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The graph of a morphism $X \rightarrow Y$ is closed in $X \times Y$ if $X$ and $Y$ are affine varieties. What if $X$ and $Y$ are projective varieties?

I am still not quite familiar with projective varieties. So I need some help. Thanks very much.

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I believe this is so, since projective morphisms are proper, if I remember correctly. – Zhen Lin Jul 8 '11 at 11:42
@Zhen: Thank you very much for editting and adding the tags. I can put only one tag to my question, and I will take care of it from now on. Many thanks. – ShinyaSakai Jul 9 '11 at 8:38
up vote 10 down vote accepted

The graph of a morphism $f: X \to Y$ is the pull-back under the product map $f\times 1: X \times Y \to Y \times Y$ of the diagonal $\Delta(Y) \subset Y \times Y.$ Thus for the graph to be closed, what you need is the diagonal $\Delta(Y)$ to be closed in $Y \times Y$. This is true for all quasi-projective varieties, and so in particular for projective varieties (as well as affine varieties, as you noted in the question).

In general, a variety (or more generally, a scheme) is called separated if the diagonal is closed. Although there are non-separated objects, in practice it is hard to find them if you don't deliberately go looking for them.

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Aren't varieties defined to be separated schemes (reduced, of finite type, etc.)? – Zhen Lin Jul 8 '11 at 12:32
@Zhen: Dear Zhen, This depends on who you ask! In Hartshorne, yes. In Mumford's Red Book, yes (but he requires schemes to be separated too, and uses the old terminology of preschemes and prevarieties for non-separated versions of schemes and varieties). But others would use variety to mean any finite-type scheme over a field, or perhaps any finite-type geometrically reduced scheme over a field, or perhaps ... . In my experience, the best rule of thumb is to never presume a meaning for ``variety'', but always to check a particular author/speaker's intended meaning. Regards, – Matt E Jul 8 '11 at 12:37
It seems that there are different definitions of a variety. Anyway, if I am not mistaken, the answer to my question will be yes if the varieties satisfy the Hausdorff separation axiom. Thanks to everyone. – ShinyaSakai Jul 9 '11 at 8:48
@ShinyaSakai: The Zariski topology on varieties in general do not satisfy the Hausdorff condition! Indeed, if the field is infinite, then a variety is a Hausdorff space if and only if it is a finite collection of points! – Zhen Lin Jul 9 '11 at 9:52
@Zhen: Thank you very much. I am wrong for using the term "Hausdorff separation axiom". In fact, it should be called "Hausdorff axiom", which requires a topological space $X$ to have the set $\{(x,x)| x \in X\}$ closed in $X \times X$. This is equivalent to the Hausdorff separation Axiom if the $X \times X$ is given the product topology. This is why Humphreys gave it the name "Hausdorff Axiom". In his book, a prevariety is a variety if and only if this "Hausdorff axiom" is satisfied. – ShinyaSakai Jul 12 '11 at 14:39

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