Sorry for my bad English.
I confuse about Mumford’s abelian variety p58.
Let $X$ be complete irreducible algebraic variety over algebraic closed field $k$.
And $f:X\to \mathbb{P}^n$ is not finite morphism.
In this book, he said there is irreducible curve $C$ on $X$, such that $f(C)$ is one point.
But I can’t construct it.
Please tell me solutions, hints, or references, thanks.
A morphism is finite if and only if it is (1) proper and (2) has finite fibers. We can first of all conclude that $f$ is proper by Corollary II.4.8(e) in Hartshorne: take $Y = \mathbb P^n$ and $Z = \operatorname{Spec}(k)$; since $X$ is complete, $X \to Z$ is proper, and we know that $(g:Y \to Z) = (\mathbb P^n \to \operatorname{Spec}(k))$ is separated, so we conclude that $f$ is proper.
This means that under your hypotheses, $f$ is not finite if and only if it has at least one infinite fiber. Assuming for a moment that every fiber of $f$ has finitely many irreducible components, we conclude that $f$ has at least one fiber of positive dimension. First choose an irreducible component of this fiber, and then select an irreducible curve $C$ in this component. Since $C$ lies in a fiber, $f(C)$ is a point.
Finally, the matter of the fibers having finitely many irreducible components: this should follow from some kind of Noetherian assumption that is generally baked into the definition of a variety, but I don't have a copy of Mumford's book in front of me, so I will leave this detail to you.