Let $k$ be an arbitrary field, we work with schemes $X$ of finite type over $k$. Does every irreducible projective scheme have a finite surjective morphism to a projective space $\mathbb{P}^n_k$?. What if I just assume that $X$ is equidimensional. Does the same argument work?

We know that a proper $k$-scheme with this property must be also be projective by formal properties of ample line bundles.

I would do this by projecting from sufficiently general points, and this probably works (maybe not in complete generality) but I can't help but think there is a cleaner argument (that also doesn't require possibly assuming that the field is infinite, algebraically closed or of characteristic $0$).

Feel free to assume our schemes have basic niceness properties.

  • $\begingroup$ Yes, this is true, and I think general linear projection is the usual argument. It works over algebraically closed fields. For CM varieties, the morphism is flat and you can obtain a nice proof of Serre duality using duality for finite flat morphisms... $\endgroup$ Dec 30, 2012 at 21:34
  • $\begingroup$ Piotr, can this be said in a way that is obvious in complete generality as above? (Since it's not clear to me.) $\endgroup$
    – LMN
    Dec 30, 2012 at 21:37
  • $\begingroup$ Embed $X$ in $\mathbb{P}^N$. If $N>\dim X$, find a point $p$ not on $X$ and project from that point. You will get a finite morphism from $X$ onto $X'\subseteq \mathbb{P}^{N-1}$, and so on. Am I missing something? $\endgroup$ Dec 30, 2012 at 21:47
  • 1
    $\begingroup$ Piotr, I think you're right. Projection from any point $P$ not on $X$ defines a morphism. (If there is no such $k$-valued point then I think we're still ok, ...). To check this is a finite morphism we can base change to $\bar{k}$. Now, WLOG $X = X_{red}$, and there are only finitely many points of $X$ on a line through $P$ (since $X$ is Zariski closed, and infinitely many pts on this line would imply $P \in X$). A quasi-finite proper morphism is finite. $\endgroup$
    – LMN
    Dec 30, 2012 at 22:32
  • $\begingroup$ Does anybody have a different argument to show that projective varieties have finite surjective maps to projective spaces? $\endgroup$
    – LMN
    Dec 30, 2012 at 23:00

2 Answers 2


Let $X$ be any projective scheme of dimension $n$ over an arbitrary field $k$. Then there exists a finite surjective morphism from $X$ to $\mathbb P^n_k$.

Embed $X$ in some $P=\mathbb P^N_k$. By the homogeneous prime avoidance lemma, there exists a hypersurface $H_0$ of $P$ which doesn't contain any generic point of $X$. Then $\dim (X\cap H_0)=n-1$. Similarly there exists a hypersurface $H_1$ of $P$ which doesn't contain any generic point $X\cap H_0$. We have $\dim (X\cap H_0\cap H_1)=n-2$. Repeating the argument we find $n+1$ hypersurfaces $H_0, \dots, H_n$ such that $$X\cap H_0\cap \dots \cap H_n=\emptyset.$$ Each $H_i$ is defined by a homogeneous polynomial $F_i$. Remplacing $F_i$ with some positive power (this doesn't change the property of the intersection being empty), we can suppose they all have the same degree $d$. These $n+1$ sections of $O_P(d)$ don't have commun zeros in $X$, so they define a morphism $f: X\to \mathbb P^n_k$.

It remains to show $f$ is finite. I just repeat the argument from Lemma 3 in Kedlaya: More étale covers of affine spaces in positive characteristics, J. Alg. Geometry (2005): let $z\in \mathbb P^n_k$. There exists $i\le n$ such that $z\in P\setminus H_i$. So the fiber $X_z$ is projective over $k(z)$ and also affine because it is closed in the affine scheme $P\setminus H_i$. This implies that $X_z$ is finite. So $f$ is quasi-finite and projective, so it is finite. It is surjective because $\dim X=n$.


I assembled the comments above into an answer:

Assume that the field of definition is infinite (or allow finite extensions of the base field in the construction). Let $X$ be a projective scheme over a field $k$ and let $n$ be the dimension of $X$. Let $N$ be smallest such that there exists a finite morphism $X\to \mathbb{P}^N$. I claim that $N=n$. Suppose otherwise; let $f:X\to \mathbb{P}^N$ be finite. Since $N>n$, we can find a point $p\in\mathbb{P}^n$ not lying on $X':=f(X)$ (possibly after passing to a finite extension of $k$). Then $p$ defines a projection $\pi: \mathbb{P}^{n}\setminus\{p\} \to \mathbb{P}^{n-1}$ (can assume that $p=(0:\ldots:0:1)$, then $\pi(x_0:\ldots:x_{N-1}:x_N) = (x_0:\ldots:x_{N-1})$). This $\pi$ restricted to $X'$ is quasi-finite (otherwise, $X'$ would contain a line through $p$, hence $p$) and proper, hence finite. The composition $\pi\circ f: X\to \mathbb{P}^{N-1}$ is then finite as well - a contradiction.


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