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

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 $$k$$.

• I’m aware it’s a bit overkill, but I don’t know of a simpler argument than the first lemma linked there stacks.math.columbia.edu/tag/05DS . Mostly the key points are as follows: 1) tensoring with a separable extension of $k$ preserves the reducedness; 2) if $K$ is a finitely generated field over $k$ such that $K \otimes \overline{k}$ is reduced (so it is satisfied as soon as $k$ is perfect), then $K$ is a separable extension of a purely transcendental extension of $k$ and thus tensoring $K$ with any reduced $k$-algebra is reduced. Nov 11, 2019 at 17:48
• @Mindlack I think you mean Lemma 10.42.6? Nov 11, 2019 at 21:04
• @red_trumpet: That lemma is my point 1) (and the conclusion after the point 2)). But it’s not obvious that any finitely generated field extension of a perfect field is separably generated, is it? Nov 11, 2019 at 22:17

If $$k$$ is a perfect field and $$A,B$$ are reduced $$k$$-algebras, then $$A\otimes_k B$$ is reduced.
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 $$B$$.
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."