At the beginning of section 8.3.1 of Liu's "Algebraic geometry and arithmetic curves", he defines a fibered surface $\pi\colon X \to S$ (I am only interested in the case when the base scheme is $S=\mathrm{Spec}\mathbb{Z}$) and immediately remarks that flatness and surjectivity of $\pi$ are equivalent. Why is that so? And does he mean surjectivity onto the underlying space or as a morphism of sheaves or both?
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3Flatness does not implies surjectivity (the converse does not hold either). But there are three additional assumptions : $S$ is a Dedekind domain, $X$ is integral and $\pi$ is projective. With these condition, this is indeed equivalent : a non constant morphism from an integral scheme to a Dedekind scheme is flat. A flat map is open, a projective map is closed. The result follows. – Roland Feb 15 '18 at 11:43
1 Answers
This answer assumes the hypothesis of the definition in Liu mentioned in the question.
$\textbf{Let $\pi$ be surjective}$.
There is a local homomorphism of local rings $\pi^* : \mathcal{O}_{S,f(x)} \rightarrow \mathcal{O}_{X,x}$ induced by $\pi$.
$\textit{Claim}$ : $\pi$ is flat.
We know that $S$ is integral. Hence there is an inclusion of function fields that is $$k(S) \hookrightarrow k(X)$$ This in particular implies that the map $\pi^*$ induce on rings is an inclusion for all $x \in X$ as the local rings of both the schemes are subrings of the local ring at the generic point of respective schemes($X$ and $S$ are integral schemes).
; i.e., The above $\pi^{*}$ is injective.
Recall that for module over Dedekind domain, flatness is equivalent to the module being torsion free.
Since $\mathcal{O}_{x,X}$ is of course torsion free on account of being an integral domain ( C.f. Here we used the injectivity of $\pi^{*}$ ), we get that the map $\pi$ is flat.
Now for the other part. $\textbf{Assume $\pi$ is flat}$.
Then the image of $\pi$ is open since $\pi$ is flat ( C.f. https://stacks.math.columbia.edu/tag/01UA, https://stacks.math.columbia.edu/tag/01TX ). Also the book assumes $\pi$ to be a projective morphism and hence proper and thus has closed image. Thus Image($\pi$)(is non-empty) is both open and closed and since $S$ is irreducible in particular connected, this implies that $\pi$ is surjective.
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Let me follow your comment. First Let $\eta \in X$ and $\zeta \in S$ be its generic point. Since $\pi$ is surjective, it is dominant map and so $\pi(\eta) = \zeta$. So it induces an inclusion of function field $K(S) \hookrightarrow K(X)$. And note that for each $x\in X$, $\mathcal{O}{S,f(x)} \subset K(S)$ and $\mathcal{O}{X,x} \subset K(X)$. And by some commutative diagram involving $\pi^{}$ and $K(S) \hookrightarrow K(X)$, we show that the $\pi^{}$ is injective for all $x\in X$. – Plantation Jun 09 '22 at 05:47
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And you wrote, "Recall that for module over DVR's flatness is equivalent to the module being torsion free". I think that the property 'Dedekind domain' is more proper than the 'DVR'.(c.f. https://stacks.math.columbia.edu/tag/0AUW). In the Liu's book, p.115, Def. 1.2., the Dedekind domain may have dimension $0$ ; i.e., the above $\mathcal{O}_{S, f(x)}$ cannnot be DVR. – Plantation Jun 09 '22 at 09:27
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And my question is, 1) why $X$ is integral? In the defitnition 8.3.1. , he defined a fibered surface as integral, projective, flat S-scheme $π:X→S$ of dimension 2. Here the 'integral' means the integrality of the $X$? not instead of the integral morphism? – Plantation Jun 09 '22 at 09:28
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And 2) why the image of flat morphism is open? (I know that a flat morphism which is locally of finite presentation is universally open.) Is there any reference? – Plantation Jun 09 '22 at 09:28
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1@Plantation I was under the impression that $X$ is integral scheme. That is what I understood from the definition. For your last comment perhaps one should argue as follows : Since $\pi$ is flat, its image contains the generic point. Since $\pi$ is projective it must be a closed morphism. SInce the image contains the generic point it must be the (integral) scheme $S$. – random123 Jun 10 '22 at 16:01