Let $A$ be a finite-dimensional linear associative algebra over some field $F$. Then denote by $M_n(A)$ the set of $n \times n$ matrices with entries in $A$ and the usual operations. Then $M_n(A)$ itself is a linear associative algebra over $F$. I have a question on a proof that $M_n(A)$ is semisimple if and only if $A$ is semisimple. The proof uses the fact that $M_n(A)$ may be regarded as the direct product $M_n(F) \times A$, surely when I accept this fact then the result follows, but I do not see in what way $M_n(A)$ corresponds to $M_n(F) \times A$?
This claim is taken from W.D. Munn, On semigroup algebras, and it is Lemma 4.5 in this article.
I'll do a proof by example as I'm too lazy to write it down in general, but you'll get the idea.
Consider the map $$M_2(A)\rightarrow M_2(F)\otimes A: \begin{pmatrix} a&b\\c&d \end{pmatrix}\mapsto \begin{pmatrix} 1&0\\0&0 \end{pmatrix}\otimes a + \begin{pmatrix} 0&1\\0&0 \end{pmatrix}\otimes b+\dots $$
This is your desired isomorphism.
More generally you can see that semisimplicity is a Morita invariant and $A$ and $M_n(A)$ are Morita equivalent algebras.