Proper Non Constant Morphism of Curves has Finite Fibers

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.

• 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. – KReiser Mar 27 at 5:39
• @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. – KarlPeter Mar 27 at 12:10
• @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? – 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.