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
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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 $V$).



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 $d$?)
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.


