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I am trying to show the following: if $X$ is an integral proper $k$-scheme, $k$ a field, then $O_X(X)$ is a finite field extension of $k$.

I have succeeded to show that $O_X(X)$ is a field but I don't see why it must be a finite field extension.

(To show it is a field I used that a global section s corresponds to a morphism $X \to \operatorname{Spec} k[x]$, one can show the image is a closed point, so if $s \neq 0$ there is an irreducible polynomial $g \in k[x]$ such that $g(s)=0$, so it is invertible.)

I would like to avoid using cohomology/Grothendieck's finiteness result for proper morphisms. a similar question was asked here but I am not assuming $X$ is geometrically integral.

KReiser
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steedsnisps
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  • What if $k=\mathbb{Q}$ and $X$ is the spectrum of the algebraic closure of $k$? – Mohan Nov 21 '20 at 17:28
  • This is a nice question - via restriction to an affine open in $X$, you need to show that the only possible subfields of a $k$-algebra of finite type are finite extensions of $k$. First, $O_X(X)$ is a finitely generated field (it's a subfield of the finitely-generated extension $k(X)/k$ - this is a potentially difficult graduate algebra exercise), but then we need to show that a transcendental field can't embed in to a fg $k$-algebra. That's harder - this MO post does it, but I'm not sure this is any simpler than what you want to avoid. – KReiser Nov 22 '20 at 06:47
  • @Mohan X is not finite type over Q. – steedsnisps Nov 23 '20 at 10:30
  • @KReiser would the following work: let $m$ be a maximal ideal of $A$, a k-algebra of finite type containing $O_X(X)$. Then $O_X(X)$ embeds into $A/m$. $A/m$ is a field which is a finitely generated $k$-algebra, so it is a finite field extension over $k$. So $O_X(X)/k$ is a finite field extension – steedsnisps Nov 23 '20 at 10:36
  • @steedsnisps You didn't say finite type . – Mohan Nov 23 '20 at 15:03
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    @Mohan finite type is part of the definition of proper. – steedsnisps Nov 23 '20 at 15:49
  • @steedsnisps Yes, I do think that would work and I feel a little silly I didn't see that earlier. – KReiser Nov 23 '20 at 21:51

1 Answers1

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The solution has been worked out in the comments. As restriction maps for the structure sheaf on an integral scheme are injective, $O_X(X)$ embeds in to $O_X(\operatorname{Spec} A)=A$ for $\operatorname{Spec} A$ any affine open subscheme of $X$. Let $\mathfrak{m}$ be a maximal ideal of $A$. Then $O_X(X)\subset A$ does not intersect $\mathfrak{m}$, so the map from $O_X(X)$ to its image in $A/\mathfrak{m}$ is an injection. Thus $O_X(X)$ embeds in to a residue field of a finite type scheme over $k$, and all such residue fields are finite extensions of $k$ by Zariski's lemma.

KReiser
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  • Actually, why $O_{X}(X)$ is a field? I understood that an element s corresponds to a morphism $X\rightarrow Spec k[t]$ which is constant and the image is a closed point, but then why $g(s)=0$? And how does this help to prove that s has an invertible element? – T. Wildwolf Apr 15 '22 at 16:19
  • @T.Karawolf $k[t]$ is a PID, so a maximal ideal corresponding to a closed point is generated by a monic irreducible polynomial $f$. If $s\neq 0$, then we can write $f=x^n+\cdots+c$ for $c\in k^\times$, and so $s(s^{n-1}+\cdots)=-c$, thus $s$ is a unit. – KReiser Apr 15 '22 at 17:19
  • Thank you but s "corresponds" to the morphism that sends any $x\in X$ to the $V(f)=(f)$ . Why should be vanished by f? I cannot see this clearly. Because we have that V(f) is the set of primes. – T. Wildwolf Apr 15 '22 at 17:29
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    @T.Karawolf maps $X\to\operatorname{Spec} k[t]$ are in bijection with ring maps $k[t]\to \mathcal{O}_X(X)$ by $t\mapsto s$ for $s\in\mathcal{O}_X(X)$ and the closure of the image is $V$ of the kernel of this ring map. See for instance 01I1 and surrounding discussion, or this MSE question. – KReiser Apr 15 '22 at 17:53
  • A quick argument on why the image is a closed point, since I got stuck on this myself: $X \to \operatorname{Spec } k[t]$ is proper, since the composite map to $\operatorname{Spec }k$ is proper and $\operatorname{Spec }k[t]$ is separated. So the image of $X$ is a proper irreducible subscheme of $\operatorname{Spec }k[t]$ which must be a point. – peco Feb 15 '24 at 16:27