Is every finite dimensional semisimple algebra over $k$ isomorphic to a direct sum of finitely many matrix algebras over $k$?

Question

Let $k$ be a field , let $R$ be a finite dimensional semisimple algebra over $k$ ; is it true that $\exists t \in \mathbb N$ and $n_1,...,n_t \in \mathbb N$ such that $R$ is isomorphic with $\oplus_{i=1}^t M(n_i,k)$ as a $k$- algebra ? If not true in general , can we characterize such semisimple algebras over a given field $k$ which can be written as a direct sum of finitely many matrix algebras over $k$ ?

Note that this requires the semisimple algebra to have an apparently more "nice" structure than is implied by the Artin-Wedderburn Theorem .

Answer 1

is it true that $\exists t \in \mathbb N$ and $n_1,...,n_t \in \mathbb N$ such that $R$ is isomorphic with $\oplus_{i=1}^t M(n_i,k)$ as a $k$- algebra ?

No, the best example being, IMO, the one given by Geoff in the comments: $\mathbb H$. It is not a matrix ring over $\mathbb R$ because a nontrivial matrix ring over a field always has nontrivial right ideals (but $\mathbb H$ does not.)

can we characterize such semisimple algebras over a given field $k$ which can be written as a direct sum of finitely many matrix algebras over $k$ ?

This may not be satisfying, but one criterion could be to check that $End(S_R)\cong k$ for every simple right module $S_R$. That's a necessary and sufficient condition that all matrix rings in the Wedderburn decomposition can be taken to be $k$.

Answer 2

No, in general $R$ is a finite direct sum of matrix algebras over division rings each of which is a finite-dimensional $k$-algebra.

Answer 3

You can reduce to the case the algebra is simple (a semisimple finite dimensional $k$-algebra is the product of simple ones).

A simple finite dimensional $k$-algebra is $M_n(D)$ where $D$ is a finite dimensional division $k$-algebra (Wedderburn). If $k$ is algebraically closed, then $D=k$.

If $k$ is not algebraically closed and $K$ is a proper finite extension of $k$, you have a counterexample: $K$ is a simple $k$-algebra and not a full matrix ring over $k$.

I don't think you can get a better characterization than the one in rschwieb’s answer.