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

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 .

• The quaternions $\mathbb{H}$ give a counterexample of a finite dimensional simple (and hence semisimple) $\mathbb{R}$-algebra that is not isomorphic to a direct sum of finite dimensional matrix rings $\operatorname{Mat}_{n}\mathbb{R}$. In fact, the quaternions give a nice (and the only, up to isomorphism) example of a finite dimensional central simple $\mathbb{R}$-algebra. – Geoff Jun 19 '17 at 15:29
• @Geoff "the quaternions give a nice (and the only, up to isomorphism) example of a finite dimensional central simple $\mathbb R$-algebra." Something there is mistyped, right? You already mentioned $M_n(\mathbb R)$, not to mention even $\mathbb R$ itself are other examples. Moreover, matrix rings over $\mathbb H$ are examples. So it is definitely not "up to isomorphism." I imagine you meant to say "the only f.d. central simple $\mathbb R$ algebras other than $\mathbb R$ -- up to Morita equivalence." – rschwieb Jun 19 '17 at 16:28
• @Geoff “finite dimensional central division $\mathbb{R}$-algebra”. – egreg Jun 19 '17 at 17:13
• This is always true iff $k$ is algebraically closed. – Qiaochu Yuan Jun 19 '17 at 19:29
• @rschwieb Yes, you are indeed correct. Mornings are very unkind to me and I mistyped my comment :/. Ah well, thank you for pointing out my error to me. – Geoff Jun 19 '17 at 19:29

## 3 Answers

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 $k$.

• what is $S_R$ ? – user Jun 19 '17 at 17:14
• @users I spelled it out more explicitly. – rschwieb Jun 19 '17 at 19:14

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

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.

• Sorry for reopening, but why is $K$ a simple $k$-algebra? Why is it not a full matrix ring over $k$? – M.C. Nov 4 '20 at 19:09
• @M.C. A field is a simple ring, isn't it? And a matrix ring is noncommutative unless the size of the matrices is one, but this can't be the case if $K$ is a proper extension. – egreg Nov 4 '20 at 22:30
• Oops, I was about delete that comment. Thank you anyway! – M.C. Nov 5 '20 at 7:08