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.
EDIT: The reason that finite implies closed is not because it's proper (finite implies proper precisely BECAUSE finite are universally closed). To show this essentially follows from the Going Up Lemma (which finite maps satisfy)--see this for details.