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?