# Closed proper subvarieties of curves are finite sets of points

Why is it true that a closed proper subvariety of a curve is a finite set of points?

I had the following lines of thinking:

Let $C$ be a curve and $X$ a proper closed subvariety of $C$. Write $C$ as a union of its irreducible components $C = C_0 \cup \ldots \cup C_r$. Then $X = X_0 \cup \ldots \cup X_r$ where $X_i$ = $X \cap C_i$ and each $X_i$ is closed in $C_i$. Since $C$ is a curve, each of the $C_i$ have dimension $1$. So each $X_i$ must be a single point (is this true?) since a single point is always closed. Thus $X$ consists of $r+1$ points.

If my reasoning is correct, then great. The reason I think it isn't is because in my notes I have the following Lemma:

Let $\alpha : X \to Y$ be a non-constant morphism of irreducible curves. Then

i) $\forall y \in Y$, $\alpha^{-1}(y)$ is a finite set

ii) ...

Using the same reasoning as before, I'd show that $\alpha^{-1}(y)$ consists of just $1$ point; rather than "finitely many".

Thanks

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Dear Jonathan, if the curve $C$ is not irreducible, the result you are asking about is false, since any irreducible component of $C$ is a closed, infinite, proper subset of $C$. –  Georges Elencwajg May 3 '12 at 22:00
@GeorgesElencwajg Okay, so why is it true if $C$ is irreducible? –  Jonathan May 3 '12 at 22:31
I'm just saying that you have to assume that $C$ is irreducible if you want the result to hold. –  Georges Elencwajg May 3 '12 at 22:38
@GeorgesElencwajg I mean: "why is a proper subvariety of an irreducible curve a finite set of points?" –  Jonathan May 3 '12 at 22:48
I have written an answer to your question. –  Georges Elencwajg May 3 '12 at 22:57

As I explained in a comment, you must assume that $C$ is irreducible if you want the result to hold.
So, suppose that $C$ is an irreducible curve and let $X_0$ be an irreducible component of your closed, strict subset $X\subsetneq C$ . Of course you have a fortiori $X_0 \subsetneq C$
If $x_0\in X_0$ is a point , you have the chain of irreducible closed subsets $\lbrace x_0\rbrace \subset X_0 \subsetneq C$.
Since $C$ has Krull dimension $1$, the above chain cannot be strict and thus $\lbrace x_0\rbrace = X_0$ .
So, irreducible components of $X$ have now been proved to be points.
Since each irreducible component of $X$ is a point and since $X$ has only finitely many irreducible components, $X$ is finite.