Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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... –  Piotr Achinger Dec 30 '12 at 21:34
    
Piotr, can this be said in a way that is obvious in complete generality as above? (Since it's not clear to me.) –  LMN Dec 30 '12 at 21:37
    
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? –  Piotr Achinger Dec 30 '12 at 21:47
1  
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. –  LMN Dec 30 '12 at 22:32
    
Does anybody have a different argument to show that projective varieties have finite surjective maps to projective spaces? –  LMN Dec 30 '12 at 23:00
show 2 more comments

2 Answers 2

up vote 4 down vote accepted

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 $\dim X=n$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.