# 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.

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.

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.