Tensor of finite-dimensional algebra over perfect field is semisimple Let $K$ be a field and let $\Lambda_{1}$ and $\Lambda_{2}$ be two finite-dimensional $K$-algebras with Jacobson radicals $J_{1}$ and $J_{2}$ respectively. How to show or where can I find the proof of the following statement?

$\Lambda_{1} / J_{1} \otimes_{K} \Lambda_{2} / J_{2}$ is always semisimple if $K$ is perfect or if $\Lambda_{1}$ and $\Lambda_{2}$ are path algebras of quivers factored by admissible ideals.

Thank you.
 A: We have $gldim A \otimes_K B= gldim A + gldim B$ if A and B are seperable algebras over the field $K$, see https://www.cambridge.org/core/journals/nagoya-mathematical-journal/article/on-the-dimension-of-modules-and-algebras-viii-dimension-of-tensor-products/58116B52E52F0F6165E84AE11284CCF6 corollary 18.
Now being semisimple for finite dimensional algebras is equivalent to global dimension zero.
A: Here is an alternate version of @Mare's answer.   First recall that a $K$-algebra $A$ is separable if it is semisimple under all base extensions; its enough to check over an algebraic closure of $K$.
Let us write $L\otimes_K A$ as $A^L$ for a $K$-algebra $A$ and field extension $L/K$.
Any semisimple algebra over a perfect field is separable. By Wedderburn-Artin, it suffices to show that if $D$ is a finite dimensional division algebra over $K$, then $D^{\overline K}=\overline{K}\otimes_K D$ is semisimple, where $\overline{K}$ is an algebraic closure of $K$.  Let $F$ be the center of $D$; then $F/K$ is a finite field extension and hence separable because $K$ is perfect.  Then $D^{\overline K}\cong (\overline K\otimes_K F)\otimes_F D$.  But since $F/K$ is separable, basic field theory says $\overline K\otimes_K F\cong \overline K^{[F:K]}$.  Therefore, $D^{\overline K}\cong (\overline K\otimes_F D)^{[F:K]}$.  But $D$ is central simple over $F$ and so by a basic result in the theory of central simple algebras, $\overline K\otimes_F D\cong M_n(\overline K)$ where $n^2=[D:F]$. Thus $D^{\overline K}$ is semisimple.
Also, if $A$ is a split semisimple $K$-algebra (so isomorphic to a direct product of matrix algebras over a field), then $A$ is separable.  In particular, if $\Lambda = KQ/I$ where $Q$ is a quiver and $I$ is an admissible ideal, then $\Lambda/J(\Lambda)\cong K^{|Q_0|}$ and hence is a separable $K$-algebra.
Thus your question really boils down to proving that if $A$ and $B$ are separable $K$-algebras, then so is $A\otimes_K B$.   It is enough to show that $(A\otimes_K B)^{\overline K}$ is semisimple where $\overline K$ is an algebraic closure.  But $(A\otimes_K B)^{\overline K}\cong A^{\overline K}\otimes_{\overline K}B^{\overline K}$ as they both have the same universal property. The right hand side is a tensor product of direct sums of matrix algebras over $\overline{K}$ and hence is a direct sum of matrix algebras over $\overline{K}$ (as $M_n(\overline K)\otimes_{\overline K} M_m(\overline {K})\cong M_{mn}(\overline K)$) and thus semisimple.  Therefore, $A\otimes_K B$ is separable and hence semisimple.
