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My question arose out of the following question in Qing Liu's Algebraic Geometry and Arithmetic Curves:

Let $X, Y$ be schemes over a locally Noetherian scheme $S$, with $Y$ of finite type over $S$. Let $x\in X$. Show that for any morphism of $S$-schemes $f_{x}:Spec(\mathcal{O}_{X,x})→Y$, there exist an open subset $U$∋$x$ of $X$ and a morphism of $S$-schemes $f:U→Y$ such that $f_{x}=f\circ i_{x}$, where $i_{x}:Spec(\mathcal{O}_{X,x})→U$ is the canonical morphism.

A solution for which is given here: Extending a morphism of schemes. The solution given reduces to the case that $X$, $Y$ and $S$ are affine and thus we can assume that, if $X = \text{Spec} A$, $Y = \text{Spec} B$ and $S = \text{Spec} R$, then $A$ and $B$ are $R$-algebras. I can't, however, seem to convince myself that we can reduce to $S$ affine, as that should require us to be able to restrict to open subsets of $X$ and $Y$ so that their respective structural morphisms both map into the same affine subscheme of $S$. Is it not possible that the images of the structural morphisms are totally disjoint?

RobR
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1 Answers1

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When you say $f: X \to Y$ is map of $S$-schemes, that means that it respects the map to $S$, or in other words the map $X \to Y \to S$ is the same as $X \to S$, so even if you restrict to affine subschemes the images won't ever be disjoint.

Ashwin Iyengar
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  • Ah I see, my confusion was that there was no map between $X$ and $Y$, but I see now you can use the S-maps from $Spec\mathcal{O}_{X,x}$ to $X$ and to $Y$ to ensure that you can reduce to $S$ affine. – RobR Mar 26 '18 at 12:10