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?
