SOLUTIONS Math B4900 Homework 9 4/18/2018

SOLUTIONS Math B4900 Homework 9 4/18/2018 1. Show that if G is a finite group and F is a field, then any simple F G-modules is finitedimensional. [This is not a consequence of Maschke s theorem; it s just a really useful lemma for working with F G-modules in general.] Proof. An F G-module M is simple if and only if M F Gm for all m P M 0. So if M is simple, # + ÿ M F Gm αgm ˇ α P F. gpg So tgm g P Gu is a finite spanning set for M (as a vector space over F ), and therefore is a finite-dimensional vector space. 2. Let P Cte 1, e 2, e 3 u be the permutation representation of CS 3, and let T Cte 1 ` e 2 ` e 3 u be the module inside P. Following our proof of Maschke s theorem, pick C any basis of P that contains v e 1 ` e 2 ` e 3 ; let T 1 be the linear span of C tvu, so that P T T 1 as vector spaces (no need to make this decomposition work as modules); compute π 1 and π with respect to this choice of T 1 ; and verify that kerpπq is tc 1 e 1 ` c 2 e 2 ` c 3 e 3 c 1 ` c 2 ` c 3 0u (the orthogonal complement to T, which we ve studied before). Finally, make a different choice of basis C (containing v), and repeat this process. Did you get the same π in the end? Does it matter? Proof. Try 1: Let C tw 1, w 2, vu, where w 1 e 1 and w 2 e 1 ` e 2. Let T 1 Ctw 1, w 2 u tpα 1, α 2, 0q α i P F u. With respect to the decomposition P T T 1, the projection π 1 : P Ñ T is given by π 1 : α 1 e 1 ` α 2 e 2 ` α 3 e 3 ÞÑ α 3 v (since α 1 e 1 ` α 2 e 2 ` α 3 e 3 α 3 v pα 1 α 3 qe 1 ` pα 2 α 3 qe 2 P T 1 ). Next, we must compute gπ 1 g 1 applied to u α 1 e 1 ` α 2 e 2 ` α 3 e 3 pα 1, α 2, α 3 q. To that end, note that for all g P G 1

2 and x P T, we have gx x. So gπ g 1 1 π g 1 1. So 1π 1 1 1 u π 1 puq α 3 v; p12qπ p12q 1 1 u π 1 pα 2, α 1, α 3 q α 3 v; p23qπ p23q 1 1 u π 1 pα 1, α 3, α 2 q α 2 v; p13qπ p13q 1 1 u π 1 pα 3, α 2, α 1 q α 1 v; p123qπ p123q 1 1 u π 1 pα 2, α 3, α 1 q α 1 v; and p132qπ p132q 1 1 u π 1 pα 3, α 1, α 2 q α 2 v. So Therefore πpuq 1 6 pα 3 ` α 3 ` α 2 ` α 1 ` α 1 ` α 2 qv 1 3 pα 1 ` α 2 ` α 3 qv. kerpπq tpα 1, α 2, α 3 q α 1 ` α 2 ` α 3 0 R. Try 2: Let C tw 1, w 2, vu, where w 1 e 1 e 3 and w 2 e 2 e 3. Let T 1 Ctw 1, w 2 u tpα 1, α 2, α 1 α 2 q α i P F u. To compute π 1, we need to compute c 3 such that α 1 e 1 ` α 2 e 2 ` α 3 e 3 c 1 w 1 ` c 2 w 2 ` c 3 v so that (π 1 pα 1 e 1 ` α 2 e 2 ` α 3 e 3 q c 3 v). Continuing this computation, we have So Thus α 1 e 1 ` α 2 e 2 ` α 3 e 3 c 1 w 1 ` c 2 w 2 ` c 3 v c 1 pe 1 e 3 q ` c 2 pe 2 e 3 q ` c 3 pe 1 ` e 2 ` e 3 q pc 1 ` c 3 qe 1 ` pc 2 ` c 3 qe 2 ` p c 1 c 2 ` c 3 qe 3. α 1 c 1 ` c 3, α 2 c 2 ` c 3, α 3 c 1 c 2 ` c 3. c 3 1 3 pα 1 ` α 2 ` α 3 q (add all three equations and divide by 3; note that we don t care about c 1 or c 2 ). So with respect to the decomposition P T T 1, the projection π 1 : P Ñ T is given by π 1 : α 1 e 1 ` α 2 e 2 ` α 3 e 3 ÞÑ 1 3 pα 1 ` α 2 ` α 3 qv. Next, we must compute gπ 1 g 1 applied to u α 1 e 1 ` α 2 e 2 ` α 3 e 3 pα 1, α 2, α 3 q. To that end, note that for all g P G, since 1 3 pα 1 ` α 2 ` α 3 q is symmetric in α 1, α 2, and α 3, we have π 1 g 1 u π 1 u; and for all and x P T, we have gx x. So gπ 1 g 1 π 1. So for all g P S 3, we have So gπ 1 g 1 u 1 3 pα 1 ` α 2 ` α 3 qv. πpuq 1 6 p6 1 3 pα 1 ` α 2 ` α 3 qvq 1 3 pα 1 ` α 2 ` α 3 qv.

3 Therefore kerpπq tpα 1, α 2, α 3 q α 1 ` α 2 ` α 3 0 R. Both times, we did get the same π; but what was important was that π T id and kerpπq R. 3. Let G be a finite abelian group. Show that every simple CG-module is 1-dimensional. Deduce that if ρ : CG Ñ EndpV q is a (not necessarily simple) finite-dimensional representation of CG, then ρpcgq is simultaneously diagonalizable (there is a choice of basis for V so that the matrix corresponding to ρpaq is diagonal for every a P CG). [Hint: Let V be a simple CG-module (so that V is finite-dimensional), and let ρ be the associated representation. The trick is to show for any g P G, the Jordan canonical form of ρpgq is actually just a constant diagonal matrix (which is useful because a constant matrix is diagonal in any basis). Along the way, you ll need to show that the eigenspaces of ρpgq are CG-modules, i.e. if v is an eigenvector for ρpgq of weight λ, then so is h v ρphqv for all h P G.] Proof. Fix x P G, let v be a non-zero eigenvector with respect to the action of x, and let λ P C be the associated eigenvalue (we know every linear map on C n has at least one non-trivial eigenvector). In other words, we have Then for all g P G, x v λv. x pg vq pxgq v pgxq v g px vq g pλvq λpg vq. So g v is also an eigenvector for x of eigenvalue v. Therefore Gv Ď V pλq ; and thus F Gv Ď V pλq. But if V is simple, then F Gv V. So V pλq V. Therefore x acts by the constant λ on all of V. However, there was nothing special about x. Namely, every element g of G acts by a constant λ g. So V F Gv F tg v g P Gu F tλ g v g P Gu F v, which is a one-dimensional vector space. 4. Let R be a ring with 1. Prove that if every R-module is completely decomposable, then every R-module is injective. [Hint: Recall Zorn s lemma, which says that if A is a non-empty partially ordered set in which every chain has an upper bound, then A has a maximal element. Then review the proof of Prop 11 of 7.4, where we used Zorn s lemma to prove the existence of maximal ideals in rings with 1. Now, your goal is to show that if Q is a submodule of an R-module N, then it has a direct sum complement. Use Zorn s lemma to show that there is a submodule M maximal (with order given by containment) with respect to the condition Q X M 0. If Q ` M N in every instance, then 2 holds. Otherwise, let X be a simple submodule of N N{M that is not contained in

4 Q ` M (which exists since N is completely decomposable), and consider X π 1 pxq. You ll need the 4th isomorphism theorem.] Proof. Let N be an R-module, and fix a submodule Q of N. Let A tu submodule of N Q X U 0u, and consider the partial order on A given by U 1 ď U 2 if U 1 Ď U 2. Now, let C be a chain in A (a subset of A where every two elements are comparable), and let V ď U. UPC Claim: V is an upper bound of C. Proof: V is non-empty: 0 is in every module, so 0 P U for all U P C. So 0 P Ť UPC U V. V is closed: For all u 1, u 2 P V, there is some U 1 and U 2 in C such that u 1 P U 1 and u 2 P U 2 ; but since U 1, U 2 P C, we have U 1 Ď U 2 or U 1 Ě U 2. Without loss of generality, suppose U 1 Ď U 2, so that u 1, u 2 P U 2. Then for all r P R, u 1 ` ru 2 P U 2 Ď ď UPC U V. V is in A: We just showed that V is a submodule of N. Now, if a P V X Q, then a is in some U P C. But U X Q 0, so that a 0. Thus V X Q 0. V is an upper bound: For all U P U, we have U Ď Ť UPC U V, so that U ď V. {{ So every chain in A has an upper bound; and therefore by Zorn s lemma, A has at least one maximal element. Let M be a maximal element of A (meaning M is a submodule of N intersecting trivially with Q; and if L is a submodule of N containing M and trivially intersecting with A, then L M). If N M ` Q, then N M Q, and we re done. Otherwise, if N M ` Q, consider N N{M. Since Q ` M N, we have Q ` M N. And since every R-module is completely decomposable, N has at least one simple module X that is not a submodule Q ` M. Since X is simple, this implies that X X Q ` M 0. Abusing notation, using the fourth isomorphism theorem, let X be the unique submodule of N containing M such that πpxq X (where π : N Ñ N is the natural projection). So, again, by the fourth isomorphism theorem X X pq ` Mq π 1 px X Q ` Mq π 1 p0q M. But M X Q 0; so X X Q 0. Since M Ĺ X, this contradicts the maximality of M. So we must have been in the case where N M Q.

5. Suppose R is a ring with 1 satisfying the 4th condition of the Artin-Wedderburn theorem: The left-regular module decomposes as R À λpλ Rλ, where each R λ is a simple R-module. More specifically, R λ Re λ for some idempotent e λ P R, and the idempotents te λ λ P Λu satisfy ÿ e λ e µ δ λµ e λ and e λ 1. λpλ Let N be an R-module, and let x be a non-zero element of N. (a) Show that there is some λ P Λ for which e λ x 0. (b) Show that if e λ x 0, then Re λ x is a simple submodule of N. [Hint: We have previously verified that X Re λ x tre λ x r P Ru is a submodule of N, we just need to show that it s simple. Do so by showing that for all non-zero y P X, there is some r P R such that ry e λ x, and conclude that Ry X. You will need to use the fact that Re λ is a simple R-module.] Proof. Let λ P Λ be such that e λ x 0, and let y be any non-zero element of Re λ x. So there s some non-zero s P R such that y se λ x. But as an R-module, Re λ is simple! Namely, se λ is a non-zero element of Re λ, so that Therefore, there is some r P R satisfying Rpsr λ q Re λ Q e λ. e λ rpse λ q, so that e λ x rpse λ qx ry. Thus e λ P Ry, so that Re λ Ď Ry Ď Re λ, showing that Ry Re λ (for all non-zero y P Re λ x). Thus Re λ x is simple as desired. (c) Show that N is completely decomposable. [Hint: In the previous part, you showed that N contains simple submodules (i.e. in contrast to Z s left regular module). Now use Zorn s lemma applied to the set of submodules of N that are direct sums of simple submodules (i.e. the set of completely decomposable submodules), and let M be the maximal such submodule. If M N, then again let Y be a simple submodule of N N{M and consider Y π 1 py q. ] Proof. By the previous part, N has at least one simple submodule; and so it has at least one completely reducible submodule. Let A tu submodule of N Q X U 0u, and consider the partial order on A given by U 1 ď U 2 if U 1 Ď U 2 and there are simple submodules t µ P Ωu of U 2 à such that U 2 U 1. Namely, U 1 ď U 2 if U 2 has a decomposition that contains any decomposition of U 1. 5

6 Now, let C be a chain in A (a subset of A where every two elements are comparable), and let V ď UPC U. Claim: V is an upper bound of C. Proof: The proof that V is a submodule follows exactly as in the previous question. It remains to show (1) V P A, and (2) V ě U for all U P C. Fix U P C, and let C 1 be all the elements in C above U (i.e. C 1 tw P C W ą Uu). For each W P C 1, let Ω W be the indexing set for the simple submodules comprising the decomposition of the direct sum complement of U in W. Since W ď W 1 implies that there is a decomposition of W 1 that contains any decomposition of W ; and there is a decomposition of W that contains any decomposition of U; without loss of generality, we can assume that Ω W Ď Ω W 1; in other words, à (1) W 1 U à. W 1 W Ω W Then we claim that à V U, where Ω ď Ω W. W PC 1 First, we can see that (2) V U ` ÿ because every element v P V is in some W P W such that v P W, and W is contained in this sum. Next, we use our third favorite equivalent definition of direct sum to show that this sum is isomorphic to the direct sum, that every element of V has a unique finite expansion in elements of U, with µ ranging over Ω. To this end, let v P V. Then there is some W P C 1 such that v P W. Since à W U, W there are unique u P U and tw µ P µ P Ω W u, with finitely many being non-zero, such that v u ` ÿ w µ. W Now assume, for the sake of contradiction, that there are some other u 1 tx µ P µ P Ωu such (3) v u 1 ` ÿ x µ. P U and

By (2), the set Ω 1 tµ P Ω x µ 0u Ď Ω is finite. Moreover, this second decomposition can t possibly happen in W (because we had a direct sum), so Ω 1 Ę Ω W. But since Ω 1 is finite, there must be some W 1 ě W such that Ω 1 Ď Ω W 1. So x µ P Ď W 1 for each µ P Ω 1. But by (2), any such expansion is unique; and (3) was already one such expansion. Therefore, v has a unique expansion of the form (3) across all W 1 P C 1 above W ; and therefore has a unique expansion in V. À Therefore V U. And thus V P A and V ě U. {{ 7 So every chain has an upper bound, and thus A has at least one maximal element. Let M be one such element. Either M N, in which case N P A and we re done; or we can consider a (non-zero) element x P N M. We showed above that there s some λ P Λ for which e λ x 0; and for any such λ, that Re λ x is a simple submodule of N. Slightly more, since ÿ M S x 1 x e λ x ÿ e λ x, λpλ there must at least one λ P Λ for which e λ x R M and X Re λ x is simple. In this case, since X is simple, we have X Ă M or X X M 0. But since But then λpλ e λ x R M and e λ x P X, we must have X X M 0. X ` M X M; so that X ` M P A and X ŋ M, a contradiction. So M N as desired.