# The Stacks Project

## Tag: 0747

This tag has label brauer-theorem-wedderburn and it points to

The corresponding content:

Theorem 10.3.3. 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 10.3.2. By Lemma 10.3.1 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 $A \cong \text{Mat}(n \times n, K^{op})$. $\square$

\begin{theorem}
\label{theorem-wedderburn}
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.
\end{theorem}

\begin{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 \ref{lemma-simple-module}.
By
Lemma \ref{lemma-rieffel}
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
$A \cong \text{Mat}(n \times n, K^{op})$.
\end{proof}


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