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Suppose $X$ is a smooth, projective curve, $Y$ is an arbitrary curve(may be singular), and both curves are over an algebraically closed field $k$ with character 0. Let $f: X \to Y$ be a morphism between curves. Is $f$ a projective morphism? Here, projective morphism is in the sense of Hartshorne, i.e. $X \to Y$ factors through $X \to \mathbb{P}^{n}_{Y} \to Y$, with $X \to \mathbb{P}^{n}_{Y}$ a closed embedding, $\mathbb{P}^{n}_{Y} \to Y$ the the natural projection to $Y$ factor.

I guess it is projective by the following general heuristic argument:

Statement:Suppose $X \subset \mathbb{P}^{n}$ is a closed subvariety, $Y$ is another variety, then any morphism $f: X \to Y$ is projective.

One can define $f' :X \to \mathbb{P}^{n}_{Y}$ by $x \mapsto (x,f(x))$, and this is an injective map. Moreover, because $X$ is proper, its image must be closed. I guess these guarantee $f'$ is a closed embedding, and the projection $\mathbb{P}^{n}_{Y} \to Y$ is easy to define.

I am not quite sure about the above argument, especially $f'$ being a closed embedding. Any suggestions?

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The answer is yes. Your proof looks fine to me. If you want a way to do it with less explicit computation, are you familiar with the notion of a base change? If you are, The map you define is the composition of $X \rightarrow X \times Y$ and $X \times Y \rightarrow P^n \times Y.$ The first one is a base change from $Y \rightarrow Y \times Y$, and is thus a closed embedding. The second one is a base change of $X \rightarrow P^n.$ This proves that $f'$ is a closed embedding. –  only Aug 26 '12 at 3:55
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There is also a more general statement (Which I know as the cancellation theorem) that is very useful: Let $X$, $Y$ be $Z$-schemes, with structure morphisms $g$ and $h$, and let $f:X \rightarrow Y$ be a map of $Z$-schemes. Then $f$ is just the composition of the graph morphism $X\rightarrow X \times_Z Y$ and the projection $X \times_Z Y \rightarrow Y.$ The most important application of this is that if we have some property $P$ of morphisms closed under composition/base change, then to prove $f$ is $P$, it suffices to show $g$ and the diagonal morphism of $h$ have property $P$. –  only Aug 26 '12 at 3:59
    
Dear Only, That makes a lot sense! I would be very happy to accept your comments as the answer to my question. Thank you again!! –  Li Zhan Aug 26 '12 at 17:22
    
@only: you need to assume the curve $Y$ is separated. –  user18119 Aug 26 '12 at 21:13
    
Hmm, now that I think of it, I've always treated curve as just "reduced separated scheme of dimension 1", which would imply separated. I don't think I've seen a formal definition of curve... –  only Aug 27 '12 at 2:25

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up vote 3 down vote accepted

The answer is yes. Your proof looks fine to me. If you want a way to do it with less explicit computation, are you familiar with the notion of a base change? If you are, The map you define is the composition of X→X×Y and X×Y→Pn×Y. The first one is a base change from Y→Y×Y, and is thus a closed embedding. The second one is a base change of X→Pn. This proves that f′ is a closed embedding.

There is also a more general statement (Which I know as the cancellation theorem) that is very useful: Let X, Y be Z-schemes, with structure morphisms g and h, and let f:X→Y be a map of Z-schemes. Then f is just the composition of the graph morphism X→X×ZY and the projection X×ZY→Y. The most important application of this is that if we have some property P of morphisms closed under composition/base change, then to prove f is P, it suffices to show g and the diagonal morphism of h have property P.

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