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.ÿþpage 1 of Chapter 9CHAPTER 9 INTRODUCING NONCOMMUTATIVE ALGEBRAWe will discuss noncommutative rings and their modules, concentrating on two funda-mental results, the Wedderburn structure theorem and Maschke s theorem.Further insightinto the structure of rings will be provided by the Jacobson radical.9.1 Semisimple ModulesAvector space is the direct sum of one-dimensional subspaces (each subspace consistsof scalar multiples of a basis vector).Aone-dimensional space is simple in the sense that itdoes not have a nontrivial proper subspace.Thus any vector space is a direct sum of simplesubspaces.We examine those modules which behave in a similar manner.9.1.1 Definition An R-module M is simple if M = 0 and the only submodules of M are0 and M.9.1.2 Theorem Let M be a nonzero R-module.The following conditions are equivalent,and a module satisfying them is said to be semisimple or completely reducible.(a) M is a sum of simple modules;(b) M is a direct sum of simple modules;(c) If N is a submodule of M, then N is a direct summand of M, that is, there is a submoduleN of M such that M = N •" N.Proof.(a) implies (b).Let M be the sum of simple modules Mi, i " I.If J †" I, denote Mjj"Jby M(J).By Zorn s lemma, there is a maximal subset J of I such that the sum definingN = M(J) is direct.We will show that M = N.First assume that i " J.Then N )" Mi is/a submodule of the simple module Mi, so it must be either 0 or Mi.If N )" Mi = 0, thenM(J *"{i}) is direct, contradicting maximality of J.Thus N )" Mi = Mi, so Mi †" N.Butif i " J, then Mi †" N by definition of N.Therefore Mi †" N for all i, and since M is thesum of all the Mi, we have M = N.(b) implies (c).This is essentially the same as (a) implies (b).Let N be a submodule of M,where M is the direct sum of simple modules Mi, i " I.Let J be a maximal subset of I suchthat the sum N + M(J) is direct.If i " J then exactly as before, Mi )" (N •" M(J)) = Mi,/so Mi †" N •" M(J).This holds for i " J as well, by definition of M(J).It follows thatM = N •" M(J).[Notice that the complementary submodule N can be taken as a directsum of some of the original Mi.](c) implies (a).First we make several observations.(1) If M satisfies (c), so does every submodule N.[Let N d" M, so that M = N •" N.If V is a submodule of N, hence of M, we haveM = V •" W.If x " N, then x = v + w, v " V, w " W, so w = x - v " N (using V d" N).But v also belongs to N, and consequently N =(N )" V ) •" (N )" W) =V •" (N )" W).](2) If D = A •" B •" C, then A =(A + B) )" (A + C).[If a + b = a + c, where a, a " A, b " B, c " C, then a - a = b - c, and since D is a directsum, we have b = c = 0 and a = a.Thus a + b " A.](3) If N is a nonzero submodule of M, then N contains a simple submodule.[Choose a nonzero x " N.By Zorn s lemma, there is a maximal submodule V of N such thatx " V.By (1) we can write N = V •" V , and V = 0 by choice of x and V.If V is simple,/we are finished, so assume the contrary.Then V contains a nontrivial proper submoduleV1, so by (1) we have V = V1 •" V2 with the Vj nonzero.By (2), V =(V + V1) )" (V + V2).Since x " V , either x " V + V1 or x " V + V2, which contradicts the maximality of V.]/ / /To prove that (c) implies (a), let N be the sum of all simple submodules of M.By (c)we can write M = N •" N.If N = 0, then by (3), N contains a simple submodule V.But page 2 of Chapter 9then V d" N by definition of N.Thus V d" N )" N = 0, a contradiction.Therefore N =0and M = N.c&9.1.3 Proposition Nonzero submodules and quotient modules of a semisimple module aresemisimple.Proof.submodule case follows from (1) of the proof of (9.1.2).Let N d" M, whereTheM = Mi with the Mi simple.Applying the canonical map from M to M/N, we haveiM/N = (Mi + N)/N.iThis key idea has come up before; see the proofs of (1.4.4) and (4.2.3).By the secondisomorphism theorem, (Mi + N)/N is isomorphic to a quotient of the simple module Mi.But a quotient of Mi is isomorphic to Mi or to zero, and it follows that M/N is a sum ofsimple modules.By (a) of (9.1.2), M/N is semisimple.c&Problems For Section 9.11.Regard a ring R as an R-module.Show that R is simple if and only if R is a divisionring.2.Let M be an R-module, with x a nonzero element of M.Define the R-module homo-morphism f : R ’! Rx by f(r) =rx.Show that the kernel I of f is a proper ideal of R,and R/I is isomorphic to Rx
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