Jacobson radical of group algebra

Question

For a finite group G and a finite field $\mathbb{F}_p$ of characteristic $p$, J($\mathbb{F}_{p^k} \otimes_{\mathbb{F}_p }\mathbb{F}_p G ) = J(\mathbb{F}_{p^k}G)$? where $J(\mathbb{F}_{p^k}G)$ is the Jacobson radical of the group algebra $\mathbb{F}_{p^k}G$ and $\mathbb{F}_{p^k}$ is an extension field of $\mathbb{F}_p$.

Answer 1

Assuming that the question is the one of YCor: $J(F_{p^k}\otimes F_p G) = F_{p^k}\otimes J(F_pG)$, the answer is yes. The reason is the following: For a finite dimensional algebra $A$ over a field $K$ one can describe $J(A)$ as the minimal ideal such that $A/J(A)$ is a semisimple algebra. It is also the maximal nilpotent ideal of $A$. Let us show that if we have a finite separable extension of fields $L/K$, then $J(L\otimes_K A) = L\otimes J(A)$. On the one hand, $L\otimes_K J(A)$ is a nilpotent ideal, and is therefore contained in $J(L\otimes A)$. On the other hand, $A/J(A)$ is semisimple and can therefore be written as a direct sum of matrix algebras over division algebras. But then $$(L\otimes_K A) / (L\otimes_K J(A)) \cong L\otimes_K (A/J(A)).$$ The last algebra is semi-simple due to the fact that if $D$ is a division algebra over $K$, then $L\otimes_K D$ is a semi-simple $L$-algebra (here we use in an essential way the fact that $L/K$ is finite and separable). This implies the inclusion in the other direction, and the two ideals are therefore equal.

Answer 2

More generally, let $k$ be any field, let $K/k$ be any separable algebraic extension, and let $R$ be any $k$-algebra. Then $J(K\otimes_k R)=K\otimes_k J(R)$. [See Theorem 5.17 in Lam's "First Course in Noncommutative Rings," for instance.]