Let $A$ be a finite-dimensional semisimple $k$-algebra where $k$ is a field. If $F/k$ is a finite separable field extension, then show that $A \otimes _ k F$ is a semisimple $F$-algebra.
Our definition for a finite field extension $F/k$ being separable is that for any $L/k$ algebraic, the $L$ algebra $L \otimes _k F$ is semisimple.
Any help is appreciated.
Another edit As requested, here is the 'reduced' solution extracted from the general picture described below.
By your definition, $F\otimes_k F$ is semisimple. In particular, the multiplication map $\mu: F\otimes_k F\to F$ splits as a map of $F\otimes_k F$ modules. Let $\sigma: F\to F\otimes_k F$ be an $F\otimes_k F$-linear splitting of $\mu$ and write $\sigma(1) = \sum_i e_i\otimes f_i$. Then $$(\dagger)\qquad\sum_i e_i f_i\ =\ 1$$ by $\mu\circ\sigma=\text{id}$, and $$(\ddagger)\qquad\sum_i x e_i\otimes f_i\ =\ \sum_i e_i\otimes f_i x\quad\text{ for all } x\in F$$ by the $F\otimes_k F$-linearity of $\sigma$.
Let $\pi: M\to N$ be an epimorphism of $A\otimes_k F$-modules. We need to show that it splits. As $A$ is semisimple, $\pi$ splits as a morphism of $A$-modules. Let $s: N\to M$ be an $A$-linear splitting and consider the map $\tilde{s}: N\to M$ given by $$\tilde{s}(n) := \sum_i e_i s(f_i n)\quad\text{ for } n\in N.$$ By $(\dagger)$, we still have $\pi\circ\tilde{s} = \text{id}_N$, and by $(\ddagger)$, $\tilde{s}$ is $F$-linear. Finally, the $A$-linearity of $s$ transfers to $\tilde{s}$. Hence $\tilde{s}$ is an $A$-linear and $F$-linear, hence $A\otimes_k F$-linear splitting of $\pi$ as required.
Edit The following is an alternative solution that is less general but more direct than the one sketched below:
$A/k$ being semisimple, it is a product of matrix algebras over division algebras, so we may assume $A=D$ to be a division algebra - this is because base extension commutes both with finite products and with passing to matrix algebras, and matrix algebras over semisimple algebras are semisimple.
So suppose $D$ is a finite-dimensional division algebra over $k$ and $F/k$ is separable and your sense. We wish to prove that $D\otimes_k F$ is semisimple. Consider the center $L := \text{Z}(D)$, a finite field extension of $k$. To show that $D\otimes_k F$ is semisimple, it suffices to check that $D^{\text{opp}}\otimes_L (D\otimes_k F)$ is semisimple (the general statement is: If $B\otimes_k A$ is semisimple, then so is $A$). Now $$D^{\text{opp}}\otimes_L (D\otimes_k F)\cong (D^{\text{opp}}\otimes_L D)\otimes_k F\stackrel{(\ast)}{\cong}\text{Mat}_{\dim_L D}(L)\otimes_k F\cong\text{Mat}_{\dim_L D}(L\otimes_k F).$$ By your definition of separability, $L\otimes_k F$ is semisimple, hence so is $\text{Mat}_{\dim_L D}(L\otimes_k F)$ and we're done.
The crucial step here is $(\ast)$, see e.g. Corollary 2.9 in Milne's Notes on Class Field Theory
The following is a bit more general than what you asked for but hopefully enlightens the general relation between semisimplicity, separability and stability of semisimplicity under base extension. As an exercise, you might write everything out explicitly in your concrete situation.
Let ${\mathbb k}$ be a commutative ring. Call an extension $A/B$ of ${\mathbb k}$-algebras semisimple if a short exact sequence of $A$-modules is split over $A$ if and only if it is split over $B$ (this does not involve ${\mathbb k}$). Call $A/B$ universally semisimple over ${\mathbb k}$ if for any ${\mathbb k}$-algebra $C$ the extension $A\otimes_{\mathbb k} C / B\otimes_{\mathbb k} C$ is semisimple. Call $A/B$ separable over ${\mathbb k}$ if the multiplication map $\mu_{A/B}: A\otimes_{B} A^{\text{opp}}\to A$ is split as a morphism of $A\otimes_{\mathbb k}A^{\text{opp}}$-modules.
If $B$ is (absolutely) semisimple, (relative) semisimplicity of $A/B$ is equivalent to the (absolute) semisimplicity of $A$. Also, if $A/B$ and $B/C$ are semisimple, then so is $A/C$.
You can now check the following:
For $\text{(1)}$, you can '$A$-linearize' any $B$-linear map between $A$-modules by multiplying it with a separability idempotent of $A/B$, i.e. the image of $1$ under an $A\otimes_BA^{\text{opp}}$-linear splitting of $\mu_{B/A}$. This is reminiscent of the proof of Maschke's Theorem on the semisimplicity of group algebras of finite groups.
Conversely, you have:
Indeed, $\text{(1)}\Rightarrow\text{(2)}$ follows by taking $C := A^{\text{opp}}$ in the definition of universal semisimplicity, while $\text{(2)}\Rightarrow\text{(3)}$ is true since $\mu_{A/B}$ is split over $B\otimes_{\mathbb k} A^{\text{opp}}$ through $a\mapsto 1\otimes a$.
Since $\text{(3)}\Rightarrow\text{(1)}$ by Lemma 1, we see that in fact $\text{(1)}\Leftrightarrow\text{(2)}\Leftrightarrow\text{(3)}$. Also, we conclude that the above definition of separability over ${\mathbb k}$ of a finite field extension $F/{\mathbb k}$ is equivalent to your definition of separability: By definition, the above version is at least as strong as your definition, and conversely, taking $L := F$ in your definition verifies property $\text{(2)}$.
Finally you need to put everything together. Suppose $F/{\mathbb k}$ is a separable field extension in your sense, then:
Then $F/{\mathbb k}$ is separable in the above sense, hence universally semisimple over ${\mathbb k}$, so $F\otimes_{\mathbb k} A/ A$ is semisimple for all $A/{\mathbb k}$.
In particular, for $A$ (absolutely) semisimple, both $F\otimes_{\mathbb k} A/ A$ and $A/{\mathbb k}$ are semisimple.
In this case, $F\otimes_{\mathbb k} A / {\mathbb k}$ is semisimple, too, by transitivity of semisimplicity. Since ${\mathbb k}$ is absolutely semisimple, $F\otimes_{\mathbb k} A$ is absolutely semisimple, too, as claimed.