Does hold the isomorphism$ (A \otimes_R B)/J(A \otimes_R B) \cong (A/J(A)) \otimes_R (B/J(B))$ where $R$ is a commutative ring?

Question

while reading an article, I saw the isomorphism $(A \otimes_K B)/J(A \otimes_K B) \cong (A/J(A)) \otimes_K (B/J(B))$ where $A$ and $B$ are two algebras over the field $K$ and $J$ denotes Jacobson radical. Why is this isomorphism true? Also, if $A$ and $B$ are two algebras over the commutative ring $R$ then can we say this isomorphism true?

Answer 1

This is false over any field $K$ that isn't perfect. For example, if $K = \mathbb{F}_p(t)$ and $A = B$ are both the purely inseparable extension $\mathbb{F}_p(\sqrt[p]{t})$, then $A \otimes_k B \cong A[x]/x^p$ has a nontrivial nilpotent, so $J(A \otimes_k B) \cong (x)$ is nontrivial even though $J(A)$ and $J(B)$ are both trivial (since $A, B$ are fields).