# Open Morphism of Schemes

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?

• 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. – user138661 Apr 27 '19 at 15:04