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I have looked through all my standard algebraic geometry texts and tried many tricks using Zariski's main theorem and Noether normalization, but remain stuck by the following:

Let $\pi:X\to S$ be a morphism of finite type between integral, Noetherian schemes and let $x$ be a point of $X$. Does there exist an open neighbourhood of $X$ which admits a finite, surjective morphism onto a smooth $S$-scheme?

In this generality I think that the answer is 'no', though I do not have a counterexample. What if we impose additional assumptions such as $\pi$ being flat or proper (or even projective)?

A related question, an affirmative answer to which would imply the same for the previous question in the projective case, is the following: if $X$ is a projective scheme over a local ring $A$, then does $X$ admit a finite surjection to $\mathbb{P}_A^d$ for some $d\ge 1$?

(I am imagining that $A=\mathbb{Z}_p$, so please do not assume that the residue field of $A$ is infinite!)

Thank you!


Update

With Brian's help (thank you), the interesting remaining problem is the following: does every projective variety $V$ over a finite field $k$ admit a finite surjective morphism to $\mathbb{P}_k^d$ for some $d$? I have a gnawing suspicion that the answer is 'no'.

It is useful to remember Noether normalisation in this case: if $I$ is a non-zero ideal of $k[X_1,\dots,X_n]$, then one can find a finite morphism $k[Y_1,\dots,Y_{n-1}]\to k[X_1,\dots,X_n]/I$ by sending $Y_i$ to $X_i-X_n^e$ for some big enough $e\ge 1$. Unfortunately, projectivising this construction produces a morphism $\mathbb{P}_k^n\setminus C\to\mathbb{P}_k^{n-1}$ where $C$ is quite a large closed subscheme of $\mathbb{P}_k^n$ (unless I have made a mistake); so if $V$ is our variety inside $\mathbb{P}_k^n$, then it is difficult to ensure that $V$ doesn't meet $C$. Therefore we can't successively project down to smaller dimensional spaces. (In contrast with the case when $k$ is infinite, for then we use changes of variables looking like $Y_i=X_i-\alpha_iX_n$ and the resulting morphism between projective spaces is defined everywhere except for one point, which we can assume doesn't lie on $V$).

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For first question must avoid $\pi$ a closed immersion. But false even for $\pi$ flat . Let $K$ be non-Galois cubic number field and $p$ rat'l prime over which there is ramified $P$ and another prime $Q$. (Trivial to make $K$, $p$.) Let $S={\rm{Spec}}(\mathbf{Z}_ {(p)})$ and $X = {\rm{Spec}}(O_ {K,P})$. The "problem" is gap between quasi-finite and "locally finite" for Zariski top. No such gap in analytic theory, and likewise if use etale top.then for flat $\pi$ it is affirmative (target an affine space), via Noether normalization on a fiber and Zariski's Main Theorem (in EGA form). –  BCnrd May 5 '10 at 14:03
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For the 2nd question, again assume $\pi$ flat to avoid dumb counterexamples. Then by fiber-dimension reasons, any map $X \rightarrow \mathbf{P}^d_ A$ over $A$ that is finite on special fiber is finite surjective. As you expect, by fixing projective embedding for $X$, using linear projections away from some suitable linear space disjoint from $X$ (on special fiber, thus over $A$ by properness) thereby does the job when residue field is infinite. I don't remember the specific formulas (in Red Book?) used to handle finite field case, but is there a problem lifting those formulas? –  BCnrd May 5 '10 at 14:36
    
The interesting question of projective Noether normalization over finite fields has Bjorn "Bertini over finite fields" Poonen's name written all over it. I eagerly await his clever solution. –  BCnrd May 5 '10 at 18:47
    
@BCnrd Thank you very much for all your comments on my question. I sure hope that the answer to the finite field question is yes, because otherwise my original question remains open under what you have shown to be the only reasonable hypotheses, namely $X\to S$ being flat and projective. Poonen's paper certainly looks relevant, so I will wait hear what he says. –  Matthew Morrow May 5 '10 at 19:52
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Over finite fields, see Kedlaya's paper arxiv.org/abs/math/0303382. –  Qing Liu May 5 '10 at 22:32

1 Answer 1

up vote 5 down vote accepted

I claim that every projective scheme $V$, over a finite field $k$, all of whose components have dimension $\leq d$, admits a finite morphism to $\mathbb{P}^d_k$. Let $q=|k|$.

Key Lemma: Let $V_1$, $V_2$, ... $V_s$ be a finite collection of subvarieties of $\mathbb{P}_k^N$. Then there is a homogenous polynomial $f$ in $k[x_0, x_1, \ldots, x_N]$ such that $f|_{V_i}$ is nonzero for every $i$.

Proof: (improved thanks to comments below)

Choose a closed point $v_i$ on each $V_i$; we will force $f$ not to vanish at any of the $v_i$. Choose $M$ large enough that every $V_i$ is in $\mathbb{P}^N(\mathbb{F}_{q^M})$. Let $a_0$, $a_1$, ... $a_N$ be a basis for $\mathbb{F}_{q^{(N+1)M}}$ over $\mathbb{F}_{q^M}$. Then the linear form $$a_0 x_0 + a_1 x_1 + \cdots a_N x_N$$ does not vanish on any $Vvi$. This, of course, does not have coefficients in $k$. But the product $$\prod_{i=0}^{(N+1)M-1} \left( a_0^{q^i} x_0 + a_1^{q^i} x_1 + \cdots a_N^{q^i} x_N \right)$$ is similarly nonzero, and does have coefficients in $k$. This concludes the proof of the key lemma.


Now, let $V$ be the projective variety, for which we want to prove the result. By the key lemma, there is some polynomial $f_0$, of degree $r_0$, which does not vanish on $V$. Every irreducible component of $V \cap \{ f_0 =0 \}$ is of dimension $\leq d-1$. Apply the lemma again to these irreducible components to find a polynomial $f_1$, of degree $r_1$, so that every component of $V \cap \{ f_0 = f_1 = 0 \}$ has dimension $\leq d-2$. Continuing in this manner, we construct polynomials $f_0$, $f_1$, ..., $f_d$ such that $V \cap \{ f_0 = f_1 = \cdots = f_d = 0 \}$ is empty.

Set $R=\prod r_i$. Our map $V \to \mathbb{P}^d_k$ will be given by $$(f_0^{R/r_0} : f_1^{R/r_1} : \cdots : f_d^{R/r_d})$$. By our construction, this map has no base points, and is thus well defined.


By the "standard argument", this map is finite. I now recall the standard argument.

Since $V$ is proper, we just need to check that the map has finite fibers. Suppose, for the sake of contradiction, that the the fiber above $$(a_0: a_1 : \cdots a_{i-1} : 1 : a_{i+1} : \cdots : a_d)$$ has positive dimension. Let $C$ be a curve in this fiber, so $f_j^{R/r_j} - a_j f_i^{R/r_i}$ vanishes on $C$ for all $j \neq i$. The curve $C$ must meet the hypersurface $f_i=0$, say at $z$. Then all the $f_i$ vanish at $z$, contradicting our construction.

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@David Speyer. Thanks very much! That is a nice proof. Taking into account BCnrd's comments, I think we reach the following satisfactory variant of my original question: if $\pi: X\to S$ is a flat, projective morphism, then each point of the base has an open neighbourhood $U$ for which there exists a finite morphism of $S$-schemes $\pi^{-1}(U)\to\mathbb{P}_U^d$. I am very pleased with how my first MO question has turned out :) –  Matthew Morrow May 6 '10 at 10:26
    
If $V$ is the special fibre of a nice scheme over a local ring then it will be equidimensional, so I am happy with that assumption, and I've checked that it lifts up to the local ring. Finally, surely the $d=0$ case of the key lemma implies the $d>0$ case? Just pick a closed point on each $V_i$? –  Matthew Morrow May 6 '10 at 11:27
    
Thanks for the comments; I have edited accordingly. –  David Speyer May 6 '10 at 12:54
    
I think I now have eliminated every reference to pure dimensionality. –  David Speyer May 6 '10 at 14:43

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