## 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).

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 $M$.

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 $A''$ as an algebra so that $M$ is 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

1. $A$ has a simple module,

2. any nonzero module contains a simple submodule,

3. a simple module over $A$ has finite dimension over $k$, and

4. 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 $A \cong \text{Mat}(n \times n, K^{op})$. $\square$

[1] This means that given $a'' \in A''$ and $m \in M$ we have a product $m a'' \in M$. In particular, the multiplication in $A''$ is the opposite of what you'd get if you wrote elements of $A''$ as endomorphisms acting on the left.

Comment #180 by Andreas on

There is a misspelling in the label of the section on Wedderburn's theorem. \label{section-weddenburg} should be \label{section-wedderburn}

Comment #3811 by Hua WANG on

There is a slight mistake in Lemma 0745---$A$ is isomorphic to the opposite of $A''$ instead of $A''$ itself, as $R(mn) = R(n)R(m)$ in the usual order of function composition. Similarly, later in the proof $M$ is viewed as a right-$(A'')^{op}$ module since it is a left-$A''$ module, and $R(M)$ is a right ideal of $(A'')^{op}$ instead of $A''$.

Rieffel's article uses left modules. Since we use the right modules here, it might be better to define the bicommutant of a right ideal $M$ of $A$ as the opposite of ${\rm End}_{A'}(M)$.

Comment #3920 by on

Good catch! I fixed the problem in a different (but essentially equivalent manner). See changes here.

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