# Algebraic Geometry – Open Morphism of Schemes

Let $$f: X to S$$ a finite morphism between affine schemas $$X = Spec (A), S = Spec (R)$$. Note by $$phi: R to A$$ the corresponding circular card.

I am looking for pure theoretical / algebraic ring tools / criteria to define if $$f$$ is a open the map (in the topological sense). More concretely in the sense of what conditions the ring morphism $$phi$$ and resp induced morphisms $$phi_p: R_p to A_p$$ on the family of locations to $$p$$ implies that $$f$$ is open.

The bottom of my question is the following thread: The Finite and Locally Free map is open.

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

Obviously, the problem is local so we can work with the above setting 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 author observes that, due to local weakness, the stems of $$f _ * mathcal {O} _X$$ are nonzero on an open subset of $$S$$.

What does he mean? That at every point $$s in S$$ there is a rod in $$( phi _ * mathcal {O} _X) _s cong mathcal {O} ^ n_ {S, s}$$ which can be extended to a section on an open subset $$U subset S$$? Is it not regulated by definition stems as direct limit representatives?

Again, since the problem is local so wlog $$U = D (f)$$ or $$f in R$$. Why does this stem condition imply that $$f$$ is open? Does this stem from a more general criterion of openness based on switched algebra methods?