Proposition (D& F, Thm )

Transcription

1 Proposition (D& F, Thm ) Let Q be an R-module. The following are equivalent. (1) If 0 ãñ A ' Ñ B Ñ C Ñ 0 is exact, then so is 0 ãñ Hom R pc, Qq Ñ '1 1 Hom R pb,qq Ñ Hom R pa, Qq Ñ0. (2) For any injective ' : A Ñ B and f P Hom R pa, Qq, thereisa lift F of f in Hom R pb,qq: ' A B f (3) If Q is isomorphic to a submodule of B, thenitisadirect summand of B, i.e.everyses0 ãñ Q ' Ñ B Ñ C Ñ 0 splits. (Compare to [D& F, Thm ].) An R-module is called injective if it satisfies any of 1 3. Essentially, property 3 is our favorite again; it says that any module B that Q injects into has Q as a direct summand. Q F 2. For any injective ' : A Ñ B and f P Hom R pa, Qq, thereisa lift F of f in Hom R pb,qq: ' A B Proposition (Baer s Criterion) f An R-module Q is projective if and only if for every left ideal I of R, everyr-module homomorphism I Ñ Q lifts to an R-module homomorphism R Ñ Q. (See [D&F, Prop (1)] for proof.) Q F

2 Defn. An R-module M is divisible if for all r P R, wehave rm M. Ex. Z is not divisible as a Z-module since nz Z for n 1. Ex. F 3 is divisible as a F 3 -module. Ex. Every field F is a divisible module over itself: Fˆ F 0 is a multiplicative group and therefore acts transitively on itself by left multiplication. Proposition (D&F, Prop (2)) If R is a PID, an R-module is injective if and only if it is divisible. Ex. Z is not divisible over itself, so it is not injective. (But again, 0 ãñ Z Ñ n Z Ñ Z{nZ Ñ 0 does not split.) Ex. Q is divisible over Z, soq is an injective Z-module. Theorem (D&F, Thm ) Every R-module M is contained in an injective R-module. Moreover, the minimal such Q is well-defined. Minimality: Fix M. ThereexistsaninjectiveQ such that for all injective Q 1 Ö M, wehaveq Q for some M Ñ Q Ñ Q 1 i.e. up to isomorphism M Ñ Q Ñ Q 1. We call Q the injective hull or injective envelope of M.

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4 Semisimple rings: decompositions galore AringR is semisimple if every R module is isomorphic to the direct sum of simple modules. Equivalently, R is semisimple if R is a direct sum of simple rings. Non-ex: R Z. The let regular module M Z is not simple (it has submodules nz for all n), but does not split. So Z is not semisimple. Example: We will show FG is semisimple for all finite groups G and fields F whose characteristic of F does not divide G. For example, CS n is semisimple because C has characteristic 0. Non-ex: F 3 has characteristic 3, whichdivides S 3 6; andf 3 S 3 is not semisimple. For example, the permutation module is not simple but does not split. Theorem (Maschke s theorem [D&F, Thm ]) Let G be a finite group and let F be a field whose characteristic does not divide G. LetV be an FG-module, and U be a submodule. Then there is also a submodule W satisfying V U W. (There s a submodule W in V satisfying U X W 0 and U ` W V.) Before we get to the proof... Big example: Let G be a group of order n, F be a field of char not dividing n, and consider the module V FG over R FG. Given the linear algebra that we know, we can decompose FG using inner products and projections as follows...

5 Big example: every submodule of FG is a direct summand Let G be a group of order n, F be a field of char not dividing n, and consider the module V FG over R FG. Define the sym. bilinear form x, y on FG by and extend linearly. xg, hy g,h for all g, h P G, Claim 1: x, y is non-degenerate. If a gpg gg satisfies So xa, F Gy 0, then xa, hy 0 for all h P G. C G ÿ 0 g g, h ÿ g xg, hy h gpg gpg for all h P G. Soa 0. X Big example: every submodule of FG is a direct summand Let G be a group of order n, F be a field of char not dividing n, and consider the module V FG over R FG. Define the sym. bilinear form x, y on FG by and extend linearly. xg, hy g,h for all g, h P G, Claim 2: x, y is invariant. (meaning xgu, vy xu, g 1 vy@g, u, v) Fix an order on G tg 1,...,g n u,yieldinganorderedbasisforfg as a vector space. Since every g P G Ñ R acts on G Ñ V by a permutation, the associated matrix M MG G p pgqq for the action of g on FG is a permutation matrix. (Ch4: G acts on itself by left mult., which induces a hom into S G.) In particular, M 1 M t. So, identifying a P FG with MB 1 paq, we have xga, by pmaq t b a t M t b a t M 1 b xa, g 1 by for all a, b P FG. X

6 Big example: every submodule of FG is a direct summand Let G grp of order n, F field of char - n, and consider V FG over R FG.FixorderG tg 1,...,g n u,anddefinex, y on FG by xg, hy g,h and extending linearly. Then x, y is symmetric, bilinear, invariant, and nondegenerate. Let U be a submodule of FG,anddefine U K tv P FG xu, vy 0u. Nondegenerate ñ dimpu K q dimpfgq dimpuq. Bilinear ñ U K is a subspace. Invariant ñ U K is closed under the FG-action: If xu, vy 0 for all u P U, thenforallg P G, xu, gvy xg 1 u, vy 0, since U is closed under the G-action. So gv P U K for all g P G, andthus gpg g v P U K. X Remains to show: U ` U K FG and U X U K 0. Note: The first implies second by a dimension count. Let G grp of order n, F field of char - n, and consider V FG over R FG.Definex, y on FG by xg, hy g,h and extending linearly. Showing U ` U K FG: Recall, for any a, b P V,wecanwritea b {{ ` b K where since b {{ xa, by xb, by b P Fb and b K a b {{ PpFbq K, xb K by,by xa, by xa, xb, by 0. xb, by Similarly, in any vector space V F n with subspace U, anyvector v can be decomposed as v u ` u K where u P U and u K P U K as follows...

7 Let G grp of order n, F field of char - n, and consider V FG over R FG.Definex, y on FG by xg, hy g,h and extending linearly. Showing U ` U K V : Goal: for each v P V,findu P U and u K P U K such that v u ` u K. Let B tu 1,...,u m u be a basis of U, andwrite u 1 u 1 ` ` m u m for some i P F. Solving v u P U K is the same as solving 0 xu i,v uy xu i,vy` 1 xu i,u 1 y` ` m xu i,u m y for all i 1,...,m. But since x, y is the dot product on V,thisis the same as finding p 1,..., m q t such that 0 M t pv M q, where M u 1 u m P M n,m pf q. Let G grp of order n, F field of char - n, and consider V FG over R FG.Definex, y on FG by xg, hy g,h and extending linearly. Showing U ` U K V : Goal: for each v P V,finduPU and u K P U K such that v u ` u K ;whichisequivalenttofinding p 1,..., m q t such that 0 M t pv M q, where M u 1 u m P M n,m pf q. Note: M is not square, but M t M P M m pf q is. And since the columns of M are linearly independent, M t M is invertible. So 0 M t pv M q ô M t M M t v ô pm t Mq 1 M t v. Then u 1 u 1 ` ` m u m M MpM t Mq 1 M t v P U, and u K v u P U K. X

8 Result: Every v P V can be written as v u ` u K,where u MpM t Mq 1 M t v P U and u K v u P U K (where M has column vectors corresponding to a basis of U). Sanity check 1: When m 1, we rebacktom u and MpM t Mq 1 M t v upu t uq 1 u t v upxu, uyq 1 xv, uy xu, vy xu, uy u as desired. Sanity check 2: If you started with an orthonormal basis of V (pairwise orthogonal and xv i,v i y 1), and considered the space spanned by the first m basis vectors, then M m So M t M I m, and MpM t Mq 1 M t I m 0 n m. In summary: G is a grp of order n, F is a field of char - n, andv FG is a module over R FG.Definex, y on FG by xg, hy g,h and extending linearly. Then x, y is symmetric, bilinear, invariant, and nondegenerate; and for every submodule U Ñ V,eachv P V can be written as v u ` u K,where u MpM t Mq 1 M t v P U and u K v u P U K (where M has column vectors corresponding to a basis of U). So (1) U K is a submodule, (2) V U ` U K,and(3) U X U K 0. Therefore V U U K. Finally, note that MpM t Mq 1 M t : V Ñ U is a FG homomorphism: We have U id u, which is a homomorphism on U. And for v P V,writingv u ` u K,wehave so that gu K P U K (since x, y is G-invariant), pgvq pgu ` gu K q gu g puq.

9 What made this proof work? From the perspective of the ring: FG is an algebra over a field, so fin-dim l modules are all vector spaces (isomorphic to F d for d dimpv q). Using the standard dot product on F d, the projection produces an orthogonal complement U K to any subspace U Ñ V. So V U U K as vector spaces. What s missing: under what circumstances is U K asubmodule? In our example, invariance was a result of Mp pgqq 1 Mp pgqq t for all g P G. Not always true! In particular, it requires tools specific to the module, not the ring. And, in fact, the complementary direct summand is often not the orthogonal complement. Further, if you use an inner product other than dot product, then you might get U ` U K à V.

10 Example Let R V CS 2.Fixbasisv 1 1 and v 2 p12q, andidentifyv with C 2 and EndpV q with M 2 pcq. Consider the bilinear form x, y x, y J where J Nondegenerate X Symmetric X. 0 1 Not invariant: ˆ1 ˆ1 xv 1,v 1 y p1, 0q p1, 0q ˆ ˆ0 0 xv 2,v 2 y p0, 1q p0, 1q 1, So xp12qv 1, p12qv 1 y xv 2,v 2 y xv 1,v 1 y. Example Let R V CS 2.Fixbasisv 1 1 and v 2 p12q, andidentifyv with C 2 and EndpV q with M 2 pcq. Consider the bilinear form x, y x, y J where J. 0 1 Nondegenerate X Symmetric X Not invariant ˆ Now try to decompose CS 2... We have the submodule T Ce`, wheree` v 1 ` v 2. Further, T K tv P CS 2 xv 1 ` v 2,vy 0u So v v 1 ` v 2 P T K whenever ˆ1 0 xv 1 ` v 2,vy p 1, 2 q But that implies that T K T,sothatT ` T K T V. (Compare to the known decomposition CS 2 T S, wheres Ce.)

11 Example Again, let R V CS 2.Fixbasisv 1 1 and v 2 p12q, and identify V with C 2 and EndpV q with M 2 pcq. Now consider x, y x, y J where J 0 2 Nondegenerate X Symmetric X Not invariant: ˆ1 xv 1,v 1 y p1, 0q ˆ0 xv 2,v 2 y p0, 1q 2 1, so that x, y is not S 2 -invariant. Example: Again, let R V CS 2.Fixbasisv 1 1 and v 2 p12q, andidentifyv with C 2 and EndpV q with M 2 pcq. Now consider x, y x, y J where J 0 2 Nondegenerate X Symmetric X Invariant ˆ Again, let T Ce` and consider T K : v v 1 ` v 2 P T K whenever ˆ1 0 xv 1 ` v 2,vy p 1, 2 q ` 2 2. So T K Cp2v 1 v 2 q. This time, we do have T X T K 0 and T ` T K V,sothat V T T K as vector spaces. However, this is still the wrong decomposition of V as modules, since T K is not closed under the S 2 action: p12qp2v 1 v 2 q v 1 ` 2v 2 R T K. So V fl T T K.