2
$\begingroup$

Let $L/K$ be a field extension. Let $R,S$ be two local commutative $K$-algebras and let $\varphi : R \to S$ be a homomorphism of $K$-algebras, not assumed to be local. Let's call a prime ideal $\mathfrak{p} \subseteq R \otimes_K L$ good when $\mathfrak{p} \cap R = \mathfrak{m}_R$. Notice that good prime ideals correspond to prime ideals in the tensor product of fields $R/\mathfrak{m}_R \otimes_K L$.

I wonder if the following is true:

Question. Is there some good prime ideal $\mathfrak{p} \subseteq R \otimes_K L$ such that for all $f \in R \otimes_K L$ with $f \notin \mathfrak{p}$ the image of $f$ in $S \otimes_K L$ is not contained in every good prime ideal of $S \otimes_K L$? Equivalently, the image of $f$ in $S/\mathfrak{m}_S \otimes_K L$ is not nilpotent.

In terms of the local $K$-schemes $X=\mathrm{Spec}(R)$, $Y=\mathrm{Spec}(S)$ and the morphism $Y \to X$, the question is the following: Is there some point in $X_L$ over the closed point of $X$ such that every open neighborhood of it pulls back to an open subset in $Y_L$ which contains a point over the closed point of $Y$?

I have checked some special cases, but either they were trivial or too hard to understand, since tensor products of fields can be nasty. Of course it is true when $L=K$, and it is also true when $\varphi$ is local. My feeling is that the statement is false in general, but I might be wrong. Maybe counterexamples can be constructed from localizations of $R$.

The context for this question is to prove a certain result for locally ringed spaces, and for locally ringed spaces with exactly two points it comes down to the question above.

$\endgroup$
2
$\begingroup$

Let $x$, resp $y$ be the closed point of $X$, resp $Y$. Denote $f : Y \to X$ the given morphism. Then $f(y) \leadsto x$ (specialization). Let $x_L$ be any point of $X_L$ mapping to $x$. The morphism $X_L \to X$ is flat. Hence there is a specialization $z \leadsto x_L$ in $X_L$ such that $z$ maps to $f(y)$. Since $Y_L = Y \times_X X_L$, there is a point $w$ of $Y_L$ which maps to $y$ in $Y$ and $z$ in $X_L$. This proves what you want as $w$ will be in the inverse image of any open neighborhood of $x_L$ you pick.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.