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In commutative algebra the classic Noether normalization lemma says that every ring finitely generated over a field is a finitely generated module over a polynomial ring with coefficients in this field. The geometric interpretation of this statement is that if $X$ is an affine variety of dimension $n$ then there is a surjective finite map from $X$ to the affine $n$-space $\mathbb{A}^n$.

What about projective varieties? Does an analogous statement hold? That is, if $X$ is a closed subset of $\mathbb{P}^n$ of dimension $m$, is there necessarily a finite surjection $X \to \mathbb{P}^m$?

user115940
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    Yes! Geometrically, take a general linear subspace $L$ in $\mathbb P^n$ of codimension $m+1$ disjoint from $X$ and $\Lambda =\mathbb P^m$ disjoint from $L$. Projection away from $L$ onto $\Lambda$ gives the finite map you want. –  Feb 17 '14 at 19:15
  • Is there an elegant proof that this map is finite? – user115940 Feb 17 '14 at 20:16
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    Dear user115940, Asal's morphism is proper (since $X$ is projective) and has finite fibers (obvious geometrically: a line not included in $X$ has finite intersection with $X$), so it is finite. An algebraic proof is given in Shafarevich's Basic Algebraic variety, Volume 1, page 64. – Georges Elencwajg Feb 17 '14 at 20:51
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    Dear @Asal: your idea is too nice to remain a comment. I strongly encourage you to promote it to an answer. – Georges Elencwajg Feb 17 '14 at 20:55
  • Dear @Georges, done! –  Feb 18 '14 at 09:11

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(Making my comment into an answer...)

Yes! Geometrically, take a general linear subspace $L$ in $\mathbb P^n$ of codimension $m+1$ disjoint from $X$, and $\Lambda \cong \mathbb P^m$ disjoint from $L$. Projection away from $L$ onto $\Lambda$ gives the finite map you want.

As Georges explained in the comments, a proper morphism with finite fibres is finite, as proved in Shafarevich, Basic Algebraic Geometry Vol. 1. This projection is proper because $X$ is projective, and has finite fibres because $X$ intersects any linear subspace of codimension $m$ containing $L$ in finitely many points. So this morphism is indeed finite.

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    This argument works perfectly over an infinite field. Over finite fields, this is proved in a paper of Kedlaya in J. Alg. Geometry. – Cantlog Feb 18 '14 at 09:32
  • Dear @Cantlog, thanks for that interesting comment. –  Feb 18 '14 at 09:36
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    Thanks for the answer. But I am not sure I can make your first argument precise, i.e why is there a linear subspace of the right codimension disjoint from $X$? – user115940 Feb 18 '14 at 09:40
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    Dear @user115940, good question. One way to prove it is by induction on the codimension of $X$: it is certainly true when $X$ has codimension 1, and if we know it for codimension $k$, then we can prove it for codimension $k+1$ simply by projecting from a point not in $X$. –  Feb 18 '14 at 10:43
  • I just wanted to add a(nother) reference for the projection being a finite map: This is Mumford's Red Book, Chapter II, §7, Proposition 6. – windsheaf Jul 27 '17 at 10:30