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

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Let ϕ:X→S be a finite and locally free morphism of schemes.

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

I don't understand the author's argument that the image \phi(X) is (as topological map) open.

Indeed, since \mathcal{O}_{S,s} contains non zero stalks also (\phi_*\mathcal{O}_X)_s \cong \mathcal{O}^n_{S,s} as free module has this property too. But why does it imply the openness of \phi?

Remark: I know that observing that locally free implies flatness this statement can be swapped to "flat and finite presentation implies openness". But I'm keen interested concretely in author's argument presented in the excerpt. Does anybody see how it works?

user267839
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  • @Denis Nardin: yes yes that's true (see my remark). The point is that it seems that the author didn't use the route over flatness and contructable sets and argues more "directly" with the stalk argument given in the excerpt. the main issue of this thread is not to understand why the statement is true but the concrete argument of the author – user267839 Apr 27 '19 at 14:54
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    The author does not claim that the map is open (which is nevertheless true), only that its image is. For this the argument is correct. – Laurent Moret-Bailly Apr 27 '19 at 16:17
  • @LaurentMoret-Bailly: oh yes. I see. But I still unsure about this argument with stalks. If f(x) :=s \in im(f) then by locally freeness there exist a non zero stalk t \in \mathcal{O}_{S,s} beeing non zero on an open s \in U (if I inderstood it correctly then this is what the author means with this stalk condition). The aim is to show that this U (maybe after shrining) it is contained in f(X). I don't see how it follows. – user267839 Apr 27 '19 at 16:53
  • ...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 $(f_*\mathcal{O}X)_y \cong \mathcal{O}^n{Y,y} \neq 0 for all y \in S by his considerations in the excerpt. If y \not \in f(X) we know that the complement of f(X) is open so we can find an direct system (U_i)_i \subset S containing y such that f^{-1}(U_i) = \emptyset$. – user267839 Apr 27 '19 at 20:02
  • Then $(U_i)i is cofinal towards the system of system of all open neighbourhoods of y and therefore (f*\mathcal{O}_X)_y \cong \varinjlim_i \mathcal{O}_X(f^{-1}(U_i))= 0$. a contradiction – user267839 Apr 27 '19 at 20:02
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    More simply, putting \mathcal{A}:=f_*\mathcal{O}_X, a point s is in the image of \phi if and only if the \kappa(s)-algebra \mathcal{A}(x) is nonzero. So the image of \phi is the (open and closed) subset of S where the rank of \mathcal{A} is positive. – Laurent Moret-Bailly Apr 28 '19 at 09:55

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