Lemma 11.3.1. Let A be a possibly noncommutative ring with 1 which contains no nontrivial two-sided ideal. Let M be a nonzero right ideal in A, and view M as a right A-module. Then A coincides with the bicommutant of M.
11.3 Wedderburn's theorem
The following cute argument can be found in a paper of Rieffel, see [Rieffel]. The proof could not be simpler (quote from Carl Faith's review).
Proof. Let A' = \text{End}_ A(M), so M is a left A'-module. Set A'' = \text{End}_{A'}(M) (the bicommutant of M). We view M as a right A''-module1. Let R : A \to A'' be the natural homomorphism such that mR(a) = ma. Then R is injective, since R(1) = \text{id}_ M and A contains no nontrivial two-sided ideal. We claim that R(M) is a right ideal in A''. Namely, R(m)a'' = R(ma'') for a'' \in A'' and m in M, because left multiplication of M by any element n of M represents an element of A', and so (nm)a'' = n(ma'') for all n in M. Finally, the product ideal AM is a two-sided ideal, and so A = AM. Thus R(A) = R(A)R(M), so that R(A) is a right ideal in A''. But R(A) contains the identity element of A'', and so R(A) = A''. \square
Lemma 11.3.2. Let A be a k-algebra. If A is finite, then
A has a simple module,
any nonzero module contains a simple submodule,
a simple module over A has finite dimension over k, and
if M is a simple A-module, then \text{End}_ A(M) is a skew field.
Proof. Of course (1) follows from (2) since A is a nonzero A-module. For (2), any submodule of minimal (finite) dimension as a k-vector space will be simple. There exists a finite dimensional one because a cyclic submodule is one. If M is simple, then mA \subset M is a sub-module, hence we see (3). Any nonzero element of \text{End}_ A(M) is an isomorphism, hence (4) holds. \square
Theorem 11.3.3.slogan Let A be a simple finite k-algebra. Then A is a matrix algebra over a finite k-algebra K which is a skew field.
Proof. We may choose a simple submodule M \subset A and then the k-algebra K = \text{End}_ A(M) is a skew field, see Lemma 11.3.2. By Lemma 11.3.1 we see that A = \text{End}_ K(M). Since K is a skew field and M is finitely generated (since \dim _ k(M) < \infty ) we see that M is finite free as a left K-module. It follows immediately that A \cong \text{Mat}(n \times n, K^{op}). \square
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