# Finite Morphism is Closed and Open

I have a question about a statement used in Szamuely's "Galois Groups and Fundamental Groups" in the exerpt below (look up at page 152):

Let $$f:X \to Y$$ be a finite and locally free morphism of schemes.

Remark: "locally free" means that the $$\mathcal{O}_Y$$-module $$f_* \mathcal{O}_X$$ is locally free

The author claims that $$f$$ is (as topol map) closed and open.

Why?

My ideas:

Firstly, regarding closeness it suffice to show that $$f$$ is proper. The high tech weapon would be the valuative criterion for properness conbining with the fact that finite maps are integral and for integral ring maps $$R \to A$$ the going up works.

Is there exist a more "elementary" argument without valuative criterion for "finite implies proper"?

Secondly the author claims that the map is also open by local freeness of $$f_* \mathcal{O}_X$$. This sould imply that $$f_* \mathcal{O}_X$$ has non zero stalks. Why does this imply the openness?

UPDATE: Thanks to Alex Youcis' great answer below meanwhile several questions were solved. To only problem I'm still faced is Szamuely's argument that local freeness of $$f_* \mathcal{O}_X$$ implies that the image $$f(X)$$ is open. One way - as Alex stated - is to use the fact that locally free imply flat and then argue that flat and finite presentation implies open. That's fine.

But from the didactical point of view I'm still interested in Szamuely's argument concretely: he makes the to observation that for all $$y \in Y$$ the stalk $$(f_*\mathcal{O}_X)_y \cong \mathcal{O}^n_{Y,y}$$ isn't zero. How does this already imply that the map $$f$$ is open?

Finite implies closed and flat and finite presentation implies open. Since $$f$$ is finite locally free it's therefore both flat and finite so both open and closed.
• yes, this way by passing from locally free to flat seems to work (according to Stacks ). The only unsolved aspect is that in the excerpt to author seems to work with the property that the stalks open open set of $Y$ not vanish. Do you see which argument Szamuely here had in mind? – KarlPeter Apr 27 '19 at 2:55
• I think that I have overseen one cruical point. namely the author not claims that his argument with stalks proves that the map is open but only that the image $f(X)$. Do you see how this argument works? If $f(x):=s∈im(f)$ then by locally freeness there exist a non zero stalk $t∈\mathcal{O}_{S,s}$ beeing non zero on an open $s∈U$ (if I inderstood it correctly then this is what the author means with this stalk condition ). The aim might be to show that this $U$ (maybe after shrining) it is contained in $f(X)$. I don't see how it follows. – KarlPeter Apr 27 '19 at 17:46
• @KarlPeter I don't understand what you're concerned about. Whatever Szamuely meant to say, the reason it's open is for the reasons I cited. I can't divine the mind of Szamuely unfortunately. Also, notice that a posteriori that $f$ is actually surjective since $f(X)$ is open and closed. – Alex Youcis Apr 27 '19 at 17:49
• ...think I got it: This is just if we take a $f(x)$ then by considerations above we can find an open neighbourhood $S$ of $f(x)$ such that $(f_*\mathcal{O}_X)_y \cong \mathcal{O}^n_{Y,y} \neq 0$ for all $y \in S$ by his considerations in the excerpt. If $y \not \in f(X)$ we know that the complement of $f(X)$ is open so we can find an direct system $(U_i)_i \subset S$ containing $y$ such that $f^{-1}(U_i) = \emptyset$. – KarlPeter Apr 27 '19 at 19:46