Counter-example: Radical of a k-algebra

Question

I have the following question. Let $k$ a field, $A$ a $k$-algebra, and $k\hookrightarrow K$ a field extension. It´s well known that $$Rad(A)\otimes_{k}K\subset Rad(A\otimes_{k}K)$$

($Rad(A)$=radical of $A$).

When is it $Rad(A)\otimes_{k}K= Rad(A\otimes_{k}K)$?

Is it always true?

Answer 1

The equality does not always hold. A counter-example is given by $k = \mathbb{F}_2(t^2)$ and $A=K=\mathbb{F}_2(t)$. Then

• $Rad(A) \otimes_k K = 0$;
• $Rad(A \otimes_k K)$ is not zero, since $A \otimes_k K = \mathbb{F}_2(t) \otimes_{\mathbb{F}_2(t^2)} \mathbb{F}_2(t)$ contains a non-zero nilpotent element, namely $1\otimes t + t\otimes 1$.

Thus the inclusion $Rad(A) \otimes_k K \subset Rad(A \otimes_k K)$ is strict in this case.