SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB. Chapter 8

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1 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB CİHAN BAHRAN I changed the notation in some of the questions. Chapter 8 1. Prove that if G is any finite group then the only idempotents in the integral group ring ZG are 0 and 1. [If e is idempotent consider the rank of the free abelian group ZGe and also its image under the homomorphism ZG F p G for each prime p dividing G, which is a projective F p G-module. Show that rank Z ZGe is divisible by G. Deduce from this that if e 0 then e = 1.] Let e be an idempotent in ZG and fix a prime p. Let ϕ : ZG F p G be the suggested ring homomorphism. Write f = ϕ(e), now as e and f are both idempotents, we have ZG = ZGe ZG(1 e) and F p G = F p Gf F p G(1 f). Note that ϕ maps ZGe onto F p Gf. Therefore if we pick a Z-basis {a 1,..., a n } for ZGe, then {ϕ(a 1 ),..., ϕ(a n )} generate F p Gf as a Z-module; hence as an F p -vector space. Thus Similarly, On the other hand, we have dim Fp F p Gf n = rank Z ZGe. dim Fp F p G(1 f) rank Z ZG(1 e). G = rank Z ZG = rank Z ZGe + rank Z ZG(1 e) G = dim Fp F p G = dim Fp F p Gf + dim Fp F p G(1 f). Thus the above inequalities can t be strict. Denoting the p-part of G by G p, by Corollary 8.3, G p divides dim Fp F p Gf = rank Z ZGe. Since this happens for every prime p, we deduce that G divides rank Z ZGe G. Therefore either rank Z ZGe = 0 or rank Z ZGe = G. The former yields e = 0 and the latter yields e = (a) Let H = C 2 C 2 and let k be a field of characteristic 2. Show that (IH) 2 is a one-dimensional space spanned by H = h H h. Write H = a, b a 2 = b 2 = 1, ab = ba. So IH is generated by a 1 and b 1 as a left kh-module. Hence pairwise products of these generators generate (IH) 2. Note that 1

2 since char k = 2, SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 2 (a 1) 2 = a 2 1 = 0 (b 1) 2 = b 2 1 = 0 (a 1)(b 1) = ab a b + 1 = ab + a + b + 1 = H (b 1)(a 1) = ba b a + 1 = ab + a + b + 1 = H. Thus (IH) 2 is generated by H as a left kh-module, but also as a k-vector space by Exercise 6.2(a). (b) Let G = A 4 = (C 2 C 2 ) C 3 and let F 4 be the field with four elements. Compute the radical series of each of the three indecomposable projectives for F 4 A 4 and identify each of the quotients Rad n P S / Rad n+1 P S. Now do the same for the socle series. Hence determine the Cartan matrix of F 4 A 4. [Start by observing that F 4 A 4 has 3 simple modules, all of dimension 1, which one might denote by 1, ω and ω 2. This exercise may be done by applying the kind of calculation which led to Proposition 8.9] By Proposition 8.8(a), the simple F 4 G-modules are the simple F 4 C 3 -modules. Write C 3 = t t 3 = 1 and F 4 = {0, 1, ω, ω 2 }. Since F 4 = C 3, there are three group homomorphisms from C 3 to F 4. They send t to 1, ω and ω 2, respectively. Each of these homomorphisms yield a one-dimensional F 4 C 3 -module, which we also denote by 1, ω and ω 2. Clearly, here 1 is the trivial representation and the other two are nontrivial. Also ω ω 2, as the element ωt ω 2 F 4 C 3 annihilates the module ω but not ω 2. Thus 1, ω and ω 2 are non-isomorphic. Since dim F 4 C 3 = 3, they form a complete list of simple F 4 C 3 -modules (because every simple occurs at least once in the composition series of the regular representation). Write H = C 2 C 2 as in (a). Note that a 1 and b 1 both annihilate H, so (IH) 3 = 0. By Proposition 8.8, the radical series of P 1 (P 1 is the projective cover of the trivial module 1, it is P k in the notation of the proposition) is 0 (IH) 2 = H IH F 4 H = P 1. The action of C 3 on the above modules is by conjugation, we may choose the conjugation by t on H acting as a b ab a. To determine the radical layers of P 1, note that C 3 and H both permute the elements of H, so they fix H. Hence the bottom layer is trivial, that is, isomorphic to 1. By the virtue of being a projective cover of 1, the top layer is also 1. Now we investigate the middle layer IH/ H. Note that 1 + ωa + ω 2 b IH H and we have a (1 + ωa + ω 2 b) = a + ω + ω 2 ab ω + a + ω 2 (a + b + 1) (mod H) = ω + a + ω 2 a + ω 2 b + ω 2 = 1 + ωa + ω 2 b

3 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 3 b (1 + ωa + ω 2 b) = (b + ωab + ω 2 ) ω 2 + b + ω(a + b + 1) (mod H) = ω 2 + b + ωa + ωb + ω = 1 + ωa + ω 2 b t (1 + ωa + ω 2 b) = (1 + ωb + ω 2 ab) 1 + ωb + ω 2 (a + b + 1) (mod H) = 1 + ωb + ω 2 a + ω 2 b + ω 2 = ω + ω 2 a + b = ω(1 + ωa + ω 2 b). Thus the one-dimensional F 4 -vector space spanned by the coset (1 + ωa + ω 2 b) + H is an F 4 A 4 -submodule of IH/ H which is isomorphic to ω (Note that checking the action of t would have been enough above because as the radical layers are semisimple, H must act trivially on them). Now consider the element 1 + ω 2 a + ωb. We have t (1 + ω 2 a + ωb) = 1 + ω 2 b + ωab 1 + ω 2 b + ω(a + b + 1) (mod H) = 1 + ω 2 b + ωa + ωb + ω = ω 2 + ωa + b = ω 2 (1 + ω 2 a + ωb). Thus the one-dimensional F 4 -vector space spanned by the coset (1 + ω 2 a + ωb) + H is an F 4 A 4 -submodule of IH/ H which is isomorphic to ω 2. Since dim IH = dim F 4 H 1 = 3, we have dim Ä IH/ H ä = 2. Thus we conclude that IH/ H = ω ω 2. Using Proposition 8.8 again, we also conclude that the radical layers of P ω are ω, ω 2 1, ω and the radical layers of P ω 2 are ω 2, 1 ω, ω 2. Thus the Cartan matrix of F 4 A 4 is (with respect to any ordering of 1, ω, ω 2 ). The socle series of these indecomposable projectives coincide with their radical series because of the following: Proposition 1. Let k be a field and G a finite group. If S is a simple kg-module such that the Loewy length of P S is at most 3, then the socle and radical series of P S coincide. Proof. We give a proof when the Loewy length is 3, other cases are similar or trivial. We have 0 Rad 2 P S Soc P S. Since Soc P S = S is simple, we get Rad 2 P S = Soc P S. We also have Rad P S Soc 2 P S P S. As P S / Rad P S = S is simple, we get Rad PS = Soc 2 P S.

4 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 4 (c) Now consider F 2 A 4 where F 2 is the field with two elements. ñ Prove ô that the dimensional F 2 -vector space on which a generator of C 3 acts via is a simple 1 1 F 2 C 3 -module. Calculate the radical and socle series for each of the two indecomposable projective modules for F 2 A 4 and hence determine the Cartan matrix of F 2 A 4. ñ d We are asked to show that S = : d, e F 2 is a simple F 2 C 3 -module where the eô ñ ñ ô d e generator t acts via t =. Suppose V is a proper nonzero submodule eô d + e of S. Since dim F2 S = 2, we have dim F2 V = 1. Because there is only one group homomorphism ñ from ñ C 3 to (F ñ 2 ), V ñ must ô be a trivial module. So there exists 0 d d d e V such that = t =, which yields d = e = 0; a contradiction. eô eô eô d + e So we have at least two simple F 2 C 3 -modules: the trivial module k and S. Since dim k + dim S = 3 = dim F 2 C 3, there are no other simples. These are also the simple modules of F 2 A 4. Similar to part (b), the radical series of P k is 0 H IH P k and the top and bottom radical layers are k. We give two ways to show that the middle layer is isomorphic to S: First, if it wasn t S, it would have to be k, making all composition factors of P k isomorphic to k. This is impossible by Theorem 8.10 because A 4 does not have a normal 2-complement. Second approach is by direct calculation. Consider the elements a + 1 and b + 1 in IH. We claim that the cosets corresponding to these elements in IH/ H are F 2 - independent. So assuming there exists c, d, e F 2 such that c(a + 1) + d(b + 1) = eh, we show that c and d must be zero. Indeed, ca + c + db + d = ea + eb + eab + e (c + d + e) + (c + e)a + (d + e)b + eab = 0. Hence c + e = d + e = e = 0, so c = d = 0. So these cosets form a basis for IH/ H. Moreover, the action of t C 3 on these basis elements is given by t (a + 1) = b + 1 t (b + 1) = ab + 1 a + b (a + 1) + (b + 1) (mod H). This is the same as the action of ñ ô 0 1, therefore IH/ H 1 1 = S as F 2 A 4 -modules. By Proposition 8.8, the radical layers of P S are S, S S and S from bottom to top. We give two ways two decompose the semisimple module S S into a direct sum of simples. First way is computational, dealing with basis. Note that S S is a 4-dimensional F vector space where t acts via the matrix A =. By calculation, the Smith

5 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 5 X X 1 1 normal form of XI A = XI + A = 0 1 X 1 is X X X Thus as an F 2 [X]-module where X acts as A, S S is isomorphic to F 2 [X] (X + 1) F 2[X] (X 3 + 1) = F 2[X] (X + 1) F 2[X] (X + 1) F 2 [X] (X 2 + X + 1) Hence A is similar to the matrix F 2 C 3 -module (hence as an F 2 A 4 -module), S S = k k S. From here we conclude that as an The second way to see is more conceptual. As a 2-dimensional simple, the dual module S must be isomorphic to S. And we have Hence Hom F2 C 3 (S F2 S, k) = Hom F2 C 3 (S, Hom F2 (S, k)) = Hom F2 C 3 (S, S ) = Hom F2 C 3 (S, S). dim Hom F2 C 3 (S S, k) = dim End F2 C 3 (S) = 2 because the multiplicity of S in F 2 C 3 is 1. Thus we conclude that the multiplicity of k in S S is 2 and hence S S = k k S. In summary, the radical layers of P k are k,s,k and those of P S are ñ S, kô k S, S. Thus 2 2 the Cartan matrix of F 2 A 4, with the order k, S of the simples, is. The socle series 1 3 of P k and P S coincide with their radical series by Proposition 1. We see that the Cartan matrix is not symmetric here, and the reason is because the field F 2 is not big enough. 3. Let G = H K where H is a p-group, K is a p -group, and let k be a field of characteristic p. Regard kh as a kg-module via its isomorphism with P k, so H acts as usual and K acts by conjugation. (a) Show that for each n, (IH) n is a kg-submodule of kh, and that (IH) n /(IH) n+1 is a kg-module on which H acts trivially. (IH) n is evidently preserved under the action of H, so it suffices to show that K (IH) n (IH) n. Employ induction on n. Now an arbitrary element of IH is of the form a h h where a h = 0. So for g K, we have h H h H g a h h = a h ghg 1 h H h H = h H a g 1 hg h. Since h g 1 hg permutes H, we have a g 1 hg = a h = 0. This finishes the basis h H h H step.

6 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 6 Note that the action of K on H respects the group multiplication, that is g (hh ) = (g h)(g h ) for g K. Thus using the induction hypothesis, we have K (IH) n+1 = K ((IH)(IH) n ) = (K IH)(K (IH) n ) (IH)(IH) n = (IH) n+1. Since H is a p-group, as a kh-module, (IH) n /(IH) n+1 = Rad n kh/ Rad n+1 kh is semisimple, hence has trivial H-action by Proposition 6.3. (b) Show that P k = kh IH (IH 2 ) (IH) 3 is the radical series of P k as a kg-module. This follows from Proposition 8.8(e). (c) Show that there is a map IH/(IH) 2 k (IH) n /(IH) n+1 (IH) n+1 /(IH) n+2 x + (IH) 2 y + (IH) n+1 xy + (IH) n+2 which is a map of kg-modules. Deduce that (IH) n /(IH) n+1 is a homomorphic image of (IH/(IH) 2 ) n. It suffices to show that IH/(IH) 2 (IH) n /(IH) n+1 (IH) n+1 /(IH) n+2 (x + (IH) 2, y + (IH) n+1 ) xy + (IH) n+2 is a well-defined k-bilinear map. The multiplication map IH (IH n ) (IH) n+1 is clearly k-bilinear, so we only need to show well-definition. So suppose x, x IH and y, y (IH) n such that x x (IH) 2 and y y (IH) n+1. Then xy x y = xy xy + xy x y = x(y y ) + (x x )y (IH) n+2. For the last claim, induct on n. For n = 1, the claim is trivial via the identity map. Suppose that there is a surjective map (IH/(IH) 2 ) n (IH) n /(IH) n+1. Tensoring with the identity map on IH/(IH) 2, we get a surjective map (IH/(IH) 2 ) n+1 = (IH/(IH) 2 ) n IH/(IH) 2 (IH) n /(IH) n+1 IH/(IH) 2. Finally post composing this with the (clearly surjective) map defined in the first map we get a surjective map (IH/(IH) 2 ) (n+1) (IH) n+1 /(IH) n+2. (d) Show that the abelianization H/H becomes a ZG-module under the action g xh = gxg 1 H. Show that the isomorphism IH/(IH) 2 k Z H/H specified by (x 1) + (IH) 2 1 xh of Chapter 6 Exercise 17 is an isomorphism of kg-modules. Suppose xh = yh where x, y H. So x 1 y H and for g G we have (gxg 1 ) 1 (gyg 1 ) = gx 1 g 1 gyg 1 = gx 1 yg 1 gh g 1 = H

7 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 7 because H is characteristic in H and H is normal in G, which implies that H is normal in G. So the action is well-defined. And since conjugation by g is a group homomorphism on H, the induced action on the abelian group H/H is Z-linear. Denote the specified map by ϕ. It suffices to show that ϕ preserves both the H-action and the K-action. The H-action on IH/(IH) 2 is trivial by part (a) and the H acts trivialy on H/H too (H absorbs conjugation by H). And for a K, ϕ Ä a Ä (x 1) + (IH) 2ää = ϕ Ä (axa 1 1) + (IH) 2ä = 1 axa 1 H = 1 a (xh ) = a (1 xh ) = a ϕ ÄÄ (x 1) + (IH) 2ää. 4. The group SL(2, 3) is isomorphic to the semi direct product Q 8 C 3 where the cyclic group C 3 acts on Q 8 = {±1, ±i, ±j, ±k} by cycling the three generators i, j and k. Assuming this structure, compute the radical series of each three indecomposable projectives for F 4 SL(2, 3) and identifyy the radical quotients [Use Chapter 6 Exercise 15. ] Rad n P S / Rad n+1 P S. Will come back to this after I fully understand the exercise in Ch Let G = P S 3 be a group which is the semi direct product of a 2-group P and the symmetric group of degree 3. (Examples of such groups are S 4 = V S 3 where V = (1, 2)(3, 4), (1, 3)(2, 4) and GL(2, 3) = Q 8 S 3 where Q 8 is the quatertnion group of order 8.) (a) Let k be a field of characteristic 2. Show that kg has two non-isomorphic simple modules. By Corollary 6.4, the normal 2-subgroup P of G acts trivially on the simple kg-modules and consequently the simple kg-modules coincide with the simple ks 3 -modules via the projection G S 3. Since the abelianization of S 3 is C 2 and there is no element of order 2 in k, there is no group homomorphism from S 3 to k other than the trivial one. Hence the only one-dimensional (simple) ks 3 -module is the trivial module k. There is a group homomorphism S 3 GL 2 (k) ñ ô 0 1 (1, 2) 1 0 ñ ô 0 1 (1, 2, 3) 1 1 (see page 2 noting that char k = 2) which yields a two-dimensional ks 3 -module V. It is straightforward to check that V has no nonzero elements fixed by G, hence does not

8 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 8 contain a copy of k. Since k is the only one-dimensional ks 3 -module, we deduce that V is simple. Let r the multiplicity of V in the radical quotient of ks 3. Then by Theorem 7.14, r = dim D (V ) where D = End ks3 (V ñ ). Observe ô ñ that by ô our description of ñ V, D ô V is precisely a b the centralizer of the matrices and in M (k). Let A = D. First, c d we have Çñ ô ñ ôå Çñ ô ñ ôå A = A ñ ô ñ ô A = A ñ ô ñ ô 0 b 0 0 = 0 d c d ñ ô a 0 so b = c = 0 and A =. Second, we have 0 d ñ ô ñ ô A = A ñ ô ñ ô 0 a 0 d =, d 0 a 0 ñ ô a 0 so a = d and A =. Therefore D consists of scalar matrices, hence is isomorphic 0 a to k. Consequently r = dim k (V ) = 2. It follows that the radical quotient ks 3 / Rad kg contains a copy of k V V. But this already covers 5 dimensions and since ks 3 is not semisimple, the radical quotient can t have dimension more than 5. Thus ks 3 / Rad ks 3 = k V V and so k, V are the only simple ks 3 -modules. (b) Let e 1, e 2, e 3 be the orthogonal idempotents which appeared in Example 7.5. Show that each e i is primitive in F 4 G and that dim F 4 Ge i = 2 P for all i. [Use the fact that F 4 Ge i are projective modules.] Consider the (natural) surjective ring homomorphism ϕ : F 4 G F 4 [G/P ] = F 4 S 3. Note that ϕ fixes e i s. Using Exercise 10 and 3 in Chapter 6, we have that ker ϕ = F 4 G IP = F 4 G Rad F 4 P Rad F 4 G is nilpotent. Since e i s were shown to be primitive in F 4 S 3 in Example 7.5, they are also primitive in F 4 G by Theorem Since each F 4 Ge i is a projective F 4 G-module, by Corollary 8.3 G 2 = 2 P divides dim F 4 Ge i. In particular, as e i is nonzero we have 2 P dim F 4 Ge i. Because e is are orthogonal, we get 6 P = G = dim F 4 G dim(f 4 Ge 1 + F 4 Ge 2 + F 4 Ge 3 ) = dim F 4 Ge 1 + dim F 4 Ge 2 + dim F 4 Ge 3 6 P.

9 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 9 Thus the above inequalities can t be strict; so F 4 G = F 4 Ge 1 F 4 Ge 2 F 4 Ge 3 and dim F 4 Ge i = 2 P. (c) Show that if e 1 = () + (1, 2, 3) + (1, 3, 2) then F 4 S 4 e 1 is the projective cover of the trivial module and that F 4 S 4 e 2 and F 4 S 4 e 3 are isomorphic, being copies of the projective cover of a 2-dimensional module. Since S 4 = V S 3, the largest semisimple quotients of F 4 S 4 and F 4 S 3 are the same, which is k S S by Example 7.5. Thus by Theorem 7.14, the multiplicity of P k and P S in F 4 S 4 are 1 and 2, respectively. Now let n = g S 4 g F 4 S 4. Since ne 1 = 3n = n, the indecomposable projective F 4 S 4 e 1 contains a trivial submodule. Thus by Theorem 8.15, F 4 S 4 e 1 = Pk. And by Theorem 7.14, F 4 S 4 e 2 = PS = F4 S 4 e 3. (d) Show that F 4 Ge i = F4 (1, 2, 3) e i G (1,2,3) for each i. Write H = (1, 2, 3). Note that each e i lies in F 4 H. Since F 4 H is a semisimple ring, F 4 He i are projective F 4 H-modules, hence F 4 He i G H are projective F 4 G-modules. By orthogonality of the e i and the fact that dim F 4 H = 3, we deduce that dim F 4 He i = 1. Thus dim F 4 He i G H= G : H = G 3 = S 3 P 3 = 2 P, which yields, by (c), that F 4 He i G H are indecomposable. And again by (c), we only need to distinguish which one is the projective cover of the trivial module. Now, F 4 He 1 is generated by h H h, hence is a copy of the trivial F 4 H-module. Thus F 4 He 1 G H = F 4 [G/H] is a permutation F 4 G-module. Since permutation modules surject on the trivial module, it must be isomorphic to P k. 6. Let A be a finite-dimensional algebra over a field k, and let A A be the right regular representation of A. The vector space dual (A A ) = Hom k (A A, k) becomes a left A-module via the action (af)(b) = f(ba) where a A, b A A and f (A A ). Prove that the following two statements are equivalent: (a) (A A ) = A A as left A-modules. (b) There is a non-degenerate associative bilinear pairing A A k. An algebra satisfying these conditions is called a Frobenius algebra. Prove that, for a Frobenius algebra, projective and injective modules are the same thing. By basic set theory, there is a bijection Φ : {functions from A A to k} {functions from A to k A } defined by Φ(f)(a)(b) = f(a, b). Now we investigate step by step how Φ preserves the various structures. Let f : A A k. Φ(f) is k-linear if and only if f is linear in the first argument. k-linear if and only if For Φ(f) is Φ(f)(λa + a ) = λφ(f)(a) + Φ(f)(a )

10 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 10 for every a A and λ k. And that happens if and only if Φ(f)(λa + a )(b) = (λφ(f)(a) + Φ(f)(a )) (b) Φ(f)(λa + a )(b) = λφ(f)(a)(b) + Φ(f)(a )(b) f(λa + a, b) = λf(a, b) + f(a, b) for every b A. Φ(f) maps A to A if and only if f is linear in the second argument. Indeed, for a A, Φ(f)(a) A if and only if for every λ k and b A. Hence Φ restricts to a bijection Φ(f)(a)(λb + b ) = λφ(f)(a)(b) + Φ(f)(a)(b ) f(a, λb + b ) = λf(a, b) + f(a, b ) Φ : {f : A A k f is a k-bilinear form} Hom k (A, A ). Now, we observe the following: Assume Φ(f) is an isomorphism. First, suppose there exists a such that f(a, b) = 0 for every b A. So Φ(f)(a)(b) = 0 for every b A and this means Φ(f)(a) = 0. Thus a = 0 since Φ(f) is injective. Second, assume that there exists b such that f(a, b) = 0 for every a A. Suppose b is not zero. Then there exists δ A such that δ(b) = 1 and since Φ(f) is surjective, δ = Φ(f)(a 0 ) for some a 0. Hence 0 = f(a 0, b) = Φ(f)(a 0 )(b) = δ(b) = 1, a contradiction. We conclude that f is non-degenerate. Conversely, assume f is non-degenerate. We show that ker Φ(f) = 0, which is equivalent to showing Φ(f) is an isomorphism because A and A have the same dimension: Suppose Φ(f)(a) = 0 for some a. This means 0 = Φ(f)(a)(b) = f(a, b) for all b A; hence by non-degeneracy a = 0. Thus, Φ further restricts to a bijection Φ : {non-degenerate k-bilinear forms on A A} Iso k (A, A ). Finally, we observe that Φ(f) : A A ( A A) is a right A-module homomorphism ( A A is a right A-module via (g c)(d) = g(cd)) if and only if Φ(f)(ac) = Φ(f)(a) c for every a, c A. And that happens if and only if Φ(f)(ac)(b) = (Φ(f)(a) c) (b) Φ(f)(ac)(b) = Φ(f)(a)(cb) f(ac, b) = f(a, cb) for every b A. That is, Φ(f) is a right A-module homomorphism if and only if f is associative. Note that in this case the dual map of Φ(f) yields a left A-module homomorphism from A A to (A A ). That is, Φ establishes the equivalence of (a) and (b) up to a dualization. Alternatively, defining Ψ : {functions from A A to k} {functions from A to k A } by Ψ(f)(a)(b) = f(b, a) would directly lead to the desired isomorphism in a similar fashion with Φ.

11 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 11 Now we show the coincidence of projectives and injectives. Although the coincidence holds in general (T.Y. Lam s Lectures on Modules and Rings have material related with this), I can show it only for finitely generated modules. Let I be a finitely generated injective left A-module. Then I is a finitely generated projective right A-module (the right action of A on the dual of an arbitrary left A-module is defined by (f a)(m) = f(am)). Hence I is a summand of (A A ) n = ((A A) ) n = ((A A) n ) for some n. Thus I is a summand of ( A A) n, hence projective. Conversely, let P be a finitely generated projective left A-module. Hence P = P is a summand of ( A A) n = ((AA ) ) n = ((AA ) n ) for some n. Thus P is a summand of (A A ) n, so is projective. Therefore P = P is injective. 7. Let A be a finite-dimensional algebra over a field k and suppose that the left regular representation A A is injective. Show that every projective module is injective and that every injective module is projective. Again I only show the conclusion for finitely generated modules. Let P be a finitely generated projective left A-module. So P is a summand of ( A A) n for some n. But ( A A) n is injective, hence so is P. Conversely??? Need a better understanding of quasi-frobenius rings first. 8. Let S and T be simple kg-modules, with projective covers P S and P T, where k is an algebraically closed field. (a) For each n prove that Hom kg (P T, Soc n P S ) = Hom kg (P T / Rad n P T, Soc n P S ) = HomkG (P T / Rad n P T, P S ) This is true quite in general. Let A be a finite dimensional algebra and let L and M be finite dimensional (left) A-modules. We show that Hom A (L, Soc n M) = Hom A (L/ Rad n L, Soc n M) = HomA (L/ Rad n L, M). For the first isomorphism, it suffices to show that for any ϕ Hom A (L, Soc n M), we have Rad n L ker ϕ. Indeed, by the descriptions of the higher radicals and socles discussed after Proposition 6.9, ϕ(rad n L) = ϕ((rad A) n L) = (Rad A) n ϕ(l) (Rad A) n Soc n (M) = 0. And for the second isomorphism, it suffices to show that given ψ Hom A (L/ Rad n L, M), we have im(ψ) Soc n M. Indeed, since (Rad A) n annihilates the domain of ψ, it also annihilates im(ψ) and we are done. (b) Deduce Landrock s theorem: the multiplicity of T in the nth socle layer of P S equals the multiplicity of S in the nth radical layer of P T.

12 Write SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 12 ν n ST = the multiplicity of S in Rad n 1 T/ Rad n T, µ n T S = the multiplicity of T in Soc n S/ Soc n 1 S. Note that since k is algebraically closed, dim Hom kg (P T / Rad n P T, P S ) is the multiplicity of S as a composition factor in P T / Rad n P T, which is equal to ν 1 ST + ν n ST. On the other hand, dim Hom kg (P T, Soc n P S ) is the multiplicity of T as a composition factor in Soc n P S, which is equal to µ 1 T S + + µ n T S. Thus by using (a), we have ν 1 ST + ν n ST = µ 1 T S + + µ n T S for every n. Therefore ν n ST = µ n T S for every n. (c) Use Exercise 6 of Chapter 6 to show that these multiplicities equal to the multiplicity of T in the nth radical layer of P S, and also the multiplicity of S in the nth socle layer of P T. Using Exercise 6.6 and Corollary 8.16(2), we have Ä Hom kg Rad n 1 P S / Rad n ä Ä P S, P T = HomkG (Rad n 1 (P S ) / Rad n (P S ), (P T ) ä = HomkG ÄÄ Soc n P S / Soc n 1 P S ä, (PT ) ä = HomkG Ä PT, Soc n P S / Soc n 1 P S ä. Thus taking the k-dimensions of both sides in the above isomorphism, we get νt n S = µ n T S. Similarly, the isomorphism Ä Hom kg Soc n P T / Soc n 1 ä Ä P T, P S = HomkG Soc n (P T ) / Soc n 1 (P T ), (P S ) ä yields µ n S T = νn ST. = HomkG ÄÄ Rad n 1 P T / Rad n P T ä, (PS ) ä = HomkG Ä PS, Rad n 1 P T / Rad n P T ä 9. Let U be an indecomposable kg-module, where k is a field of characteristic p, and let P k be the projective cover of the trivial module. Prove that dim Ä G U ä 1 if U = = P k 0 otherwise. For an arbitrary finite dimensional module V, show that dim(g V ) is the multiplicity with which P k occurs as a direct summand of V. [Observe that kg G = Pk G = k G. Remember that P k is injective and has socle isomorphic to k.] Let s start with observing what Webb wants us to observe. Since trivial kg-modules are in particular semisimple, M G Soc(M) for any kg-module M. Therefore if S is a nontrivial simple kg-module, then PS G = 0 since Soc(P S ) = S. Thus if we write kg = (P S ) n S, simple S as the decomposition of the regular module into indecomposable projectives, we have kg G = (P G k ) n k.

13 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 13 Note that an element g G a g g of the group algebra lies in kg G if and only if a g g = t a g g = a g tg g G g G g G for every t G. Comparing the coefficient of t, we get a t = a 1 for every t. Hence kg G = k G is one-dimensional and so n k = 1. I am able to show the result only for finite-dimensional modules. First a lemma: Lemma 2. Let A be a finite-dimensional algebra over a field k. Let f : L M be a homomorphism of finite-dimensional left A-modules where Soc(L) is simple. Write f : Soc(L) Soc(M) for the map f induces between the socles. TFAE: (i) f is injective. (ii) f is nonzero. (iii) f is injective. Proof. (i) (ii) is clear. And (ii) (iii) holds because Soc(L) is simple. Now assume (iii). Then we have a commutative diagram Soc(L) L i f Soc(M) f Suppose ker f 0. Then, being a nonzero Artinian module, ker f has nonzero socle, hence we have 0 Soc(ker f) Soc(L) which implies Soc(ker f) = Soc(L). In particular Soc(L) ker f, so j f = f i = 0. This is a contradiction because j f is injective and Soc(L) 0. Corollary 3. Let f : P U be a homomorphism of finite-dimensional kg-modules where P is an indecomposable projective and U is indecomposable. Write f : Soc(P ) Soc(U) for the map f induces between the socles. TFAE: (i) f is an isomorphism. (ii) f is nonzero. Proof. (i) (ii) is trivial. Conversely, assume (ii). Since Soc(P ) is simple, f is injective by Lemma 2. Since P is also an injective module, hence f is a split monomorphism. But U is indecomposable and P 0, so f must be an isomorphism. Going back to the question, let U be a finite-dimensional indecomposable kg-module. Then there is a surjection j M. ϕ : P 1 P n U where P 1,, P n are indecomposable projectives. Assume G does not annihilate U. Then there exists 1 r n such that G ϕ(p r ) 0. In particular, 0 G P r Pr G so by the first part we conclude that P r = Pk. Let f : P r U be the restriction of ϕ to P r. Then since G P r Soc(P r ) and f(g P r ) = G f(p r ) = G ϕ(p r ) 0, the induced map f : Soc(P r ) Soc(U) is nonzero. Hence by Corollary 3, f is an isomorphism.

14 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 14 Conversely, if U = P k, since G kg = k G 0 and on the other hand G annihilates every indecomposable projective other than P k, we have G P k 0. Hence G P k = Pk G = k G has dimension 1. Finally, let V be a finite-dimensional kg-module. So it has a decomposition V = U 1 U s into indecomposable modules (in a unique way by Krull-Schmidt). Hence s dim(g V ) = dim(g U j ) j=1 Ñ é s 1 if U j = Pk = j=1 0 otherwise. = number of P k s as a summand of V. 10. Let U be a finite-dimensional kg-module, where k is a field, and let P S be an indecomposable projective kg-module with simple quotient S. Show that in any decomposition of U as a direct sum of indecomposable modules, the multiplicity with which P S occurs is equal to dim Hom kg (P S, U) dim Hom kg (P S / Soc P S, U) dim End kg (S) and also to dim Hom kg (U, P S ) dim Hom kg (U, Rad P S ). dim End kg (S) Denote the multiplicity of P S as a direct sum of U by d S (U) (this is a well-defined number by Krull-Schmidt) and the multiplicity of S as a composition factor of U by n S (U). Observe that both d s and the suggested formulas as functions of U are additive, that is, they send direct sums to the sum of the corresponding numbers for the summands. Hence it suffices to verify the claims when U is a finite-dimensional indecomposable module. There are two cases: 1. U = P S. Then, dim Hom kg (P S, U) dim Hom kg (P S / Soc P S, U) dim End kg (S) = dim Hom kg(p S, P S ) dim Hom kg(p S / Soc P S, P S ) dim End kg (S) dim End kg (S) = n S (P S ) n S (P S / Soc P S ) = n S (Soc P S ) = n S (S) = 1, using Proposition 7.17(2) and its dual for symmetric algebras. Similarly, dim Hom kg (U, P S ) dim Hom kg (U, Rad P S ) dim End kg (S) = dim Hom kg(p S, P S ) dim Hom kg(p S, Rad P S ) dim End kg (S) dim End kg (S) = n S (P S ) n S (Rad P S ) = n S (P S / Rad P S ) = n S (S) = 1.

15 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 15 On the other hand, we clearly have d S (U) = U P S. So d S (U) = 0 hence we would be done if we can show Hom kg (P S, U) = Hom kg (P S / Soc P S, U) for the first equality. Indeed, let f Hom kg (P S, U). Since f is not an isomorphism, by Corollary 3 we have f(soc P S ) = 0. So f factors through P S / Soc P S. For the second equality, it suffices to show that Hom kg (U, P S ) = Hom kg (U, Rad P S ). By taking duals, this is equivalent to showing Hom kg ((Rad P S ), U ) = Hom kg ((P S ), U ) Hom kg ((P S ) / Soc(P S ), U ) = Hom kg ((P S ), U ) Hom kg (P S / Soc P S, U ) = Hom kg (P S, U ). Above, we used Exercise 6.6 to pass to the second line and Corollary 8.16(2) to pass to the third line. And indeed the isomorphism in the third line does hold by the first part since U is an indecomposable which is not isomorphic to P S = (PS ). 11. Let k be an algebraically closed field of characteristic p and suppose that G has a normal p-complement, so that G = K H where H is a Sylow p-subgroup of G. Let S 1,..., S n be the simple kg-modules with projective covers P Si. (a) Show that there is a ring isomorphism kg = n i=1 M dim Si (End kg (P Si )) where the right hand side is a direct sum of matrix rings with entries in the endomorphism rings of the indecomposable projectives. [Copy the approach of the proof of Wedderburn s theorem.] By Theorem 7.14, the regular left module kg kg has a decomposition kgkg n = (P Si ) r i i=1 such that r i = dim Di S i where D i = End kg (S i ). Since G has normal p-complement and char k = p, all the composition factors of P Si are copies of S i. Hence Hom kg Ä (PSi ) r i, (P Sj ) r j ä = 0 if i j. Thus by a similar argument to the one in the proof of Wedderburn s theorem, we get a k-algebra isomorphism End kg ( kg kg) n = End kg ((P Si ) ri ) i=1 kg = (kg) op n = M ri (End kg (P Si )). i=1 Finally, by the last part of Theorem 7.14, we have r i = dim S i if k is algebraically closed. (b) For each i, show that if P Si = Pk S i then End kg (P Si ) = kh as rings. [Show that dim End kg (P Si ) = H. Deduce that the obvious map End kg (P k )

16 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 16 End kg (P k S i ) is an isomorphism.] (c) Show that if P Si P k S i, then End kg (P Si ) has dimension smaller than H. The proof of Theorem 8.10 shows that the regular module kh, considered as a kgmodule via the surjective ring homomorphism kg kh is projective. Note that in general if R S is a surjective ring homomorphism and M is an S-module, then the S-submodule lattice and the R-submodule lattice of M coincide. Thus virtually every property of M as an R-module that depends solely on the submodule structure also holds as an S-module (finite generation, Noetherian, Artinian, indecomposable, semisimple etc.). With this remark in mind, we conclude that kh is indecomposable as a kg-module because it is indecomposable as a kh-module since H is a p-group and char k = p (Corollary 6.12). Thus kh must be the projective cover of the trivial kg-module. Hence we have a ring isomorphism End kg (P k ) = End kg (kh) = End kh (kh) = (kh) op = kh where the equality above holds because kg kh is surjective. Let S be a simple kg-module. Write n S (U) for the multiplicity of S in U as a composition factor of U. By Theorem 7.17, n S (P S ) = dim End kg (P S ). On the other hand, we have n S (P k S) = n k (P k ) = n k (kh) = H. So if P S = Pk S, we have dim End kg (P S ) = H and if P S P k S, we have dim End kg (P S ) < H since then P S is a nontrivial summand of P k S. Suppose P S = Pk S. We know that the functor k S : kg-mod kg-mod is exact. Moreover it is faithful on finite-dimensional modules: let α : M N be a map of finite-dimensional kg-modules such that α S : M S N S is the zero morphism. Consider the exact sequence which gets sent to the exact sequence 0 ker α M α N 0 ker α S M S Then ker α S = M S, which yields 0 N S. dim ker α dim S = dim M dim S. Since S 0, we get dim ker α = dim M which implies that ker α = M and α = 0. Now, the faithful functor S induces an injective k-algebra map kh = End kg (P k ) End kg (P k S) = End kg (P S ). Since both sides of the above map has dimension H, it must be a ring isomorphism. 12. Let k be an algebraically closed field of characteristic p and let G = p a q where p q. Let S be a simple kg-module. Show that dim S q with equality if and only if G is a p-group.

17 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 17 In turns out that the claim is wrong. Let G = S 3 and k be an algebraically closed field of characteristic 2. By Exercise 5 in Chapter 8, kg has a two-dimensional simple module S. So in this case the claim reads 2 3, which is nonsense. 13. (a) Show that F 3 A 4 has just two isomorphism types of simple modules, of dimensions 1 and 3, and that the simple module of dimension 3 is projective. [Eliminate modules of dimension 2 by observing that a projective cover of such a module must have dimension at least 6.] A 4 has a normal 3-complement, namely H = {(), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)} = C 2 C 2. Let k be the trivial module. We claim that k is the only one-dimensional F 3 A 4 -module. Indeed, the commutator of A 4 is H and A 4 /H = C 3 and the only group homomorphism from C 3 (hence from A 4 ) to F 3 is the trivial one. Let Ω = {ω 1, ω 2, ω 3, ω 4 } be a set of 4 letters which A 4 naturally acts (via restriction of the action of S 4 ). Consider the F 3 A 4 -module homomorphism Let S = ker ɛ. Write ɛ : F 3 Ω k ω i 1. e 1 = ω 1 + ω 2 + ω 3 e 2 = ω 1 + ω 2 + ω 4 e 3 = ω 2 + ω 3 + ω 4. Note that e is lie in S. Actually e i s are linearly independent, because 0 = a 1 e 1 + a 2 e 2 + a 3 e 3 = (a 1 + a 2 )ω 1 + (a 1 + a 2 + a 3 )ω 2 + (a 1 + a 3 )ω 3 + (a 2 + a 3 )ω 4 implies a 1 = a 2 = a 3 = 0. Since dim S = 3, e i s form a basis for S. By inspection, we observe that with respect to this basis, the action of (1, 2, 3) and (1, 2)(3, 4) correspond to the matrices and 1 0 2, respectively Suppose S is not a simple F 3 A 4 -module. Then there are two cases since dim S = 3: a 1. k embeds in S. So S contains a nonzero element, say b written with respect c to the e i basis, is fixed by A 4. But a a b = b c c a + 2c = 2c b + 2c

18 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 18 implies b = c = 0 and a a 0 = = a 0 implies a = 0, a contradiction. 2. S has a 2-dimensional simple submodule. This is also impossible, as there are no 2-dimensional simple F 3 A 4 -modules: To the contrary, suppose T is a 2- dimensional simple. Consider its projective cover P T. On one hand, by Corollary 8.3, 3 divides dim P T and 2 = dim T also divides dim P T as A 4 has a normal 3-complement (all composition factors of P T are T by Theorem 8.10). On the other hand, P k T = F 3 C 3 T is a direct sum of copies of P T, hence dim P T divides dim F 3 C 3 dim T = 6. Thus dim P T = 6. Now P T and P k cover 9 dimensions of a total 12 in F 3 A 4. We just showed that 2-dimensional simples ought to have 6-dimensional projective covers, hence T is the only 2-dimensional simple. Thus the remaining 3 dimensions in F 3 A 4 can only be covered by a 3-dimensional simple U, which must also be projective. That is, we have F 3 A 4 = Pk P T U. So dim D U = 1 where D = End F3 A 4 (U). Hence, D is a 3-dimensional F 3 -division algebra. By Wedderburn s little theorem D is a field and so D = F 27 by the classification of finite fields. Then by the solution of Exercise 11, we have an F 3 -algebra isomorphism F 3 A 4 = F3 C 3 End F3 A 4 (P T ) F 27. The center of the left hand side has dimension equal to the number of conjugacy classes in A 4 (Lemma 3.15), which is 4. However, the center of the right hand side has dimension at least 6. This is a contradiction. Thus S is a simple F 3 A 4 -module. Let s compute D = End F3 A 4 (S). By our description of S, if we let A = and B = 1 0 2, D can be identified with the set of matrices in M 3 (F 3 ) that commute with A and B (since (1, 2, 3) and (1, 2)(3, 4) generate a b c A 4 ). Let M = d e f D be arbitrary. Then g h i ABM = MAB a b c a b c d e f = d e f g h i g h i d e f c a b g h i = f d e a b c i g h

19 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 19 a b c so d = c = h, e = a = i and f = b = g. So M = c a b. Now b c a AM = MA a b c a b c c a b = c a b b c a b c a a + 2b b + 2c c + 2a a c 2a + 2b + 2c 2b 2c 2a = c b 2a + 2b + 2c. c + 2b a + 2c b + 2a b a 2a + 2b + 2c Since the last column is identical on the right hand side, we have c + 2a = 2a = b + 2a which yields b = c = 0. Thus D consists of scalar matrices and D = F 3. We conclude that the multiplicity of S in F 3 A 4 / Rad F 3 A 4 is dim D (S) = dim S = 3. Hence the number of direct summands in F 3 A 4 isomorphic to P S is also 3. But P k = F 3 C 3 already covers 3 of the 12 dimensions in F 3 A 4, so the only way for three copies of P S to fit in 9 dimensions is that P S = S, that is, S is projective. (b) Show that F 3 A 4 = F3 C 3 M 3 (F 3 ) as rings. Since D = End F3 A 4 (S) = F 3, Exercise 11 immediately yields an F 3 -algebra isomorphism F 3 A 4 = F3 C 3 M 3 (F 3 ). 14. Let k be a field of characteristic p and let G = H K where H is a p-group and K has order prime to p. Show that Rad n (kg) = Rad n (kh) G H as kg-modules. This drops out as a corollary of a general result: Proposition 4. Let G be a finite group, R a commutative ring and H G. Let J be a left ideal of RH, as a subset of RG. Let I be the left ideal of kg that J generates. Then I = J G H as left RG-modules. Proof. We have I = g G gj. Since J RH, gj tj 0 implies that gh = th. So g = th for some h H and hence gj = thj = tj. Thus I is the direct sum of the R-submodules {gj : g G}. Since H = {g G : gj = J}, by Proposition 4.8 I = J G H. In the situation of the exercise, write I = Rad kg and J = Rad kh. By Proposition 8.8(b), I is the (left) ideal of kg that J generates. Therefore I n is the ideal of kg that J n generates. Hence by the above proposition, Rad n (kg) = I n = J n G H= Rad n (kh) G H. Chapter 9 1. Let E = F p (t) be a transcendental extension of the field with p elements and let F be the subfield F p (t p ). Write α = t p, so that t p α = 0. Let A = E, regarded as an F -algebra.

20 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 20 (a) Show that A has a simple module which is not absolutely simple. Since A is a field, S = A A is a simple left A-module. Since the F -algebra End A (S) = A op = A is not isomorphic to F, S is not absolutely simple. (b) Show that E is a splitting field for A, and that the regular representation of A E is a uniserial module. Show that Rad(A E ) (Rad A) E. [Notice that R F Rad A is always contained in the radical of E F A when A is a finite-dimensional algebra, being a nilpotent ideal.] As an F -algebra, A E = E F E. Let J be the ideal of E F E generated by elements of the form 1 f f 1. Since for every f E we have f p F, (1 f f 1) p = (1 f) p (f 1) p = 1 f p f p 1 = 0. Hence J is generated by nilpotent elements. Since E F E is a commutative ring, every element in J is nilpotent (in other words, J is a nil ideal). Thus J Rad (E F E). Now consider the F -algebra homomorphism µ : E F E E f g fg. As ϕ is surjective and E is a field, ker µ is a maximal ideal of E F E. We want show that the chain of inclusions J Rad(E F E) ker µ reduces to equalities. And indeed we do have ker µ J because of the following general fact applied to the inclusion F E. Proposition 5. Let R S be a ring homomorphism. This puts an R, R-bimodule structure on S and a ring structure on S R S. The kernel of the ring homomorphism µ : S R S S a b ab is the ideal generated by {1 s s 1 : s S}. Proof. Let J be the ideal generated by such elements; clearly J ker µ. An arbitrary n n element u ker µ is of the form u = a i b i such that a i b i = 0. We want to show i=1 i=1 n u J. Write a = a 1 and b = b 1. So we have ab = a i b i and we want to show that i=2 n a b a i b i (mod J). i=2

21 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 21 (In case n = 1, the sums starting from 2 are interpreted as 0). Indeed, a b 1 ab (mod J) n = 1 a i b i i=2 n = (1 a i b i ) i=2 n a i b i (mod J) i=2 where the first and the last line follows from a b 1 ab = (1 b)(a 1 1 a) J. Back to the question. Now we have J = Rad A E. Write M for the regular representation of A E. We will show that the radical series M = E F E J J 2... of M is a composition series; this implies that M is uniserial by Exercise 5 in Chapter 6. Note that [E : F ] = p and {1, t,..., t p 1 } is an F -basis of E. Therefore {t i t j : 0 i, j p 1} is an F -basis of E F E. Therefore p 1 ( ) (1 t t 1) p 1 p 1 = (1 t) p 1 r ( 1) r (t 1) r r=0 r p 1 ( ) p 1 = (1 t p 1 r )( 1) r (t r 1) r=0 r p 1 ( ) p 1 = ( 1) r (t r t p 1 r ) r r=0 isn t equal to zero. Hence J p 1 0. Therefore the Loewy length of M is at least p and so dim E (J r /J r+1 ) 1 for 0 r p 1. But on the other hand, dim E M = dim E A E = dim F A = [E : F ] = p. This forces dim E (J r /J r+1 ) = 1 for 0 r p 1 and J p = 0. Finally, 0 J = Rad A E whereas (Rad A) E = 0 E = 0. (c) Show that A is not isomorphic to F G for any group G. Suppose it is. Then since p = dim F A = dim F F G = G. Moreover F G is a field, so the augmentation map ɛ : F G F is injective (every ring homomorphism out of a field is injective). So we conclude that F G = F, which yields G = 1, a contradiction. 2. Let G be a cyclic group, k a field and S a simple kg-module. Show that E = End kg (S) is a field with the property that S E is a direct sum of modules which are all absolutely simple.

22 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 22 Suppose char k = p divides G. Let H be the (necessarily unique and normal) Sylow p-subgroup of G. Then H acts trivially on S by Corollary 6.4, making S naturally a k[g/h]-module. Note that G/H is also cyclic and E = End k[g/h] (S). So by replacing G with G/H if necessary, we may assume that char F does not divide G. Say G = g and G = n. Since kg is generated by g as a k-algebra, there is a surjective k-algebra map k[x] kg X g whose kernel contains the ideal (X n 1). So we get a surjective k-algebra map k[x]/(x n 1) kg which must be an isomorphism since both sides have dimension n. So we may write A = k[x]/(x n 1) where n is not divisible by char k. Since X n 1 is relatively prime with its formal derivative nx n 1, it is separable. Let X n 1 = f 1 f r be a factorization of X n 1 into irreducibles in k[x]. Then f i s are distinct by separability and hence by Chinese Remainder Theorem we have a k-algebra isomorphism A = k[x]/(x n 1) r = k[x]/(f i ) i=1 Since the above is an Artin-Wedderburn decomposition of A (each k[x]/(f i ) is a field) we conclude that if we write S i for k[x]/(f i ) considered as an A-module, {S 1,..., S n } is a complete list of simple A-modules up to isomorphism. Note that E i = End A (S i ) = k[x]/(f i ) as a k-algebra which is a field. So we have a field extension k E i. We claim that f i splits in E i. Note that f i has at least one root ζ in E i, namely the coset corresponding to X in the quotient. Since k(ζ) is a subfield of a cyclotomic extension of k (ζ has finite order), the extension k k(ζ) is normal. So being irreducible, f i splits in k(ζ) and in particular in E i. Thus the A E i -module S E i i = E i k S i = E i k k[x]/(f i ) = E i [X]/(f i ) is a direct sum of one-dimensional modules which are necessarily absolutely simple. 3. Let A be a finite-dimensional k-algebra which is split. Let k F be a field extension. Prove that Rad(A F ) = (Rad A) F. [The observation at the end of question 1(b) might help here.] 4. Let A be a finite-dimensional split k-algebra an let k F be a field extension. Show that every simple A F -module can be written in k. [Bear in mind the result of question 3.] 5. Let A be a finite-dimensional k-algebra and k F a field extension where F is a splitting field for A. Suppose that every simple A-module remains simple on extending scalars to F. Show that k is a splitting field for A.

23 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB 23 I use these notes for these problems. Proposition 7 is the 3rd problem and Theorem 10 gives the 4th problem. For the 5th, let S be a simple A-module. Then by assumption S F is a simple A F -module, and hence absolutely simple since F is a splitting field for A. So by Proposition 9.2, we have End A F (S F ) = F as an F -algebra. But by Corollary 4 in the notes, we have an F -algebra isomorphism Therefore End A F (S F ) = (End A (S)) F. dim k End A (S) = dim F End A F (S F ) = 1 and hence End A (S) = k. Again by Proposition 9.2, S is an absolutely simple A-module. 6. Let A be a finite-dimensional k-algebra and k F a field extension. (a) Show that if ϕ : U V is an essential epimorphism of A-modules then ϕ F : U F V F is an essential epimorphism of A F -modules. I will assume U, V (equivalently just U, since U surjects on V and A is Noetherian) to be finitely generated. Note that (Rad U) F = F k Rad U = F k (Rad A U) = (F k Rad A) (F k U) Rad(F k A) (F k U) = Rad(A F ) U F = Rad(U F ). where the inclusion comes from the fact that F k Rad A is a nilpotent ideal of F k A. Note that the second and last equalities do use the fact that U is a finitely generated A-module. Similarly we have (Rad V ) F Rad(V F ). As ϕ is essential, the induced A-module homomorphism U/ Rad U V/ Rad V is an isomorphism by Proposition 7.7(b). Applying the exact functor ( ) F, we get an A F -module isomorphism U F /(Rad U) F V F /(Rad V ) F. Note that since (Rad U) F Rad U F, we have Rad Ä U F /(Rad U) F ä = Rad U F /(Rad U) F by the definition of the radical as the intersection of maximal submodules. Therefore taking the radical quotients in the above isomorphism, we get an isomorphism U F / Rad U F V F / Rad V F which is definitely induced by U F V F. Thus again by Proposition 7.7(b), U F V F is an essential epimorphism. (b) Show that if P U is a projective cover then so is P F U F. Since the functor ( ) F : A-Mod A F -Mod is a left-adjoint, P F is a projective A F - module. As P U is essential, by part (a) so is P F U F.

24 SOLUTIONS FOR FINITE GROUP REPRESENTATIONS BY PETER WEBB Let A be a finite-dimensional k-algebra which is split. Let P be an indecomposable projective A-module. Show that if k F is any field extension then P F is indecomposable and projective as an A F -module. Show further that every indecomposable projective A F -module can be written in k. We know that P = P S for some simple A-module S. By Exercise 6(b), since P S is a projective cover, so is P F S F. And since S is absolutely simple (A is split!), S F is a simple A F -module. Therefore being the projective cover of a simple A F -module, P F is an indecomposable projective A F -module. Let T be a simple A F -module and Q be its projective cover. We want to show that Q can be written in k. By Exercise 4, we have T = S F for some simple A-module S. Let P be the projective cover of S. By what we just showed, P F is the projective cover of S F = T. By the uniqueness of projective covers, we deduce Q = P F. 8. Let G = C 2 C 2 be generated by elements a and b, and let E be a field of characteristic 2. Let t E be any element, which may be algebraic or transcendental over F 2. Let ρ : G GL 2 (E) be the representation with ñ ô 1 1 ρ(a) =, 0 1 ñ 1 ρ(b) = t 0 1 Show that this representation is absolutely indecomposable, and that it cannot be written in any proper subfield of F 2 (t). Let U be the 2-dimensional EG-module we get out of this representation. Suppose U is decomposable. Then U must be the direct sum of two 1-dimensional EG-submodules. But since G is a 2-group and char E = 2, the only 1-dimensional EG-module is the trivial one. As the action of G (in particular the action of a) on U is not trivial, we get a contradiction. Similarly for any field extension E K, we have char K = 2 and U K is a 2-dimensional KG-module where the action of G is nontrivial; hence U K is indecomposable. Thus U is absolutely indecomposable. For the second part, suppose L is a subfield of F 2 (t) and W an LG-module such that U = W E. Then dim L W = 2 and W is not trivial. Thus Soc(W ) is one-dimensional. Pick a generator v 1 for Soc(W ) and pick v 2 W Soc(W ). Then {v 1, v 2 } is a basis for W. Being in the ñ socle, ô ña andô b fix v 1 so their action with respect to such a basis is 1 λ 1 µ given by matrices, for some λ, µ L. Note that a can t act trivially on W since it acts nontrivially on U. Therefore λ 0. So by replacing v 1 with λv 1, we may assume that λ = 1. Let u 1, u 2 be the original basis of U. Note that u 1 Soc(U). Identifying W as a subspace of U, we have that v 1 is an F 2 (t)-scalar multiple of u 1. Replace u 1, u 2 by a scalar multiple such that u 1 = v 1 (note that this doesn t change the given matrices). Write u 2 = αv 1 + βv 2 with α, β F 2 (t). Now ô. u 1 = (a 1)u 2 = (a 1)(αv 1 + βv 2 ) = β(a 1)v 2 = βv 1 = βu 1