Do morphisms locally decompose into finite surjective followed by smooth? (update: Is every projective variety over a finite field a finite cover of $\mathbb{P}^d$ for some $d$?)

Question

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!

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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 $V$).

Answer 1

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.

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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.

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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.