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I'm currently taking a first introduction to abstract algebra. At the moment, we're talking about semisimple modules and rings. We have already covered the following results

Theorem (Wedderburn): $R$ is a semisimple ring if and only if there exist $r\geq 1$, divison rings $D_1, \dots, D_r$ and $n_1, \dots , n_r \in \mathbb{N}$ such that

$$ R \simeq \bigoplus_{i=1}^rM_{n_i}(D_i) $$

Proposition: if $\mathbb{k}$ is an algebraically closed field and $D$ is a division ring with a finite $\mathbb{k}$-algebra structure, $D = \mathbb{k}$.

Now, according to my lecture notes, we can easily deduce from here that if $\mathbb{k}$ is an a.c. field and $R$ a finite dimensional $\mathbb{k}$-algebra, there exist $n_1 ,\dots, n_r \in \mathbb{N}$ such that

$$ R \simeq \bigoplus_{i = 1}^rM_{n_i}(\mathbb{k}) $$

I can see that by Wedderburn's theorem, we have $R \simeq \bigoplus_{i=1}^rM_{n_i}(D_i)$, and if we see that each division ring $D_i$ is in fact a finite dimensional $\mathbb{k}$-algebra, we get the desired result. I also understand how $\bigoplus_{i=1}^rM_{n_i}(D_i)$ has a finite dimensional $\mathbb{k}$-algebra structure by composing the morphism $i: \mathbb{k} \longrightarrow Z(R)$ with the isomorphism between $R$ and the direct sum. However, how do we prove from here that each division ring $D_i$ has a $\mathbb{k}$-algebra structure?

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From the proof of Wedderburn's theorem, each $D_i$ is the endomorphism ring of a simple $R$-module, $M_i$ say. But $M_i$ is a quotient of $R$ and so is a finite-dimensional vector space over $k$. Then each $R$-endomorphism of $M_i$ is a $k$-endomorphism of $M_i$. Moreover, $k$-linear combinations of $R$-endomorphisms of $M_i$ are $R$-endomorphisms of $M_i$. Therefore $D_i=\textrm{End}_R(M_i)$ is a $k$-subalgebra of $\textrm{End}_k(M_i)$ and so is a finite-dimensional algebra over $k$. (And as $k$ is algebraically closed will be isomorphic to $k$).

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  • $\begingroup$ Got it! Thanks, that was really clear. $\endgroup$
    – qualcuno
    Jun 4, 2018 at 5:10

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