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I am interested in seeing why the statement

(1) "If $Y$ is any variety and $Z$ a closed subset of $\mathbb{P}^n \times Y$, then the projection of $Z$ on $Y$ is closed."

implies the statement

(1) "A morphism of projective varieties takes closed sets to closed sets."

In particular, let $X,Y$ be projective varieties of $\mathbb{P}^m,\mathbb{P}^n$ respectively and let $f: X \rightarrow Y$ be a morphism. If the graph $G_f = \left\{ (x,f(x)), x \in X \right\} \subset \mathbb{P}^m \times Y$ was a closed subset of $\mathbb{P}^m \times Y$, then statement (1) would readily apply and $f(X)$ would be closed in $Y$ as desired. However, why need $G_f$ be closed in $\mathbb{P}^m \times Y$?

Reference: See Theorem 1 and above in Akhil Mathew's post here: http://amathew.wordpress.com/2010/10/23/a-projective-morphism-is-proper/#comments

PS: Thanks to Mariano for a correction.

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    $\begingroup$ $X\times f(X)$ is not the graph of $f$. $\endgroup$ – Mariano Suárez-Álvarez Jun 30 '14 at 18:41
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    $\begingroup$ Why the downvotes? $\endgroup$ – Manos Jun 30 '14 at 19:08
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I'm not sure what you have available to you, but it might be helpful to note that there's a map $f \times \operatorname{id}\colon X \times Y \to Y \times Y$, under which the inverse image of the diagonal is the graph of $f$. So if $Y$ is separated then the graph is closed.

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  • $\begingroup$ I am aware that "separated" often means "Haussdorff" in algebraic geometry. I am also aware that in general topology a topological space is Haussdorff if and only if the diagonal in the product with its self is closed in the product topology. However your answer confuses me because 1) as i understand varieties are almost never separated and also the Zariski topology of the product is not the product topology. (by the way, i was not the one who downvoted your answer, i am trying to learn here). $\endgroup$ – Manos Jun 30 '14 at 19:17
  • $\begingroup$ @Manos I see. I'll try to think of another way to say it, but in algebraic geometry a morphism $f\colon X \to Y$ is separated if the associated diagonal morphism $X \to X \times_Y X$ which is the identity on each factor is a closed immersion. Happily, all projective varieties have this property. No worries about the votes -- it matters little to me. $\endgroup$ – Hoot Jun 30 '14 at 19:22
  • $\begingroup$ So the graph of a projective variety $X$ under the morphism $f:X \rightarrow Y$ is always closed in the Zariski topology of $\mathbb{P}^n \times Y$? If yes, then this will satisfy me as an answer together with a reference (Hartshorne would do). $\endgroup$ – Manos Jun 30 '14 at 19:28
  • $\begingroup$ Of course, I hope I haven't said anything wrong. I don't know how to reference Hartshorne in this instance without talking about Chapter II and it seems to me that that wouldn't be helpful. I'll take a look later, perhaps. $\endgroup$ – Hoot Jun 30 '14 at 19:28
  • $\begingroup$ Nevermind the reference for now. But it looks like to me that there are many results (like this one) for varieties coming from schemes, so that if someone wants to read a proof, he would really have to go and read the proof of the scheme-theoretic result. Do i have that right? $\endgroup$ – Manos Jun 30 '14 at 19:45

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