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Wedderburn's classification of semisimple algebras tells us that any semisimple algebra $A$ is isomorphic to a finite direct product of matrix algebras over division algebras, say $A \cong M_{n_1}(D_1) \times \dots M_{n_k}(D_k)$.

If $A'$ is a subalgebra of $A$, then is there some subset $\{ M_{n_{i_1}}(D_{i_1}), \dots M_{n_{i_m}}(D_{i_m,})\}$ of $\{ M_{n_1}(D_1), \dots , M_{n_k}(D_k) \}$ such that $A' \cong M_{n_{i_1}}(D_{i_1}) \times \dots \times M_{n_{i_m}}(D_{i_m,})$?

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Every finite dimensional $k$-algebra is isomorphic to a subalgebra of $M_n(k)$ for some $n$ (you can take $n=\dim A$, in fact)

So no, for then all algebras would be semisimple!

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  • $\begingroup$ Could you elaborate? I don't see how that implies that all algebras would be semisimple. My claim is about subalgebras of a semisimple algebra. $\endgroup$ May 28, 2016 at 19:28
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    $\begingroup$ As I wrote, every finite-dimensional $k$-algebra $A$ is isomorphic to a subalgebra of a semisimple algebra, namely, of $M_{\dim A}(k)$. If your claim were true, then all subalgebras of $M_n(k)$ would be semisimple and that, together with my observation, would imply that all finie dimensional algebras are semisimple. $\endgroup$ May 28, 2016 at 19:29
  • $\begingroup$ Ahh, I see now. Thanks! $\endgroup$ May 28, 2016 at 19:33
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The upper triangular matrices form a subalgebra of $M_2(F)$, and they are an obvious counterexample to the proposition.

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