Jacobson Radical of the Tensor Product of Fields

Question

Let $k$ be an algebraically closed field, and take field extensions $K, L$ of $k$. The $k$-algebra $K \otimes_k L$ is a domain, and in particular it is a reduced. I am wondering if the Jacobson radical $\text{Jac}(K \otimes_k L)$ is $0$. If it isn't , then what conditions could be supplied so that it is?

Answer 1

The answer is that $\operatorname{Jac}(K\otimes_k L)=0$. The proof is not terrible, but its more than a few lines.

It looks like the first proof was by John Lawrence in 1989. I couldn't find a free PDF but it is available here. I found a shorter proof here from the same year (Theorem 2.4).

Answer 2

If you suppose $K$ and $L$ are finitely generated $k$-algebras then $K \otimes_k L$ is a finitely generated $k$-algebra, and you have $\sqrt 0 = \textrm{Jac}(K \otimes_k L)$ by the Nullstellensatz. Since you already know your algebra is reduced, it follows $ \textrm{Jac}(K \otimes_k L) = 0$.

Observation: I'm using that $\sqrt I = \cap m$ where $m$ is maximal ideal containing $I$.