We have the following theorem

Wedderburn-Artin. Let $R$ be a semi-simple ring. There are division rings $D_1,...,D_n$ and $m_1,...,m_n\in\mathbb{Z}_{>0}$ such that $$R\cong\text{Mat}_{m_1}(D_1)\times...\times \text{Mat}_{m_n}(D_n)$$ and the pairs $(D_i,m_i)$ are unique up to permutation.

Now let $K$ be a field. I want to formulate and prove a Wedderburn-Artin theorem for $K$-algebras. Here a $K$-algebra is called semi-simple if it is semi-simple as a ring.

I have already proved (for a field $K$):

  • Let $R,S$ be rings. If $R\times S$ is a $K$-algebra, so are $R$ and $S$.
  • Let $R$ be a ring. If $\text{Mat}_n(R)$ is a $K$-algebra, so is $R$.

I guess the theorem is like

Wedderburn-Artin for $K$-algebras. Let $A$ be a semi-simple $K$-algebra. There are division algebras (i.e. rings being $K$-algebras) $D_1,...,D_n$ and $m_1,...,m_n\in\mathbb{Z}_{>0}$ such that $$A\cong\text{Mat}_{m_1}(D_1)\times...\times \text{Mat}_{m_n}(D_n)$$ and the pairs $(D_i,m_i)$ are unique up to permutation.

Proof. Since A is called semi-simple if it is semi-simple as a ring, we use the Wedderburn-Artin theorem for rings to get unique pairs $(D_i,m_i)$ of division rings $D_i$ and positive integers $m_i$ such that $$A\cong\text{Mat}_{m_1}(D_1)\times...\times \text{Mat}_{m_n}(D_n)$$ as rings. If we know that $\text{Mat}_{m_1}(D_1)\times...\times \text{Mat}_{m_n}(D_n)$ is a $K$-algebra then we can conclude with the two already proved facts above that the $D_i$ are not only division rings but division algebras. The only left point is: Why is $$A\cong\text{Mat}_{m_1}(D_1)\times...\times \text{Mat}_{m_n}(D_n)$$ an isomorphism of $K$-algebras?


2 Answers 2


You can just repeat step by step the proof for rings, by observing that, if $S$ is a simple $R$-module, then $D=\operatorname{End}_R(S)$ is a $K$-algebra in a natural way and the same for the endomorphism ring of $S$ as a $D$-module.

By the way, you don't need $K$ to be a field; just a commutative ring suffices.

Anyway, you can just apply the first two parts. Since $A$ is a $K$-algebra to begin with, also each $\operatorname{Mat}_{m_i}(D_{i})$ is a $K$-algebra; then $D_i$ is a $K$-algebra. The way you build these structures of $K$-algebras ensure the initial isomorphism is an isomorphism of $K$-algebras.

Let's see why. First, assume $R\times S$ is a $K$-algebra; then also $R\times S/{0}\times S$ is a $K$-algebra. We use this structure and the obvious ring isomorphism $R\times S/{0}\times S\to R$ to endow $R$ with a structure of $K$-algebra. Similarly for $S$. Now you can consider $R\times S$ as a $K$-algebra; a priori the structure could be different from the initial one: prove it is indeed the same.

Suppose $\operatorname{Mat}_n(D)$ is a $K$-algebra, where $D$ is a division ring. Then we have a ring homomorphism $K\to\operatorname{Mat}_n(D)$, with the image contained in the center. The center is formed by the scalar matrices with entries in the center of $D$, so it is isomorphic to the center of $D$. We use this isomorphism for endowing $D$ with the structure of $K$-algebra, which induces on $\operatorname{Mat}_n(D)$ the same structure of $K$-algebra we started with.

  • $\begingroup$ The excercise I am doing has 3 sub-excercises: 1. $R\times S$ $K$-algebra $\Rightarrow$ $R$ and $S$ $K$-algebra. 2. $\text{Mat}_n(R)$ $K$-algebra $\Rightarrow$ $R$ $K$-algebra. 3. Formulate and prove W.-A. for $K$ algebras. To show 2. I have used the existing (because of being an algebra) ring homomorphism $f:K\rightarrow Z(\text{Mat}_n(R))=Z(R)I_n$ to get a ring homomorphism $K\rightarrow Z(R)$ by sending $k\in K$ to the entry $(1,1)$ of $f(k)$. Can I solve the question "Why is ... not only a ring but an algebra homomorphism?" by using this knowledge? $\endgroup$
    – user404105
    Jan 4, 2017 at 22:57
  • $\begingroup$ @mathmarseille Sorry, but I wouldn't follow this path. For instance, I agree that $R\times S$ a $K$-algebra implies $R$ and $S$ are $K$-algebras, but they can be in quite different fashions. $\endgroup$
    – egreg
    Jan 4, 2017 at 23:01
  • $\begingroup$ You seem right and your way is much better because you see what happens. But I have to do it the way I have mentioned :( $\endgroup$
    – user404105
    Jan 4, 2017 at 23:05
  • $\begingroup$ @mathmarseille I added the path you can follow. $\endgroup$
    – egreg
    Jan 4, 2017 at 23:12
  • $\begingroup$ Thank you! I will go through it tomorrow. Good night. $\endgroup$
    – user404105
    Jan 4, 2017 at 23:14

Suppose that $A$ is a $K$-algebra which is semisimple as a ring and let $\phi:A\to \prod_iM_{m_i}(D_i)$ be the isomorphism given by A-W. For each $j$ the composition $$K\to A\xrightarrow\phi \prod_iM_{m_i}(D_i)\to M_{m_j}(D_j),$$ with the last map being the projection onto the $j$th factor, has image contained in the center of $M_{m_j}(D_j)$, which is isomorphic canonically to the center of $D_j$. This turns each $D_j$ into a $K$-algebra.

Using this structure, show that $\phi$ is an isomorphism of $K$-algebras.

  • $\begingroup$ I guess this is exactly this what I have just written as a comment on egreg's answer. But I can't show the fact of being a algebra homomoprhism. $\endgroup$
    – user404105
    Jan 4, 2017 at 23:01
  • $\begingroup$ What can't you do? It is essentially tautological, in view of the form of the algebra structures I am putting on each $D_j$! $\endgroup$ Jan 4, 2017 at 23:04
  • 1
    $\begingroup$ We have to show $\phi(k a)=k\phi(a)$ for all $k\in K, a\in A$. The point is that I don't know what the products $ka$ or $k\phi(a)$ are since I have only (ring) homomorphisms given. $\endgroup$
    – user404105
    Jan 4, 2017 at 23:07
  • $\begingroup$ A map from an algebra $X$ to a direct product $Y_1\times\cdots\times Y_r$ of algebras is a map of algebras if and only if it is a morphism of rings and each composition $X\to Y_i$ with a projection is a morphism of algebras. Show that. $\endgroup$ Jan 4, 2017 at 23:09
  • $\begingroup$ Now the way I constructed the $K$-algebra structure in the ring $M_{m_j}(D_j)$ makes the composition $A\to\prod_iM_{m_i}(D_i)\to M_{m_j}(D_j)$ automatically a morphism of algebras. $\endgroup$ Jan 4, 2017 at 23:10

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