ im(k[X, Y ]p → k(X, Y ))

for all but finitely many choices of f .

Solution.

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(a) This localisation is, formally, inverting every element not in p. This means that

k[X, Y ]p ={gh

: g, h ∈ k[X, Y ], h /∈ (f) i.e. f - h}.

This is clearly contained in k(X, Y ). To see that it is a subring, we have

g1h1

+g2h2

=g1h2 + g2h1

h1h2,g1h1· g2h2

=g1g2h1h2

.

The numerator satisfies the conditions above, but h1h2 /∈ (f) is not true for general f .However, since f is irreducible and k[X, Y ] is a UFD, (f) is a prime ideal, so h1h2 ∈ (f)if and only if h1 ∈ (f) or h2 ∈ (f). But since h1, h2 /∈ (f), we have h1h2 /∈ (f),so k[X, Y ]p is closed under addition and multiplication. Further, 0 ∈ k[X, Y ]p isrepresented by 0/1 and 1 is represented by 1/1. Hence it is a subring.

(b) We can write r in terms of an irreducible factorisation of its numerator and denomi-nator:

r =g1 · · · gmh1 · · ·hn

, gi, hi ∈ k[X, Y ] irreducible,

If r is in the image of this inclusion map, then we must have h1 · · ·hn /∈ (f). But since(f) is a prime ideal, h1 · · ·hn /∈ (f) if and only if hi /∈ (f) for all i. But since hi is anirreducible polynomial, if we have that hi = f · p implies that (hi) = (f). Thereforer is in the image of this inclusion if and only if hi 6∼ f for any i, and since we maymove any offending element of k[X, Y ]× to the numerator of the polynomial, this isequivalent to hi 6= f for any i. Therefore only finitely many choices of f exclude rfrom the image of the inclusion map.

Problem 8.Let k be an algebraically closed field, R = k[X1, . . . , Xn], and f : R → Rd be given byf(p) = (pf1, . . . , pfd) with all fi ∈ R. Let m be the ideal generated by X1− a1, . . . , Xn− an,where ai ∈ k for all i. Consider for an integer r ≥ 1, the map

fmr : R⊗R (R/mr)→ Rd ⊗R (R/mr)

induced by f . Prove that ker fmr 6= 0 if and only if fj(a1, . . . , an) = 0 for all j.

Solution.We can condense these tensor products to obtain the actual map

fmr : R/mr → (R/mr)d

which sends p → p · f := (pf1, . . . , pfd). If p is in the kernel, then p · fi ∈ mr for each i. Ifwe assume p is nonzero, i.e. p /∈ mr, then we must have fi ∈ ms for some s ≥ 1. But thismeans that fi(a1, . . . , an) = 0 by definition. Conversely if ker fmr = 0, then we cannot havefi ∈ m for every i, so fi(a1, . . . , an) 6= 0 for some i. This completes the proof.

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Problem 9.Let R be a unital ring. Show that the only two-sided ideals of Mn(R) are of the form Mn(I)for some two-sided ideal in R.

Solution.First, suppose that I ⊂ R is a two-sided ideal. Then given two matrices a ∈ Mn(I) andb ∈Mn(R), we have

(ab)ij =n∑k=1

aikbkj.

Since aik ∈ I, we know that aikbkj ∈ I for all k. Therefore the sum of these elements is alsoin I, so ab ∈ Mn(I). An identical argument holds for ba since I is a two-sided ideal. Henceevery Mn(I) is a two-sided ideal of Mn(R).

Conversely, let a ⊂ Mn(R) be a two-sided ideal. Let I ⊂ R be the two-sided idealgenerated by the elements of the matrices in a. Then a ⊂ Mn(I) clearly. Notice that, sincea is a two-sided ideal, we have for any matrix a ∈ a

e1i · a · ej1 = b ∈ a

where ekl is the matrix with 1R at (k, l) and 0 everywhere else. In particular this matrix bsatisfies b11 = aij and bkl = 0 everywhere else. By repeating this again, we can find a matrixc ∈ a so that cij ∈ I for any element of I and ckl = 0 everywhere else. Therefore givenany matrix in Mn(I), we can sum up n2 matrices constructed in this fashion for each of itselements, and since a is a two-sided ideal, it will be an element of a. Therefore a = Mn(I),so we are done.

Problem 10.Let G be a group and Z[G] the integral group ring. Multiplication f · h of two elementsf =

∑i ni(gi) and h =

∑jmj(gj) is

f · h =∑i,j

nimj(gigj).

Let I be the two-sided ideal

I =

{∑i

ni(gi) :∑i

ni = 0

}.

Construct a natural map F : I/I2 → G/[G,G] and prove that this map is an isomorphismof Z-modules.

Solution.We construct a map F : I → G/[G,G] to begin such that

F

(∑i

ni(gi)

)=∏i

gnii ,

35

where we suppress the coset notation for convenience. We see that this is an abelian grouphomomorphism since

F

(∑i

ni(gi) +∑j

mj(gj)

)=∏i

gnii

∏j

gmj

j =∏k

g`kk = F

(∑k

`k(gk)

)where k parametrises the group elements and `k = nk + mk. We claim that F is surjective.Consider an element g ∈ G. Then modulo the commutator, we can always represent thecoset g[G,G] by

g · hk · h−1 · k−1.We can construct a preimage of this element, namely 1(g)+1(hk)−1(h)−1(k). This elementsatisfies

∑i ni = 0, so it is indeed an element of I ⊂ Z[G]. Hence F is surjective. Note that I

is generated by g−1 where g ranges over G. Then I2 is generated by (1(g)−1(e))(1(h)−1(e))for g, h ∈ G. Hence

F ((1(g)− 1(e))(1(h)− 1(e))) = ghg−1h−1 ∈ [G,G],

so F factors through I2. Then we let F−1 : G→ I/I2 by g 7→ 1(g)− 1(e). We see that F−1

is surjective since 1(g)− 1(e) generates I and it factors through [G,G] naturally. Thereforesince F is invertible, it is an isomorphism.

Problem 11.Show that the extension of Q generated by

√5 + 3√

2 is equal to Q(√

5, 3√

2).

Solution.We construct the following tower of fields:

Q

Q(√

5)Q( 3√

2)

Q(√

5 + 3√

2)

Q(√

5, 3√

2)Q( 3√

2,√

5 + 3√

2) Q(√

5,√

5 + 3√

2)

n

3 2

2 3

6/n

= =

We know that n = 1, 2, 3, 6. n = 1 is impossible since this element is not in Q. If we hadn = 2, that would imply Q(

√5,√

5 + 3√

2)/Q(√

5) has degree 3. But from the picture weknow that it has degree at most 2. Similarly, if n = 3, then Q( 3

√2,√

5 + 3√

2)/Q( 3√

2) hasdegree 2, but it must have degree dividing 3 from the picture. Therefore n = 6, so the topextension has degree 1, i.e. the fields are the same.

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Problem 12.Show that the multiplicative group of a finite field is cyclic and use this result to prove thatthe polynomial X4 + 1 is never irreducible over any finite field.

Solution.Let F be a finite field and let F× be its multiplicative group. This is a finite abelian group,so we may write it in terms of its invariant factors

F× ∼= Z/n1Z× Z/n2Z× · · · × Z/nrZ.

F× is cyclic if and only if r = 1. Suppose r > 1. Then since n1 | n2, there is a subgroup ofF× of the form Z/n1Z× Z/n1Z. Each of the n2

1 elements in this subgroup satisfies αn1 = 1.This implies that there are n2

1 roots of the polynomial Xn1 − 1. But since there are at mostn1 roots of this polynomial, this implies that n2

1 = n1, which is a contradiction since n1 > 1else F× is infinite. Therefore we must have r = 1 so F× is cyclic.

Now let |F | = q. If charF = 2, then X4 + 1 = (X + 1)4, so the polynomial is reducible.Now since q is an odd number, we have q ≡ 1, 3, 5, 7 mod 8. In any case, q2 ≡ 1 mod 8.Consider a quadratic extension K/F . Then

|K×| = q2 − 1 ≡ 0 mod 8.

Let q2−1 = 8n. Therefore since K× = 〈α〉, we have (αn)8 = 1 is a primitive 8th root of unity.Therefore the polynomial X8−1 splits over K. In particular, since X8−1 = (X4−1)(X4+1),X4 +1 splits in K. If X4 +1 were irreducible over F , then one of its roots would have degree4, so could not lie in K, a contradiction. Therefore X4 +1 is never irreducible over any finitefield.

9 Spring 2010

Problem 1.Show that the functor from (unitary) rings to groups sending a ring A to its group of unitsA× is representable by a ring R.

Solution.We claim that the ring R = Z[X,X−1], where X is a variable over Z, represents this functorF . Consider a homomorphism ϕ : R → A. Since A is unital, we must have ϕ(1) = 1A, sothere is no choice of the image of Z ⊂ R by linear extension. Consider ϕ(X). Since we haveϕ(X)ϕ(X−1) = 1A, we must have ϕ(X) ∈ A×. Further, there are no other restrictions sinceall sums and powers of X are independent of Z. Hence there is a unique ϕ for each a ∈ A×.As sets, this means

F (A) = A× ∼= HomRing(R,A)

for all unital rings A. This precisely means that R represents F .

Problem 2.

(a) Define what it means for two categories to be equivalent.

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(b) A groupoid G is a category such that all morphism are isomorphisms. G is calledconnected if for any two objects x and y, HomG(x, y) is nonempty. Show that anynonempty connected groupoid is equivalent to a group, that is, a groupoid with oneobject.

Solution.

(a) Two categories C and D are equivalent if there exists a functor F : C → D suchthat F is fully faithful and essentially surjective. F is fully faithful if HomC(X, Y ) ∼=HomD(F (X), F (Y )) as sets for every objects X, Y of C. F is essentially surjective iffor every object Z of D, there is an object X of C so that F (X) ∼= Z as objects of D.

(b) Let G be a nonempty connected groupoid and let G be a group. Let F : G → G byF (X) = g, the sole element of of G for every X in G. Then this functor is clearlyessentially surjective (indeed, it is surjective), since there exists an object in G. Nowtake X, Y in G. Notice that we have not defined the morphism set of G yet, but we doso now. We claim that the choice HomG(g, g) = HomG(Z,Z), for some fixed object Zin G makes the functor F fully faithful as well.

We claim that HomG(g, g) = HomG(X, Y ) for every X, Y in G. First, the right handset is nonempty since G is connected, so let ϕ be some element. Additionally, thereexist isomorphisms f : Z → X and g : Y → X by assumption. Therefore we get anisomorphism g◦ϕ◦f : Z → Z which corresponds uniquely to ϕ. But since HomG(Z,Z)corresponds uniquely with HomG(g, g) by definition, this proves our claim. HenceF : G → G is an equivalent of categories.

Problem 3.Determine, using the structure theory of abelian groups or otherwise, all finitely generatedabelian groups A such that the group AutA is finite. State clearly any basic theorems thatyou use. Determine the order of the automorphism group of A = Z× Z/4Z.

Solution.From the structure theorem, we know that

A ∼= Zr × Z/pn11 Z× · · · × Z/pnk

k Z ∼= Zr × Ator.

We examine automorphisms of Zr, since AutA ⊃ AutZr. If r = 1, then Z only has twoautomorphisms, namely 1 7→ ±1. Now we examine Z2. Treating this like a free Z-module,we see that automorphisms here correspond bijectively to GL2(Z). Given α ∈ GL2(Z), wehave infinitely many choices for the first column (do not chooce (0, 0)) and more than zerochoices for the second column, |GL2(Z)| =∞. Hence Zr has infinitely many automorphismsfor r > 1.

If r = 1, then we know there are 2|Ator| maps from Z into A which preserve the free part,corresponding to (1, a) and (−1, a) for every torsion a. Further, there are at most |Ator||Ator|

maps from Ator to A since a torsion element cannot map to a non-torsion element. We

38

conclude that the only finitely generated abelian groups with finite automorphisms groupsare of rank 1 or rank 0.

Therefore we have 8 maps from Z to A sending (1, 0) to (±1, a) for A ∈ Z/4Z, and twoautomorphisms of Z/4Z sending (0, 1) 7→ (0, 1) or (0, 3). Hence there are 16 automorphisms.

Problem 4.Show that if σ ∈ AutS4 and τ ∈ S4 is a transposition, then σ(τ) is also a transposition. Bystudying the action of σ on transpositions, show that every automorphism of S4 is inner.

Solution.We know that σ(τ) has order 2, so either σ(τ) is a transposition or σ(τ) is a 2-2-cycle, i.e.of the form (1 2)(3 4) or similar. Let V ⊂ S4 be the subgroup generated by the 2-2-cycles oforder 4. Then σ(V ) is also a normal subgroup of S4 of order 4. But there is only one suchsubgroup, so σ preserves V . Therefore a transposition cannot be sent to a 2-2-cycle.

Therefore consider an automorphism of S4. We start by mapping (1 2) to a transposition.There are 6 choices. This also defines precisely the image of (3 4), since transpositions comein commuting pairs (e.g. (1 2),(3 4)). There are therefore 4 choices for the image of (1 3), andthe image of (1 4) is obtained uniquely by the image of (3 4)(1 4)(3 4). Since these generateS4, |AutS4| = 24. Further, we know that the number of inner automorphisms of S4 is equalto |S4|/|Z(S4)|. But Z(S4) is trivial, so there are 24 inner automorphisms of S4. Hence everyautomorphism is inner.

Problem 5.Give the total number and the dimensions of the irreducible complex representations of theirreducible complex representations of S4. Prove your answer.

Solution.Using character theory, we know that the number of representations is equal to the numberof conjugacy classes of S4. Since each cycle type is a unique conjugacy class, we know thatthere are 5 irreducible complex representations. Further, the sum of the dimensions of theserepresentations equals |S4| = 24, and one of these representations is the trivial representationof degree 1.

Now we need the sum of four squares to equal 23. We see that no representation hasdimension 4, since we cannot obtain 7 as a sum of three of 1, 4, 9. Further, we must have atleast one representation of dimension 3. So we need to sum 3 squares to 14. We see this isobtained by (and only by) 1 + 4 + 9. Hence the 5 irreducible complex representations of S4

have dimensions 1, 1, 2, 3, 3.

Problem 6.State and prove Schur’s Lemma.

Solution.We state this in module terminology. Let V,W be simple F [G]-modules. Then Schur’sLemma states that if ϕ : V → W is a F [G]-module homomorphism, then either ϕ is an

39

isomorphism or ϕ = 0. The proof is straightforward. Given such a ϕ, then kerϕ is a F [G]-submodule of V . Since V simple, then kerϕ = V or 0. If kerϕ = V , then we are in the caseϕ = 0. If kerϕ = 0, then V injects into W , so ϕ(V ) is a F [G]-submodule of W . But W isalso simple, so since ϕ(V ) 6= 0, we must have ϕ(V ) = W . Hence ϕ is an isomorphism.

Problem 7.Show that the group of units of the ring Z/nZ is cyclic if and only if n is a power of an oddprime number, twice the power of an odd prime number, or 4.

Solution.First, by the Chinese Remainder Theorem, we may write

Z/nZ ∼=k∏i=1

Z/paii Z.

Suppose have a unit in Z/nZ. Then it must also be a unit in each of the factors, and theconverse also holds. Hence

(Z/nZ)× ∼=k∏i=1

(Z/paii Z)×.

We now examine particular cases of the group of units of Z/paZ. Suppose p = 2. Then(Z/2Z)× ∼= {1}, and (Z/4Z)× ∼= Z/2Z, but (Z/8Z)× ∼= Z/2Z× Z/2Z. To see this last fact,we note that

(Z/8Z)× = {1, 3, 5, 7}, 12 = 32 = 52 = 72 = 1 (mod 8).

Z/2Z × Z/2Z is the only group of order 4 without an element of order 4, so this is clear.Now for Z/2aZ for a ≥ 3, the group of units clearly contains (Z/8Z)× as a subgroup, whichis not cyclic. Since a cyclic group has cyclic subgroups, this proves that Z/2aZ is cyclic ifand only if a = 1, 2.

Now let p be an odd prime. By the general theory of cyclic groups, (Z/pZ)× ∼= Z/ϕ(p)Z =Z(p − 1)Z. The generalisation to higher powers is notoriously confusing, but it is found onpage 96 of [2]. We have a canonical ring homomorphism

Z/pr+1Z→ Z/prZ

which is induced by the inclusion of ideals pr+1Z ⊂ prZ ⊂ Z. Hence we have an inducedgroup homomorphism

λ : (Z/pr+1Z)× → (Z/prZ)×.

This is a surjective group homomorphism. Further, let λ(a) = 1. Then we have

a ≡ 1 mod pr =⇒ a ≡ 1 + xpr mod pr+1

for any x ∈ {0, . . . , p− 1}. Hence | kerλ| = p, so |(Z/pr+1Z)×| = ϕ(pr+1) = pr(p− 1). Henceby the fundamental theorem of finitely generated abelian groups, we can write

(Z/pr+1Z)× ∼= Apr × Ap−1.

We need to show that these latter two groups are cyclic. It is not hard to show thatAp−1 ∼= Z/(p−1)Z from the natural surjection (Z/pr+1Z)× → (Z/pZ)× from above. However,for the other group, we need to work with p-adics, and I’ll refer the reader to [3] for moreinformation. The proof follows after proving this.

40

Problem 8.Let F be the field with 2 elements and let R = F [X]. List up to isomorphism all R-moduleswith 8 elements that are cyclic.

Solution.A finite F [X]-module is precisely a F -vector space. In particular, such an R-module V musthave dimF V = 3 so that |V | = 23 = 8. This module is cyclic if and only if it has oneinvariant factor. Hence its invariant factor must be a cubic polynomial in F [X], and thereare 8 of these corresponding to X3 + (0/1)X2 + (0/1)X + (0/1)1. Therefore there are 8 suchR-modules.

Problem 9.Let A be a left Noetherian ring. Show that every left invertible element a ∈ A is two-sidedinvertible.

Solution.Suppose that ba = 1. Then let ra ∈ EndA be the map given by x 7→ xa. We see that rais surjective since ra(xb) = xba = x for any x ∈ A. We claim that ra is also injective. LetMi denote the kernel of ria, i.e. x 7→ xai. Then M1 ⊂ M2 ⊂ . . . is an ascending chain, so itstabilises since A is Noetherian. Let Mn be the point at which the chain stabilises, so thatMn = Mn+m for all m ∈ N.

Let x ∈ M1 Then since ra is surjective, rna is surjective, so there exists y ∈ A so thatyan = x. Then y ∈Mn+1 since yan+1 = xa = 0. But since Mn = Mn+1, we have yan = 0 = x.Therefore ker ra = 0, so ra is an isomorphism.

We see that ra(ab − 1) = aba − a = a − a = 0. By injectivity, this implies ab − 1 = 0,so ab = 1. Therefore every left invertible element is two-sided invertible, and in fact by thesame inverse.

Problem 10.Let F be a field and V a finite dimensional F -vector space. Show that R = EndF (V ) hasno nontrivial two-sided ideals.

Solution.By basic linear algebra, R ∼= Mn(F ), where n = dimV . Hence it is sufficient to prove thisresult for a matrix ring Mn(F ) of arbitrary size n. Let 0 6= a ∈ Mn(F ) be any matrix.We claim that the two-sided ideal generated by a is all of Mn(F ), which would prove theclaim since all minimal ideals are principal. Let a be this ideal. Since a 6= 0, there is some(i, j) ∈ {1, . . . , n}2 so that aij 6= 0. Let Ekl denote the elementary matrix so that

Eklij = δk(i)δl(j).

In words, the matrix is zero except for 1F in the (k, l) position. We see that

Eki · a · Ejl = aijEkl ∈ a =⇒ Ekl ∈ a

41

by multiplying by a−1ij . Since k, l were arbitrary above, we see that every elementary matrixis in a. For any matrix b ∈Mn(F ), we have

b =∑i,j

bijEij ∈ a.

Therefore a = Mn(F ), so there are no proper nontrivial two-sided ideals.

Problem 11.Let F be a field of characteristic zero containing the pth roots of unity for p a prime. Showthat the cyclic extensions of degree p of F in any algebraic closure F of F are in one to onecorrespondence with the subgroups of order p of F×/(F×)p.

Solution.Let G = F×/(F×)p, and denote elements by squarefree coset representatives. Take anelement α ∈ G of order p, so that 〈α〉 has order p as well. Let K be the splitting field ofXp − α, which is irreducible over F by assumption. Then this is a Galois extension of F ,generated by a = p

√α in F , and is of order p. Therefore its Galois group must be cyclic.

Conversely, let K/F be a cyclic extension of order p. Let σ be a generator of the Galoisgroup. Then consider the norm of the element ζp. We have

NKF (ζp) =

∏σ∈Gal(K/F )

σ(ζp) = 1,

where σ acts trivially on ζp since it lies in the base field F . By Hilbert’s Theorem 90, thereexists α ∈ K so that ζp = σ(α)/α, i.e. ζpα = σ(α). Indeed, we know this α ∈ K \ F , soK = F (α). This is because any intermediate field would correspond to a divisor of p, ofwhich there is none. Further,

σ(αp) = σ(α)p = (ζpα)p = αp,

so αp ∈ F . Therefore α ∈ F×/(F×)p, and generates a subgroup of order p. This completesthe proof.

Problem 12.Determine all fields F such that the multiplicative group of F is finitely generated.

Solution.Suppose F has characteristic 0. Then view Q ⊂ F as an embedded subfield. We claim thatQ× is not finitely generated as a multiplicative group. Suppose it were generated by some{ai/bi} where i ∈ {1, . . . , n}. Then the denominator of any product of these elements hasdenominator dividing some power of

∏ni=1 bi. Then let {pj} be the set of primes in Z diving∏n

i=1 bi. In particular, this list is finite. Therefore any 1/p ∈ Q so that p 6= pj for any jcannot be in the subgroup generated by {ai/bi}. Hence Q× is infinitely generated. SinceQ× ⊂ F× as a subgroup, suppose that F× were finitely generated. Then by the classificationtheorem of finitely generated abelian groups, any subgroup of F× is also finitely generated.But this is a contradiction since Q× was not finitely generated.

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If F is a finite field, then its multiplicative group is finite, so must be finitely generated. IfF is not finitely generated over its prime field Fp, then certainly F× is not finitely generated.If it were, then the set of generators of F× would generate F , a contradiction. Therefore letF = Fp(α1, . . . , αn) be a finitely generated extension of Fp. If each αi is algebraic, then Fis finite, so its multiplicative group is finitely generated. If some α1 is transcendental, thenFp(X) ⊂ F as a subfield. But since Fp[X] is a UFD with infinitely many primes, we are inthe same situation as Q not being finitely generated. Hence a transcendental extension ofFp does not have a finitely generated multiplicative group.

We conclude that F× is finitely generated if and only if F is finite.

10 Fall 2009

Problem 1.Let Top be the category of topological spaces. Recall that a morphism f in some category iscalled a monomorphism if, for any two morphisms g1 and g2 that can be precomposed withf , fg1 = fg2 implies g1 = g2. Dually, f is called an epimorphism if, for any g1 and g2 thatcan be postcomposed with f , g1f = g2f implies g1 = g2.

(a) Show that a continuous map f : X → Y is an monomorphism in Top if and only if fis one-to-one.

(b) Show by example that an epimorphism in Top need not be onto.

Solution.Note that we have to assume Top is the category of Hausdorff topological spaces or (b) isnot true.

(a) First, suppose f is injective. Then

fg1(x) = fg2(x) =⇒ f(g1(x)) = f(g2(x)) =⇒ g1(x) = g2(x)

for all x in the domain of g1, g2. Therefore f is a monomorphism. Conversely, supposethat that f is a monomorphism but there exists x 6= y ∈ X so that f(x) = f(y). Thenlet g1 : Z → X be the projection onto the point x and g2 : Z → X be projection ontoy. Then fg1 = fg2, but g1 6= g2, which is a contradiction.

(b) Let f : (0, 1)→ [0, 1] be an embedding. Then since every continuous function on (0, 1)extends uniquely to its limit points 0, 1, g1f = g2f implies g1 = g2. But f is notsurjective.

Problem 2.Let F : Ab→ Sets be the forgetful functor from abelian groups to sets. Show that F doesnot have a right adjoint.

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Solution.Suppose it did, and call that functor G. Then we would have for every abelian group A andset X

HomAb(A,G(X)) ∼= HomSets(F (A), X).

In particular, let A be the trivial group. Then the number of maps from F (A) to a set Xis precisely |X|. However, there is only one group homomorphism from A to any abeliangroup. Therefore since sets with more than one element exist, no such G could exist.

Problem 3.Suppose A is an abelian group that is generated by n elements. Show that any subgroup ofA also can be generated by n elements.

Solution.If A is generated by n elements, then we have a surjection Zn → A → 0 which sends basis

elements of Zn to generators of A. This yields a short exact sequence Zn f→ A → 0. LetB ⊂ A be a subgroup. Then f−1(B) ⊂ Zn is a submodule. We claim that in a PID, asubmodule of a finitely generated free module is finitely generated of lesser or equal rank.

To see this, we proceed by induction on n. First, for n = 1, we know that submodulesof Z correspond to ideals of Z, which are principal, hence generated by at most 1 elements.Suppose the n− 1 case, and let M ⊂ Zn be a submodule Zn. Then let ϕ : M → Z be givenby

ϕ((x1, . . . , xn)) =n∑i=1

xi.

Then ϕ is a Z-module homomorphism, so we obtain a short exact sequence

0→ kerϕ→M → imϕ→ 0.

kerϕ is a submodule Zn−1, so it is free of rank at most n−1. Also, imϕ ⊂ Z is a submodule,we have shown it is free, so in particular it is projective. Therefore the sequence is split, sowe have M = kerϕ⊕ imϕ, which has rank at most n.

Applying this to the particular situation, f−1(B) is free of rank at most n, so the surjec-tion f−1(B) ∼= Zn → B → 0 implies that B is generated by n elements.

Problem 4.Let p < q be primes, n ≥ 0 an integer and G a group of order pqn. Show that G is solvable.

Solution.The number of Sylow q-subgroups of G satisfies nq | p and nq ≡ 1 mod q. Since p < q,so q + 1 - p, we see that nq = 1. Therefore let Q / G be the normal Sylow q-subgroup ofG. Since |G/Q| = p, we must have G/Q ∼= Z/pZ, which is an abelian group, so is solvable.Additionally, we claim that Q is solvable. We proceed by induction on n. For n = 1,Q ∼= Z/qZ, so it is abelian group, so is solvable.

For any n, any group H of order qn has a nontrivial centre Z. Then Z is solvable, andwe have |H/Z| ≤ qn−1 is a q-group of lesser order. By induction, H/Z is solvable. Since Zand H/Z are solvable, H is solvable. Hence all q-groups are solvable.

In our case, we have Q and G/Q are solvable, so G is solvable.

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Problem 5.Let G be a finite group and ρ : G → GL(V ) a complex representation. Prove that (V, ρ)splits as a direct sum of irreducible representations of G.

Solution.This is equivalent to proving Maschke’s theorem. Consider V as a C[G]-module. Let U ⊂ Vbe a submodule. If this implies that U = 0 or U = V , then V is irreducible and we are done.Otherwise, take the orthogonal complement U⊥ = X in the vector space V . By definition,V = U ⊕ X as vector spaces. We want to show that X is actually a submodule of V , andso by induction on dimension this would prove the theorem. If V happens to be infinitedimensional, then we apply Zorn’s lemma to obtain the same conclusion.

Let ϕ ∈ End(V ) be the orthogonal projection onto U . We would like to modify ϕ so thatit is a C[G]-module homomorphism as well as a vector space homomorphism. Let ϕg denoteρ(g) ◦ ϕ ◦ ρ(g)−1. Since U is a C[G]-submodule, we have

ρG(u) = ρ(g)ϕ(ρ(g−1)u) = ρ(gg−1)(u) = u.

Therefore consider the map π : V → U given by

π(v) =1

|G|∑g∈G

ϕg(v).

Since we are working over C, the fraction 1/|G| is well defined. By the above,

π(u) =1

|G|∑g∈G

ϕg(u) =1

|G|· |G| · u = u.

Therefore we claim that V = U ⊕ kerπ as C[G]-modules. If x ∈ kerπ and h ∈ G, then wesee that

π(ρ(h)x) =1

|G|∑g∈G

ϕg(ρ(h)x) =1

|G|∑g∈G

ρ(g)ϕρ(g−1)ρ(h)x

=1

|G|∑g∈G

ρ(h)ρ(h−1g)ϕρ(g−1h)x

= ρ(h)1

|G|∑g∈G

ϕh−1g(x) = ρ(h)π(x) = 0,

where we use that left multiplication by h−1 is an automorphism of G. Therefore ker π is aC[G]-submodule of V , so we are done.

Problem 6.Let G be a finite p-group and ρ : G→ GL(V ) a representation over Fp.

(a) Show that V has a one-dimensional G-invariant subspace.

(b) Show by example that (V, ρ) need not split into a direct sum of irreducible represen-tations.

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Solution.

(a) It suffices to assume V is finite dimensional here by examining subrepresentations ofthe span of {ρ(g)(v) : g ∈ G} for any v ∈ V . Then |V | = pn. Consider the partitionof V into one-dimensional subspaces, of which there are (pn − 1)/(p − 1) since eachsubspace has order p− 1. In particular,

pn − 1

p− 1=

n−1∑i=0

pi ≡ 1 mod p.

We claim the action of G on the set of linear subspaces of V has at least one fixedpoint, which would correspond to a one-dimensional subrepresentation of V . Let Xdenote the set of linear subspaces of V , and let A denote a set of unique representativesfor the nontrivial orbits of G, and Gx the stabiliser of x ∈ X in G. Further, let XG

denote the fixed points of G on X. By the orbit-stabiliser theorem we know

|X| = |XG|+∑α∈A

|Gα| = |XG|+∑α∈A

|G : Gα|.

Since G is a p-group, we know that p | |G : Gα| for each α. Therefore

|X| = |XG| ≡ 1 mod p.

Therefore there is at least one linear subspace fixed by G, which is what we need.

(b) Let p = 2, dimV = 2, and G = Z/2Z = {1, a}. Then let ρ be the map

ρ(a) =

(1 10 1

).

We have ρ(a)2 = ρ(a2) = ρ(1) = I2, so this is indeed a representation. We see that(1, 0) ∈ V is invariant under G, and so it generates a one-dimensional irreduciblesubrepresentation. However the other vectors (1, 1) and (1, 0) are interchanged by a,so neither generates a one-dimensional subrepresentation. Therefore V is not the directsum of irreducible subrepresentations.

Problem 7.Find a homomorphism A → B of commutative rings and non-zero A-modules M,N suchthat the canonical map

B ⊗A HomA(M,N)→ HomB(B ⊗AM,B ⊗A N)

is the zero map. Prove that the map is an isomorphism if M is a finitely generated projectiveA-module.

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Solution.This canonical map is defined by

b⊗ f 7→ [x⊗m 7→ b · x⊗ f(m)].

For the first part, let A = B = Z, M = Z/3Z, and N = Z/2Z. Then HomZ(M,N) = 0,since there is no nontrivial map between these two modules. Therefore the canonical mapabove is the zero map since Z⊗Z HomZ(M,N) = HomZ(M,N) = 0.

For the second part, we use that HomA(A,N) ∼= N for any A-module N . In our case,assuming M = A, we have

B ⊗A N ∼= B ⊗A HomA(A,N)

→ HomB(B ⊗A A,B ⊗A N) = HomB(B,B ⊗A N) ∼= B ⊗A N.

Therefore the isomorphism is clear. This generalises trivially to a free module M = An.Since both tensor products and HomA behave well with respect to direct sums, it holds foran M such that M ⊕M ′ = An as well. Therefore if M is a finitely generated projectiveA-module, we have the isomorphism we require.

Problem 8.Prove the following facts:

(a) Any subring of Q sharing the identity with Q is a PID.

(b) For a subring A ⊂ Z[i] sharing the identity with Z[i], if A 6= Z and A 6= Z[i], A is nota PID.

Solution.(a) First, let R ⊂ Q be such a subring. Then Z ⊂ R since R is closed under addition.

Therefore R is a domain between Z and its quotient field, so we may realise R as S−1Zfor some set S ⊂ Z \ {0}. In particular, we see that S = {b : a/b ∈ R}.Now we know that ideals in R = S−1Z are in bijective correspondence with ideals inZ avoiding S. For a ⊂ R, we have b ⊂ Z so that a = S−1b. Since b = (b) for someb ∈ Z, we have that a = (b) in R as well. Therefore R is a PID.

(b) Since Z ( A, we must have ni ∈ A for some n > 0, and since A ( Z[i], we must haven > 1. Let n the the minimal m so that mi ∈ A. Then we claim that A = Z[ni].If there were some mi ∈ A so that n - m, then applying the Euclidean algorithm wecould obtain gcd(n,m)i, which contradicts the minimality of n.

We claim that the ideal (n + 1, ni) is not principal. Suppose that a + bni generatedthis ideal for a, b ∈ Z. Then in particular, if we let N be the usual norm on C appliedto A, we have

N(n+ 1) = (n+ 1)2 = N(α)(a2 + n2b2), N(ni) = n2 = N(β)(a2 + n2b2).

Since (a2 + n2b2) divides both n2 and (n+ 1)2, it divides their gcd. Since n, n+ 1 arecoprime, their squares are also, so their gcd is just 1. Hence a2 + n2b2 = 1. But this isa contradiction, since this implies b = 0 and a = ±1, and hence (a + bi) = A, which(n+ 1, ni) is not. Therefore this ideal is not principal, and we are done.

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Problem 9.Prove that every two-sided ideal of the ring M2(Z) is principal.

Solution.Two-sided ideals of M2(Z) correspond precisely with M2(a), where a ⊂ Z is a two-sided ideal.Since Z is a PID, we have a = nZ. Therefore we need to snow that M2(nZ) is principal inM2(Z). Indeed, it is generated by

a =

(n 00 0

).

If we let Eij be the elementary matrix with 1 in the (i, j) spot and 0 elsewhere, we see that

E11 · a · E12 =

(0 n0 0

)and similarly we can find n · Eij ∈ M2(nZ) for 1 ≤ i, j ≤ 2. Repeatedly summing theseelementary matrices generates the entire ideal, so it is principal.

Problem 10.Let B be a central simple algebra over k of dimension 4. Prove the following facts.

(a) All left ideals of B have even dimension.

(b) B ∼= M2(k) if and only if B is not a division algebra.

Solution.(a) It suffices to show that B has no left ideal of dimension 1. Suppose that a were a left

ideal. Then we may write a = Ba for some a ∈ B. Since BaB = B, a must be rightinvertible. But since B is Noetherian, right invertible elements are two-sided invertible,hence Ba = B. But this is a contradiction, hence no such a exists, and we are done.

(b) We use the Artin-Wedderburn theorem. From the classification of simple algebras, weknow that B ∼= Mn(D), where n2 ·dimD = 4, and D is a division algebra over k. First,if B ∼= M2(k), then it is not a division algebra since we see(

1 00 0

)has no inverse, since it has zero determinant. Conversely, if B is not a a divisionalgebra, then by the classification theorem it must be of the form Mn(D) with n > 1.But n2 = 4, so dimD = 1, i.e. D = k. Therefore B ∼= M2(k).

Problem 11.Prove that the multiplicative group of a field F is a cyclic group if and only if F is a finitefield.

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Solution.See Spring 2010, Problem 12. We only proved finitely generated if and only if F is finite,but if F is finite then you can prove that its multiplicative group is cyclic. Since F× is afinitely generated abelian group, we can write

F× ∼= Z/n1Z× · · · × Z/nkZ,

the invariant factor decomposition of F×. If k = 1, then F× is cyclic. Suppose that k > 1.Then we have n1 | n2, there is a subgroup Z/n1Z×Z/n1Z ⊂ Z/n1Z×Z/n2Z×· · · . ThereforeF× has a subgroup H ∼= Z/n1Z× Z/n1Z. Every element in H has order n1, i.e. an1 = 1 forall a ∈ H. Therefore the polynomial Xn1−1 has |H| distinct roots. But Xn1−1 has at mostn1 < |H| = n2

1 roots over any field, so this is a contradiction. Therefore F× is cyclic.

Problem 12.Let k = F2(t, s) be the field of fractions of the two variable polynomial ring F2[t, s]. Writeθa for a root of T 2 + T + a for a ∈ k in an algebraic closure of k.

(a) How many intermediate fields between k and k(θt, θs)?

(b) How many intermediate fields between k and k(√t,√s)?

Solution.(a) We notice that both θt and θt + 1 are roots of T 2 + T + t. Specifically,

(θt + 1)2 + (θt + 1) + t = (θ + t2 + θt + t) + (1 + 1) = 0.

Therefore T 2 + T + t is a separable polynomial. We can see the same is true forT 2 + T + s. Since k is the splitting field of (T 2 + T + t)(T 2 + T + s), it is a separablenormal extension, so is Galois. Therefore intermediate fields between k and k(θt, θs)are in bijective correspondence with subgroups of the Galois group G of this extension.We see that |G| = 4, and in particular

G ∼= Gal(k(θt)/k)×Gal(k(θs)/k) ∼= Z/2Z× Z/2Z.

There are five subgroups of G: the whole group, the trivial subgroup, the first com-ponent, the second component, and the diagonal subgroup. Therefore there are threeintermediate fields.

(b) We see that√t is not a separable element. It is a root of T 2 − t, which factors as

(T +√t)2 in k(

√t,√s). Therefore the extension k(

√t,√s) is purely inseparable. We

claim that it has infinitely many intermediate fields.

Let Kf = k(√s + f ·

√t), where f ∈ k[s, t] is a nonconstant polynomial. First, note

that Kf 6= k(√s,√t), since [Kf : k] = 2 and [k(

√s,√t) : k] = 4. We claim that for

f 6= g, we have Kf 6= Kg. Suppose there were f 6= g such that Kf = Kg. Call thisfield K. Then

√s+ g ·

√t− g/f · (

√s+ f ·

√t) = (1− g/f)

√s ∈ K.

Since g 6= f , 1− g/f ∈ K×, so√s ∈ K. Hence

√s,√t ∈ K, so K = k(

√s,√t), which

is a contradiction. Therefore since there are infinitely choices for f , there are infinitelymany nonisomorphic intermediate fields Kf .

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References

[1] This proof is taken from: The theorems of Maschke and Artin-Wedderburn.http://www.math.uni-bielefeld.de/ sek/select/rw1.pdf.

[2] Lang, Serge. Algebra (Third Edition). If you don’t know how to get a copy of this book,you’re in trouble for the qual.

[3] Jerry Shurman’s Number Theory lecture notes, found athttp://people.reed.edu/ jerry/361/lectures/lec07.pdf.

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