6
$\begingroup$

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 $k$.

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – Aphelli
    Nov 11, 2019 at 17:48
  • 1
    $\begingroup$ @Mindlack I think you mean Lemma 10.42.6? $\endgroup$ Nov 11, 2019 at 21:04
  • 1
    $\begingroup$ @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? $\endgroup$
    – Aphelli
    Nov 11, 2019 at 22:17

2 Answers 2

3
$\begingroup$

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 $B$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .