4
$\begingroup$

Let $f: X \to S$ a finite morphism between affine schemes $X=Spec(A), S= Spec(R)$. Denote by $\phi:R \to A$ the corresponding ring map.

I'm looking for pure ring theoretical/algebraic tools/criterions do deside if $f$ is an open map (in topological sense). More concretely in the sense that which conditions for ring morphism $\phi$ and resp the induced morphisms $\phi_p: R_p \to A_p$ on the family of localisations at $p$ imply that $f$ is open.

Background of my question is the following previous thread of mine: Finite and Locally Free Map is Open

Here we have the situation that $f$ is a finite and locally free morphism and I want to deduce that this already imply that $f$ is open.

Obviously the problem is local so we can work with the setting as above and assume that $X,S$ affine and $A= R^n$ as $R$-module since $\phi$ is in the given context exactly the map $R \to R^n$.

The autor observes that by local freeness the stalks of $f_*\mathcal{O}_X$ are non zero over an open subset of $S$.

What does he mean? That at every point $s \in S$ there is a stalk in $(\phi_*\mathcal{O}_X)_s \cong \mathcal{O}^n_{S,s}$ which can be extended to a section over an open subset $U \subset S$? Isn't it settled by definition of stalks as representants in direct limit?

Again, since the problem is local therefore wlog $U =D(f)$ where $f \in R$. Why does this stalk condition imply that $f$ is open? Does this arise from a more general criterion for openness based on commutate algebra methods?

$\endgroup$
  • $\begingroup$ I am not sure I completely understand the question but here (mathoverflow.net/a/20792/138661) is a ring-theoretic characterization of open immersions for affine schemes and here (mathoverflow.net/a/66901/138661) is a ring-theoretic characterization of ideals $I$ such that the complement of $V(I)$ is affine. Martin Brandenburg seems to know this stuff so you could browse through his questions and answers. $\endgroup$ – user138661 Apr 27 '19 at 15:04

Your Answer

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

Browse other questions tagged or ask your own question.