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I want to show that projective subvarieties of a quasi-projective variety are closed. One possible solution should be the following:

Let $W \subseteq \mathbb{P}^n$ be a quasi-projective variety and $V \subseteq W$ a projective subvariety. Then the natural inclusion $\iota: V \hookrightarrow W$ is a regular map. Since $V$ is projective, the image $V = \iota(V) \subseteq W$ is closed by the closed map theorem.

My question is: is there a more elementary and more direct way of proving this theorem, e.g. using only the definitions of projective varieties and closed?

I wanted to use this for proving that $\mathbb{P}^n \setminus \{x\}$ is not projective (since $\mathbb{P}^n \setminus \{x\} \subseteq \mathbb{P}^n$ is not closed).

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  • $\begingroup$ Projective subvarieties are compact, and compact things are closed. Though you did not specify the topology, which one do you want? $\endgroup$
    – Misha Verbitsky
    Commented Mar 26 at 14:40
  • $\begingroup$ @MishaVerbitsky $k$ an arbitrary (algebraically closed) field and $\mathbb{P}^n = \mathbb{P}^n(k)$ with the Zariski topology. Then this doesn't work since all quasi-projective varieties are compact but as the Zariski topology is not Hausdorff this does not imply closedness. Although the projective case is probably analog to the compact case for manifolds. $\endgroup$
    – psl2Z
    Commented Mar 27 at 0:06
  • $\begingroup$ Then it follows from the valuative criterion of properness, indeed $\endgroup$
    – Misha Verbitsky
    Commented Mar 27 at 12:29

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