From Lam's A First Course in Noncommutative Rings, section 1.3.
Let $R$ be a domain (EA: that is, a ring without zero divisors) such that $M_n(R)$ is semisimple. Show that $R$ is a division ring.
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My Answer: Break up $M_n(R)$ into left ideals $J_i = \{ \text{All columns but $i $ are zero}\}$. Now I claim without proof that left subideals of $J_i$ are all of the form $J_i \cap M_n(I)$, for $I \triangleleft R$ a left ideal. I further claim that $I$ is minimal iff $J_i \cap M_n(I)$ is minimal. Therefore to decompose $J_i$ as the sum of minimal left ideals is to decompose $R$ as the sum of minimal left ideals. So $R$ is semisimple.
Since $R$ is semisimple, $R = I_1 \oplus \dotsb \oplus I_n$, for $I_j$ minimal ideals. I claim without proof that the only way this can be a domain is if there is only one summand, that is, if $R$ has no left ideals, and so is a division ring.
I'm skeptical of my answer, because it would seem they would add as an intermediate step to prove that $R$ is semisimple. Do you see any problems? Or perhaps a more elegant way?
At a high level, one could think this way: $R$ is right Artinian iff $M_n(R)$ is right Artinian for every natural number $n$. If $M_n(R)$ is semisimple, then $R$ is an Artinian domain, and that is a division ring."
The problem is, of course, to show that the Artinian condition behaves that way. If you evaluate that proving "$M_n(R)$ right Artinian implies $R$ right Artinian" is not too much of a hardship, it would be a good way to go.
There are definitely good posts here on math.SE which can help you do this, for example:
https://math.stackexchange.com/a/26830/29335