$A$ reduced implies $A\otimes_k K$ reduced, for $k$ perfect.

Question

Let $k$ be a perfect field and $k\subset K$ any field extension. Let $A$ be any reduced $k$-algebra, if it helps we may assume it is finitely generated but the result should be true regardless. How can we prove that $A\otimes_k K$ is also reduced?

Here, reduced just means it has no nonzero nilpotent elements, i.e. $Nil(A)=0$. This question is related to the so-called geometric reducedness of $A$ over $k$.

Answer 1

If $k$ is a perfect field and $A,B$ are reduced $k$-algebras, then $A\otimes_k B$ is reduced.
The proof is in Bourbaki, Algèbre, Chapitre V, Théorème 3 d), page 119.
It is a more general version of the result you ask about, in which your $K$ is not assumed to be an extension field of $k$ but only a reduced $k$-algebra $B$.

Answer 2

Remark: Assume au contraire $A\otimes K$ is not reduced. Then one can find a nilpotent nonzero element $\sum_{i=1}^n a_i\otimes \lambda_i$. Taking $A'=k[a_1,\ldots, a_i]$ we see that $A'\subset A$ and so $A'$ is reduced but $A'\otimes K\subset A\otimes K$ and $A'\otimes K$ is not reduced because we have just displayed a nonzero nilpotent element inside it. Therefore we may assume WLOG that $A$ is finitely generated over $k$. Hence, Hilbert's Basis Theorem implies it is noetherian and so it has finitely many minimal primes: $\{\mathfrak{p}_1,\ldots, \mathfrak{p}_n\}$. Because $A$ is reduced, $\bigcap \mathfrak{p}_i=Nil(A)=0$ and so we have an injective map $A\hookrightarrow\prod_{i=1}^n A/\mathfrak{p}_i$. Tensoring with $K$ we find $$ A\otimes K\hookrightarrow \prod_{i=1}^n \left(\frac{A}{\mathfrak{p}_i}\otimes K\right).$$ If we prove that each $(A/\mathfrak{p}_i)\otimes K$ is a domain then the RHS will be a finite product of domains and thus a reduced ring. Thus $A\otimes K$ will be reduced, as desired. But each $A/\mathfrak{p}_i$ is a $k$-domain and so our claim follows from the following fact: "If $k$ is a perfect field and $A,B$ are domains then $A\otimes B$ is reduced."

Then we've reduced the problem to a special case of the theorem (Bourbaki, Algebra, Chapter V, Section 15, Theorem 3) that Georges mentions. Perhaps this remark is empty or perhaps proving this special case is easier than proving the whole theorem, I'm not sure.