# Chapter9 - Chapter 9 Introducing Noncommutative Algebra We...

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Chapter 9 Introducing Noncommutative Algebra We will discuss noncommutative rings and their modules, concentrating on two fundamen- tal results, the Wedderburn structure theorem and Maschke’s theorem. Further insight into the structure of rings will be provided by the Jacobson radical. 9.1 Semisimple Modules A vector space is the direct sum of one-dimensional subspaces (each subspace consists of scalar multiples of a basis vector). A one-dimensional space is simple in the sense that it does not have a nontrivial proper subspace. Thus any vector space is a direct sum of simple subspaces. We examine those modules which behave in a similar manner. 9.1.1 Defnition An R -module M is simple if M 6 = 0 and the only submodules of M are 0 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 submodule N 0 of M such that M = N N 0 . Proof. (a) implies (b). Let M be the sum of simple modules M i , i I .I f J I , denote j J M j by M ( J ). By Zorn’s lemma, there is a maximal subset J of I such that the sum de±ning N = M ( J ) is direct. We will show that M = N . First assume that i/ J . 1

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2 CHAPTER 9. INTRODUCING NONCOMMUTATIVE ALGEBRA Then N M i is a submodule of the simple module M i , so it must be either 0 or M i .I f N M i = 0, then M ( J ∪{ i } ) is direct, contradicting maximality of J .Thu s N M i = M i , so M i N . But if i J , then M i N by deFnition of N . Therefore M i N for all i , and since M is the sum of all the M i ,wehave 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 M i , i I . Let J be a maximal subset of I such that the sum N + M ( J ) is direct. If i/ J then exactly as before, M i ( N M ( J )) = M i ,so M i N M ( J ). This holds for i J as well, by deFnition of M ( J ). It follows that M = N M ( J ). [Notice that the complementary submodule N 0 can be taken as a direct sum of some of the original M i .] (c) implies (a). ±irst we make several observations. (1) If M satisFes (c), so does every submodule N . [Let N M , so that M = N N 0 . If V is a submodule of N , hence of M , we have M = V W f x N , then x = v + w , v V , w W w = x v N (using V 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 0 + c , where a,a 0 A , b B , c C , then a 0 a = b c , and since D is a direct sum, we have b = c = 0 and a = a 0 s a + b A .] (3) If N is a nonzero submodule of M , then N contains a simple submodule.
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Chapter9 - Chapter 9 Introducing Noncommutative Algebra We...

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