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 $A' = \text{End}_ A(M)$, and let $A'' = \text{End}_{A'}(M)$ (the bicommutant of $M$). Let $R : A \to A''$ be the natural homomorphism $R(a)(m) = ma$. 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 $(nm)a'' = n(ma'')$, that is, $(R(m)a'') (n) = R(ma'') (n)$ 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. 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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