# 12. Projective modules The blanket assumptions about the base ring k, the k-algebra A, and A-modules enumerated at the start of 11 continue to hold.

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1 12. Projective modules The blanket assumptions about the base ring k, the k-algebra A, and A-modules enumerated at the start of 11 continue to hold Indecomposability of M and the localness of End A M. In analogy with the terminology local in the commutative case, the algebra A is called local if A/ Rad A k. Lemma A is local if and only if every element of A is either invertible or nilpotent. [s:projective] [ss:localendam] [l:local] Proof. The latter condition goes down to quotients. In particular it is true for A/ Rad A which by Wedderburn is a product of matrix algebras over k. If there is more than one factor, then the element that corresponds to the identity in one factor and zero elsewhere is neither invertible not nilpotent. So there is only one factor. Moreover, if this factor is an n n matrix algebra where n 2, then it too has a non-nilpotent non-invertible element, e.g., a diaganal matrix where one of the diagonal entries is 1 and all others zero. We claim that the latter condition if it holds for a quotient by a nilpotent two sided ideal holds also for A. It suffices to prove the claim for the condition holds for A/ Rad A if A is local, and Rad A is nilpotent. To prove the claim, suppose that n is nilpotent and that every element of A/n is either nilpotent or invertible. Let a be an non-nilpotent element of A. Then ā is non-nilpotent in A/n. Thus we can find b in A such that ba ab 1 (mod n). Writing ab = 1 + x with x n, we see that ab and therefore a itself has a right inverse (because x is nilpotent). Similary ba and so a itself has a left inverse. Since a has both a left inverse and a right inverse, it is invertible. [t:local] Theorem A module M is indecomposable if and only if End A M is local. Proof. If M = N P, then the projection to N followed by the inclusion of N in M is an element of End A M which is neither invertible nor nilpotent. Conversely, suppose that M is indecomposable and let ϕ be an element of End A M. For large enough n, by the Fitting Lemma (see 6.2), ker ϕ n Im ϕ n = M. If Im ϕ n = M, then ϕ n and therefore ϕ is surjective, and M being Noetherian, this means ϕ is invertible (see (6.1)). If Im ϕ n = 0, then ϕ n = 0, which means that ϕ is nilpotent. Proposition If M, U, V are A-modules such that M U M V, then U V. Proof. This follows from the Krull-Remak-Schmidt theorem (Theorem 6.2) Uniserial modules. Proposition For a module U, the following are equivalent (U is called uniserial if these hold): (1) U has a unique composition series (2) the successive quotients in the radical series of U are simple. (3) the successive quotients in the socle series of U are simple. Proof. We make a few observations in aid of the proof. Let 0 U 1... U n 1 U be a composition series. Then U 1 soc U and Rad U U n 1. Moreover, soc U 41 [p:cancel] [ss:uniserial] [p:uniserial]

2 42 [ss:fp] [p:free] [p:projective] [ss:pims] [t:pimsimple] is the sum of the U 1 and Rad U the intersection of the U n 1, as we vary over all composition series of U. Let us now prove that (2) implies (1) by induction on the radical length. Since U/ Rad U is simple, Rad U = U n 1. By induction, (1) holds for Rad U and so also for U. The argument that (3) implies (1) is similar. Since soc U is simple, U 1 = soc U. By induction on the socle length, (1) holds for U/U 1. So (1) holds also for U. Suppose (1) holds. Then, from the observations at the start of this proof, Rad U = U n 1, so U/ Rad U is simple. By induction on the length of U, we see that the quotients in the radical series for U n 1 = Rad U are all simple. Therefore the same holds for U, and (2) is proved. The proof that (1) implies (3) is similar to that of (1) implies (2) Free modules and projective modules. Proposition An A-module F is free if and only if it has a k-subspace S such that any linear map from S to an A-module M can be extended to an A-module morphism from F to M. Proof. Suppose that F is free with basis {x i }. Take S to be the k-subspace generated by the x i. Any (set) map from {x i } to M extends uniquely to a k-linear map of S to M, and to an A-module map from F to M. Conversely, if there exists such a subspace, then choosing {x i } to be k-basis of S, it is verified readily that F is freely generated by {x i }. Proposition The following conditions are equivalent for an A-module P. 12 (We call P projective if these hold.) (1) Any short exact sequence of A-modules 0 L M P 0 splits. (2) The functor M Hom A (P, M) is exact (on finite type A-modules). (3) If ϕ : M N is a surjection of A-modules and ϑ : P N any A-module map, then there exists a lift ϑ : P M of ϑ, i,e., ϕ ϑ = ϑ. (4) P is a direct summand of a free module. Proof. (1) (4) Any module is a quotient of a free module. Write F P where F is free. Now by (1) this splits. Therefore P is a direct summand of F. (4) (3) Let P Q be free. Extend ϑ to ϑ : P Q N by defining it to be 0 on Q. Since (3) holds for a free module, we get ϑ : P Q M such that ϕ ϑ = ϑ. Set ϑ := ϑ P. (3) (2) Let 0 L M N 0 be exact. Then 0 Hom A (P, L) Hom A (P, M) Hom A (P, N) is exact in general. Since (3) holds the last map is also onto. (2) (1) 0 Hom A (P, L) Hom A (P, M) Hom A (P, P ) 0 is exact. But this means that Hom A (P, M) Hom A (P, P ) is onto. If ϕ : P M is a preimage of the identity morphism on P, then ϕ is a splitting Projective indecomposable modules. We will write PIM for a projective indecomposable A-module. Theorem Let P be a PIM. 12 It is to be understood that all modules appearing are finite dimensional k-vector spaces (equivalently finite type A-modules), for it is in that context that we will need the proposition.

5 45 (6) The two bijections P P/ Rad P and P soc P from PIMs to simples (of Theorem 12.7 (2) and item (5) above) are the same, for as we will show P/ Rad P soc P for a PIM P (Theorem 12.15). [i:multsoc] (7) The number m of times a PIM P appears in the decomposition into indecomposables of kg equals dim soc P. This follows from the bijection P/ Rad P soc P of Theorem below, but it can be deduced more easily as follows. Since kg kg by Lemma below, m equals the corresponding the number for the PIM P. Now, by Theorem 12.7 (4), m = dim(p / Rad P ). But P / Rad P (soc P ). Lemma kg kg as a kg-module. [l:dualkg] Proof. Let δ g be the element of kg defined by δ g (h) := δ g,h for h G kg, where δ g,h is the Kronecker-delta function. Extend the association g δ g to a linear isomorphism kg kg. We claim that this is a kg-module isomorphism. Indeed, δ xg (h) = δ xg,h = δ g,x 1 h = δ g (x 1 h) = (xδ g )(h). [c:dualkg] Corollary Let M be a kg-module. (1) It is free if and only if its dual is free. (2) It can be imbedded in a free module. (3) It is projective if and only if its dual is projective. Proof. (1): Since any module is naturally identified with its double dual, M (kg) n M (kg n ). Since dualization commutes with taking (finite) direct sums, M (kg n ) M (kg ) n. By the lemma, M (kg ) n M kg n. (2): Choose a surjection F M with F free. Dualizing we get M F. But F is free by (1). (3): If M N is free then, by (1), so is its dual M N. This proves the only if part, applying which we get: M is projective if M is so. But M M naturally, which proves the if part too. [c:injective] Corollary The following conditions are equivalent for a kg-module I. A projective module satisfies these conditions and any module satisfying these conditions is projective. (1) Any short exact sequence of A-modules 0 I M N 0 splits. (2) The functor M Hom A (M, I) is exact. (3) If ϕ : L M is a injection of A-modules and ϑ : L I any A-module map, then there exists an extension ϑ : M I of ϑ, i,e., ϑϕ = ϑ. Proof. (3) (2): Let 0 L M N 0 be exact. Then 0 Hom A (N, I) Hom A (M, I) Hom A (L, I) is exact in general. Since (3) holds the last map is also onto. (2) (1): 0 Hom A (N, I) Hom A (M, I) Hom A (I, I) 0 is exact. But this means that Hom A (M, I) Hom A (I, I) is onto. If ϕ : M I is a preimage of the identity morphism on I, then ϕ is a splitting. (1) I is projective: By item (2) of the previous Corollary 12.12, we can choose I F with F free. Since this splits, I is a direct summand of F, and so projective. I is projective (3): By (3) of the previous Corollary 12.12, I is projective. Dualizing ϕ and ϑ, we get ϕ : L M a surjection and ϑ : L I. Since I is projective, there exist ϑ : M I such that ϕ ϑ = ϑ. The dual ϑ of ϑ has the desired property.

6 46 [sss:form] [p:form] [t:pimradsoc] [ss:tensor] The bilinear form. On kg we have a bilinear form: for g, h in G we define (g, h) to be 1 if gh = 1 and 0 otherwise. This form is symmetric, i.e., (a, b) = (b, a) (for, clearly, (g, h) = (h, g)), and associative, i.e., (a, bc) = (ab, c) (for, clearly (g, hk) = (gh, k)). Another description of the form is as follows: (a, b) is the coefficient of the identity element of G when the product ab is expressed as a linear combination of elements of G with coefficients in k. The symmetry of the form now says that this coefficient is the same for ab and ba. As a corollary of the existence of this form, we have: Proposition Let r be a right ideal and a an element of the group ring kg. If ra = 0, then ar = 0. Proof. Suppose that ar 0 for some r r. Then ar = α g g with α g 0 for some g in G. Then (a, rg 1 ) 0, and by symmetry (rg 1, a) 0. But, on the other hand, rg 1 a belongs to ra and is therefore 0. Theorem For P a PIM for kg, we have P/ Rad P soc P. Proof. Set S := P/ Rad P, T := soc P, and assume, by way of contradiction, that S T. We know that S is simple (Theorem 12.7 (1)), and so is T (see the itemized list at the beginning of this subsection). Let s be the S-isotypic component of soc kg. Decomposing kg into indecomposables, write kg = Q R, where Q is the sum of all PIMs isomorphic to P, and R the sum of the rest. We make some observations about s and its relation to Q and R. Given the observations, it will be easy to derive a contradiction using Proposition s 0: Indeed, S occurs as the socle of some PIM: see (5) of the itemized list at the beginning of this subsection. 13 s is a two sided ideal: Indeed, it is a characteristic left ideal of kg. s R: Indeed, soc kg = soc Q soc R, and soc Q is isotypic of type T. sq = 0: Indeed, on the one hand, by the previous two observations, sq s R; on the other, sq Q since Q is a left ideal. But Q R = 0. Rs = 0: Indeed, kg-endomporphisms of kg with images in s are precisely right multiplications by elements of s. Any such endomorphism restricted to R factors through R/ Rad R (since s is semisimple), but the S-length of R/ Rad R is 0 (for P is the only PIM such that P/ Rad P S). Now, Proposition together with the fact sq = 0 yields Qs = 0. In turn, this together with the fact Rs = 0 yields s = (kg)s = (Q+R)s = 0, a contradiction Tensor Products. As in 12.6 we take A = kg throughout this subsection. The tensor product (over k) of two kg-modules is defined: g(v w) := gv gw. The standard natural isomorphisms as k-vector spaces hold also as kg-modules: e.g., U (V W ) = (U V ) W. Some of these, e.g., Hom(V, W ) V W, require the assumption that V is finite dimensional over k, which anyway we are imposing blanketly. 13 In fact, although we don t need this stronger statement for the proof, the length of s equals dim S: see (7) of the itemized list at the beginning of this subsection.

7 47 A kg-module V is G-faithful if the defining group homomorphism G GL k V is an injection. (Caution: The algebra map kg End k (V ) could have non-trivial kernel for a G-faithful V.) Theorem Let V be a G-faithful kg-module. Then given a PIM P there exists an n such that P occurs as an indecomposable component of V n. Proof. Let k[v ] denote the symmetric algebra of V. Its graded components k[v ] j are kg-modules. We claim that there is a copy of the left regular representation kg in 0 j d k[v ] j for d large enough. Let us first finish the proof of the theorem assuming the claim. Since P kg, we have P 0 j d k[v ] j. Since P is injective (see Corollary 12.13), it follows that P is a direct summand of 0 j d k[v ] j. By Krull- Remak-Schmidt, P occurs an indecomposable summand of k[v ] j for some j, 0 j d. Since k[v ] j is a quotient of V n, it follows that V n surjects onto P. Since P is projective, it follows that P is a direct summand of V n, proving the theorem. It remains therefore only to prove the claim. Consider the quotient field k(v ) of k[v ]. The group G acts on k(v ) by field automorphisms. Let F denote the fixed field: F := {f k(v ) g f = f g G}. The extension F k(v ) is Galois with Galois group G. By the normal basis theorem, there exists an element f in k(v ) such that its orbit under G forms a basis of k(v ) over F. Writing f = p/q with p, q in k[v ] and taking f = f g G g q, we see that the orbit under G of f too forms a basis for k(v ) over F Exercises. Our blanket assumptions about k, A, and A-modules are in force (unless explicitly relaxed) Prove or disprove: any quotient of an indecomposable module is indecomposable Let S be a simple A-module and P the corresponding PIM. For any A- module M, the multiplicity of S in a composition series of M equals the dimension (as a k-vector space) of Hom A (P, M) Show by means of an example that Proposition does not hold in general for a finite dimensional algebra over k Let P be a projective A-module. Set M := P/ Rad P. Then the following are equivalent: (1) M is simple. (2) M is indecomposable. (3) P is indecomposable. [t:galois] [ex:lrtwo] [sss:simproj] [sss:pformcounter]

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