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On page 54 in the Book An Invitation to Algebraic Geometry, the author said that

A quasi-projective variety in $\mathbb{P}^n$ is isomorphic to a Zariski-closed subset of some $\mathbb{P}^m$ (i.e. a projective variety) if and only if it already forms a Zariski-closed subset of $\mathbb{P}^n$.

Can anyone provide a proof?

hxhxhx88
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    It's probably too much for this point in Smith's book, but the way I would think about this is that projective varieties are proper (by the fundamental theorem of elimination theory) and it follows from some general nonsense that any map from a projective variety to a variety is also proper, and in particular closed. – Hoot Apr 24 '14 at 16:33

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I think Hoot's answer is the correct one.

If $X/k$ is projective, then any locally closed embedding $X\hookrightarrow \mathbb{P}^n_k$ must be a closed embedding. Indeed, since $X/k$ is proper, and $\mathbb{P}^n_k$ separated, we know from the Cancellation Lemma (see Vakil) that $X\to \mathbb{P}^n_k$ is proper, and in particular, closed. But, any closed embedding with closed image is a closed embedding.

Alex Youcis
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