The Stacks project

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 $M$ 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

  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.

Comments (6)

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--- is isomorphic to the opposite of instead of itself, as in the usual order of function composition. Similarly, later in the proof is viewed as a right- module since it is a left- module, and is a right ideal of instead of .

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 of as the opposite of .

Comment #3920 by on

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

Comment #7822 by Zhiyu YUAN on

What does to "view as an algebra" mean in the proof to Lemma 0745, i.e. over what field will be? Simply deleting this sentence seems to do no harm to the whole proof.

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