Context: This question arose as I was reading the proof of Application 6.1 in Mumford's *Abelian Varieties*. However, I have extracted all of the relevant information below so this question should stand alone.


Let $X$ be a complete variety over an algebraically closed field $k$. Let $D$ be an effective divisor on $X$ such that the linear system $|D|$ is without basepoints. (That is, there is no point $P \in X$ such that $P$ is in the support of every divisor on $X$ that is linearly equivalent to $D$.) Let $L$ be the invertible sheaf associated to the divisor $D$. Then $|D|$ gives rise to a morphism $\phi : X \to \mathbb{P}^N$, where $N + 1 = \dim \Gamma(X, L)$. (I will describe this morphism in more detail below.) If $\phi$ is not a finite morphism, we can find an irreducible curve $C$ in $X$ such that $\phi(C)$ is a single point. Mumford claims that if $E$ is any effective divisor on $X$ that is linearly equivalent to $D$ then either $E$ contains $C$ or $E$ is disjoint from $C$. My question is: why is this true? Why can't $E$ intersect $C$ in a finite set of points?


Construction of $\phi$:
Choose a basis $s\_0, \ldots, s\_N \in \Gamma(X, L)$. Let $X\_i$ be the set of $x \in X$ such that $s\_i \not\in \mathfrak{m}\_xL\_x$. Let $U\_i = \mathrm{Spec}\,k[y\_0, \ldots, \hat{y}\_i, \ldots, y\_N]$. Define a ring homomorphism $f\_i : k[y\_0, \ldots, \hat{y}\_i, \ldots, y\_N] \to \Gamma(X\_i, \mathcal{O}\_X)$ by $f(y\_j) = s\_j/s\_i$. Then by composing the map induced by $f\_i$ on spectra with the natural map $\alpha\_i : X\_i \to \mathrm{Spec}\, \Gamma(X\_i, \mathcal{O}\_X)$ we obtain $\phi\_i : X\_i \to \mathrm{Spec}\, k[y\_0, \ldots, \hat{y}\_i, \ldots, y\_N]$. These maps glue together to define our morphism $\phi : X \to \mathbb{P}^N$.


Small reduction: Changing $\phi$ by an automorphism of $\mathbb{P}^N$ corresponds to choosing a different basis for $\Gamma(X, L)$. By doing this we may assume that $\phi(C) = [1 : 0 : \cdots : 0]$. Thus $\phi(C) \in U\_0$ and $\phi(C) \not\in U\_i$ for $1 \leq i \leq N$. Thus $C \subset X\_0$. Now, divisors on $X$ give rise to divisors on $X\_0$ by restriction, so we can reduce our problem to showing that $C$ is either contained in or disjoint from any divisor of $X\_0$ that is linearly equivalent to the restriction of $D$. The apparent advantage of this (though I haven't been able to make good use of it) is that the map $\phi\_0$ is fairly explicit and does not involve glueing morphisms together.


One last comment: We can identify $\mathbb{P}^N$ with $|D\_0|$ via $[a\_0 : \cdots : a\_N] \leftrightarrow \sum\_{i = 0}^N a\_i (s\_i)\_0$, where $(s\_i)\_0$ is the divisor of zeros of the global section $s\_i$. Under this interpretation, $\phi(C)$ is an effective divisor linearly equivalent to $D$. Perhaps this is a useful way to approach the question? 



On morphisms to projective space arising from a linear system
$E$ can't intersect $C$ in a finite number of points because otherwise the restriction of $\phi$ to $C$ would be a finite degree morphism, which you assume is not. $E$ is the pull-back of the hyperplane divisor, as Vesselin remarked.


