non constant morphism of integral curves Let $f: C_1 \to C_2$ be a non constant morphism of integral curves. by a curve I mean a  proper one dimensional scheme over fixed base field $k$. I saw often the usage of the fact that under assumptions above $f$ is a finite morphism; especially an affine one. is there a standard way to verify it? unfortunately I could nowhere find a complete proof of this statement. 
as far as I'm concerned I would it do by proper + finite fibers imply finiteness but since this theorem imcluding it's proof seems to be a heavy gun for me therefore I'm looking with keen interest for a more "standard" proof of the statement above. additionally, does it also hold if the curves are irreducible instead of integral?
 A: Here's an argument in the projective case that I learned from Vakil's notes. The heaviest tool it uses is Grothendieck's coherence theorem for projective morphisms.
If you don't want to assume that $C_1$ is projective, you need to prove that proper curves are projective or use Chow's lemma to prove an analogous result in the case where $\pi$ is only assumed to be proper.
Theorem 18.1.8. Let $Y$ be a locally noetherian scheme, and let $\pi \colon X \to Y$ be a projective morphism. Then $\pi$ has finite fibers if and only if it is finite.
Proof.
"If" is easy (finite schemes over a field have finitely many points), so we turn to "only if". It suffices to work locally around a point $y$ of $Y$.
We may assume that $Y$ is affine, $X$ is a closed subscheme of $\mathbb{P}^n_Y$, and $\pi$ is the restriction of the canonical projection $\mathbb{P}^n_Y \to Y$ to $X$. Let $S$ be the homogeneous coordinate ring of $\mathbb{P}^n_Y$.
Because $\pi^{-1}(y)$ is finite, graded prime avoidance tells us that there exists a nonzero homogeneous element $s$ of $S$ of positive degree such that the hypersurface $H = V_+(s)$ defined by $s$ does not meet $\pi^{-1}(y)$. Note that $X \smallsetminus H$ is affine, because it is closed in the affine scheme $\mathbb{P}^n_Y \smallsetminus H$; in particular the restriction $X \smallsetminus H \to Y$ of $\pi$ to $X \smallsetminus H$ is affine.
Since $\pi$ is closed, $\pi(H \cap X)$ is closed in $Y$. Let
$$V = Y \smallsetminus \pi(H \cap X),$$
which is an open neighborhood of $y$ in $Y$. Then $\pi^{-1}(V) \subset X \smallsetminus H$, so
$$
\pi|_{\pi^{-1}(V)} \colon \pi^{-1}(V) \to V
$$
is affine as a base change of the affine morphism $X \smallsetminus H \to Y$.
By Grothendieck's coherence theorem, $\pi_* \mathcal{O}_X$ is coherent; thus $\pi|_{\pi^{-1}(V)}$ is even finite. q.e.d.
Corollary. Let $k$ be a field and let $\pi \colon C \to Y$ be a nonconstant $k$-morphism from an irreducible projective $1$-dimensional $k$-scheme to a locally noetherian separated $k$-scheme. Then $f$ is finite.
Proof. The assumptions imply that $\pi$ is projective; by the preceding result it suffices to show that its fibers are finite. Let $y$ be a point of $Y$. Then $\pi^{-1}(y)$ is a closed subset of $Y$, and it is not equal to $Y$ because $\pi$ is not constant. Thus it is finite. q.e.d.
