Chapter 2. The Structure of Semisimple Algebras In this chapter we study an important class of rings: the class of rings R which are such that any R- module can be expressed as a direct sum of simple R-submodules. We study the structure of such rings through a series of results essentially due to Artin and Wedderburn. At the end of the chapter we will assume that the ring is a finite dimension algebra over a field and start the study of its representation theory. Notation: throughout this chapter, unless otherwise specified, we let R denote a unital and associative ring. References: [CR90] C. W. C I. R, Methods of representation theory. Vol. I, Wiley Classics Library, John Wiley & Sons, Inc., New York, 1990. [Dor72] L. D, Group representation theory. Part B: Modular representation theory, Marcel Dekker, Inc., New York, 1972. [NT89] H. N Y. T, Representations of finite groups, Translated from the Japenese. Academic Press, Inc., Boston, MA, 1989. [Rot10] J. J. R, Advanced modern algebra. 2nd ed., Providence, RI: American Mathematical Soci- ety (AMS), 2010. 11 Semisimplicity of rings and modules Proposition-Definition 11.1 (Completely reducible module / semisimple module) An R-module M satisfying the following equivalent conditions is called completely reducible or semisimple“‘: P t u P I I (a) M “ P S for some family tS u P of simple submodules of M; ∞ I (b) M I S for some familyÑ S of simple submodules of M;D Ñ “ ‘ 1 2 (c) every R-submodule M M admits a complement in M, i.e. an R-submodule M M 1 2 such that M M M . 28 Skript zur Vorlesung: Modulare Darstellungstheorie WS 2019/20 29 ñ Proof : (a)ñ(b): is trivial. “ P P Ñ ∞ Ñ (b) (c): Write M I S , where S is a simple R-submodule of M for each I. Let N M be an R-submodule of M. Then consider the family, partially ordered by inclusion, of all subsets J I such thatP ∞XJ P “ (1) S is a direct sum; and ∞ (2) N J S 0. Clearly this family is non-empty since it contains the empty set. Thus Zorn’s Lemma yields the 0 1 existence of a maximal element J . Now,“ set` “ ‘ P P ÿ ÿ M : N 0 S N 0 S J J 1 “ 1 1 Ñ P P Ü where the1 second equality holds by (1) and (2). Therefore, it suffice to prove that M M , i.e. X “ that S M for every I. But if I is such that S M , the simplicity of S implies that S M 0 and it follows that 1 ` “ ‘ ˜ ¸ ‘ P ÿ 0 M S N J S S 0 ñ in contradiction with the maximality of J “. The claim follows. (b)ñ(a): follows from the argument above with N 0. Ñ 1 “ ‘ “ ‰ 2 (c) (b): Let M be the sum of all simple submodules in M. By (c) there exists a complement M M to 1 1 2 2 2 Ñ 2 M , i.e. such that M M M . If M 0, we are done. If M 0, then M must contain a “ “ 1 simple R-submodule (Exercise: prove this fact), say N. But then N M , a contradiction. Thus 2 1 M 0 and so M M . Example 6 (a) The zero module is completely reducible, but neither reducible‘ nor‘ irreducible! 1 1 (b) If S S are simple R-modules, then their direct sum S S is completely reducible by definition. (c) The following exercise shows that there exists modules which| areP not completely reducible. 1 “ ` 01˘ ( Exercise2: Let K be a field and let A be the K -algebra K . Consider the A-module V : tpK ,q| whereP Au acts by left matrix multiplication. Prove that: (1) 0 K is a simple A-submodule of V ; but (2) V is not semisimple. (d) Exercise: Prove that any submodule and any quotient of a completely reducible module is again completely reducible. Theorem-Definition 11.2 (Semisimple ring) A ring R satisfying the following equivalent conditions is called semisimple: (a) All short exact sequences of R-modules split. Skript zur Vorlesung: Modulare Darstellungstheorie WS 2019/20 30 (b) All R-modules are semisimple. (c) All finitely generated R-modules˝ are semisimple. (d) The regular left module R is semisimple, and is a direct sum of a finite number of minimal left ideals. ñ Proof : First, (a)˝ and (b) are equivalent as a consequence of Lemma 4.4. The implication (b) (c) is trivial, “ P t u P and it is also trivial that (c) implies the first claim of (d), which in turn implies the second claim of (d). À PI “ ` I ` P § § Indeed, if R L for some family L1 of minimal left ideals. Then there exists a finite number P 1 R of indices I such that 1 with L for each 1 . Therefore each R may be expressed in the form “ ¨ “ ` ` 1 ˝ R “ ` ` 1 ñ 1 “ P ¨ ˝ and hence R ¨ L L . Therefore, it remains to prove that (d) ∞(b). So, assume that R satisfies¨ (d) and let M be an arbitrary non-zero R-module. Then write M M R . Now, each cyclic submodule R of M is isomorphic to a submodule of R , which is semisimple by (d). Thus R is semisimple as well by Example 6(d). Finally, it follows from Proposition-Definition 11.1 that M is semisimple. Example 7 t ¨¨¨ u “ ‘ ‘ p q“ Fields are semisimple. Indeed, if V is a finite-dimensional vector space over a field K of dimension , 1 § § 1 K then choosing a basis of V yields V K K , where dim K 1, hence K is a simple K -module for each 1 . Hence, the claim follows from Theorem- Definition 11.2(c). Corollary 11.3 Let R be˝ a semisimple ring. Then: (a) R has a composition series; (b) R is both left Artinian and left Noetherian. ˝ Proof : (a) By Theorem-Definition˝ 11.2(d) the regular module R admits a direct sum decomposition into a finite number˝ of minimal left ideals. Removing one ideal at a time, we obtain a composition series for R . (b) Since R has a composition series, it satisfies both D.C.C. and A.C.C. on submodules by Corol- lary 8.4. In other words, R is both left Artinian and left Noetherian. Next, we show that semisimplicity is detected by the Jacobson radical. Definition 11.4 p q“ A ring R is said to be J-semisimple if J R 0. Proposition 11.5 Any left Artinian ring R is J-semisimple if and only if it is semisimple. Skript zur Vorlesung: Modulare Darstellungstheorie WS 2019/20 31 ñ ‰ ú ‰ ‰0 Proof : “ ”: Assume R 0 and R is not semisimple. Pick a minimal left ideal I R (e.g. a minimal 0 0 element of the family of non-zero principal˝ left ideals of R). Then 0 I R since I seen as an R-module is simple. 0 Claim: I is a direct summand of R . ‰ “ p q“ Indeed: since Ÿ 0 £ I 0 J R I R I m I maximal m Ÿ left ideal X “t u ˝ m m “ ‘ { 0 0 0 0 there exists a maximalm0 left ideal R which does not containm0 I . Thus I m 0 and so we 0 ‰0 0 ‰ ‰ must have R I , as R is simple. Hence the Claim. ˝ m 1 1 0 Notice that then 0, and pick a minimalm left ideal I mof . Thenm 0 I , else R would 1 1 “ ‘ 0 be semisimple. The Claim applied to I yields that I is a direct summand of R , hence also in . 1 0 1 1 Therefore, there exists a non-zero left ideal such that I . Iterating this process, we m m m obtain an infinite descending chain of idealsâ â ⨨¨ 0 1 2 ˝ contradicting D.C.C. – { p q‘ p q “ ”: Conversely, if R is semisimple,p q“ thenp Rq¨p {RpJ q‘R p J qqR “byp Theorem-Definitionq¨ p q 11.2 and so as R-modules, J R p Jq“R R J R J R J R J R so that by Nakayama’s Lemma J R 0. Z Z Z Z Exercise 11.6“ p q“ Let R . Prove that J 0, but not all -modules are semisimple. In other words, is J-semisimple but not semisimple. Proposition 11.7 { p q The quotient ring R J R is J-semisimple.“ { p q Proof : Since by Exercise 9.2 the rings R and R : R J R have the same simple modules (seen as abelian groups), Proposition-Definitionp q“ 9.1(a) yields:p q“ p q` p q“ p q{ p q“ £ R £ R ´ ´ J R V simple ann V V simple ann V J R J R J R 0 R module R module 12 The Artin-Wedderburn structure theorem The next step in analysing semisimple rings and modules is to sort simple modules into isomorphism classes. We aim at proving that each isomorphism type of simple modules actually occur as direct summand of the regular module. The first key result in this direction is the following proposition: Skript zur Vorlesung: Modulare Darstellungstheorie WS 2019/20 32 Proposition 12.1 t u P IP Let M be a semisimple R-module. Let M be a set of representatives of the isomorphism classes of simple R-submodules of M and for each “I set Ñ ÿ H : – V V M V M Then the– followingP statements hold: À (i) M I H ; (ii) everyp simple 1Rq“t-submoduleu ‰ of1 H is isomorphic to M ; R (iii) Hom “H HP 0 if ; and À J (iv) if M V is an arbitrary decomposition“ “ of M into“ a direct sum of simple submodules, then ÿP àP r – – H : V J V H J V M V M Proof : We shall prove“ severalP statements which, taken together, will establish the theorem.