Let $f: C \to D$ be a proper, non constant scheme morphism of curves. (a curve is for me a $1$-dimensional, separated $k$-scheme of finite type).

Assume futhermore that $C$ and $D$ are irreducible and proper.

Let $b \in D$ an arbitrary closed point of $D$. My question is how to see that the fiber $f^{-1}(b)$ is a finite set?

My considerations:

Since the property "finite type" is stable under base change we deduce that the scheme structure $C \times_D \kappa(b)$ of the fiber $f^{-1}(b)$ is a $\kappa(b)$-scheme of finite type. Therefore the ring of grobal sections of $C \times_D \kappa(b)$ is Noetherian by Hilbert's Basissatz.

Futhermore $f^{-1}(b)$ is discrete and a union of closed points. But here I don't see why the beeing Noetherian property for the ring $O_{C \times_D \kappa(b)}(C \times_D \kappa(b))$ imply the Noether ascending property for $C \times_D \kappa(b)$ as topological space. The problem is that $C \times_D \kappa(b)$ isn't affine.

Another approach would be to show that every complement of an non empty open set in $C$ is finite but I'm not sure why it should here hold. I can only say that every open set of $C$ is dense but not more.

  • $\begingroup$ The fiber is a closed subset of a noetherian topological space and is thus noetherian. Secondly, such a morphism will always be proper (see 01W6), so it doesn't really make sense to talk about replacing proper by surjective without further altering the question. $\endgroup$ – KReiser Mar 27 at 5:39
  • $\begingroup$ @KReiser: ok so the problem reduces the point to verify that a curve $C$ is a noetherian topological space. But here can only deduce that it is locally noetherian. Indeed, locally noetherian is clear since $C$ is $k$-scheme of finite type so we can find for each $c \in C$ a wlog affine open neighborhood $U_c = Spec(R)$ such that $R = k[x_1,...,x_n]/I$ and by Hilbert $R$ is noetherian therefore $U_c$ is noetherian (especially as topological space). The proplem is that the argument $R$ noetherian $\Leftrightarrow$ $Spec(R)$ noetherian works only for affine schemes. $\endgroup$ – KarlPeter Mar 27 at 12:10
  • $\begingroup$ @KReiser:But $C$ is in general not affine so don't know how to show that $C$ is a noetherian space. The overkill argument would be to embedd it in a $\mathbb{P}^n$. Do you see a more "elementary" argument? $\endgroup$ – KarlPeter Mar 27 at 12:10

If there were an infinite fiber, then it would have an infinite irreducible component. That is, we have an infinite closed irreducible set inside of the curve $C$. This is impossible for dimension reasons unless that component is the whole curve. But the morphism is not constant, so the component can’t be the whole curve.

  • $\begingroup$ Why this fiber would have an infinite irreducible component? Must it have a finite number of irreducible components? $\endgroup$ – KarlPeter Mar 27 at 15:33
  • $\begingroup$ Seems that it also boils down to topological Noether property for the fiber. $\endgroup$ – KarlPeter Mar 27 at 15:41
  • $\begingroup$ Curves are Noetherian because they’re locally Noetherian and finite type so in particular they are quasi compact. Subsets of Noetherian spaces are Noetherian. Therefore the fiber has only finitely many irreducible components. $\endgroup$ – Samir Canning Mar 27 at 15:42
  • $\begingroup$ Do you know a reference /sketch of the argument that locally Noetherian + finite type imply Noetherian? $\endgroup$ – KarlPeter Mar 27 at 15:50
  • 1
    $\begingroup$ Finite type by definition includes quasi compact. Locally Noetherian plus quasicompact by definition means Noetherian as a scheme. Noetherian schemes are Noetherian topological spaces. $\endgroup$ – Samir Canning Mar 27 at 15:51

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