When does a projective morphism give an etale morphism (into affine space)? (Finite field) (normalization) Does anyone know of generalizations on what Mumford (Red Book) calls "uniformizing parameters"?  For example, given a regular quasi-projective scheme over a finite field $\mathbb{F}$, is there an etale morphism into affine space over $\mathbb{F}$?. 
 A: [EDIT: I re-arrange the answer to make the statements clearer]. 
The answer to the question is no, for any base field $k$. 
First, we can characterize smooth affine varieties $X$ which are étale over an affine space $A=\mathbb A^n_k$.


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* A smooth affine variety $X$ over $k$ is étale over an affine space $A=\mathbb A^n_k$ if and only if the sheaf of differentials $\Omega_{X/k}$ is free of rank $n$, generated by $n$ closed differential forms $df_1, \ldots, df_n$.
Proof: (1) If there exists an étale morphism $f: X\to A={\rm Spec}k[t_1,\ldots, t_n]$, then we have an isomorphism $\Omega_{A/k}\otimes_{O_A} O_X=f^{\star}\Omega_{A/k}\simeq \Omega_{X/k}$. Then take $f_i$ equal to the image of $t_i$ in $O(X)$. 
(2) If $df_1,\ldots, df_n$ as above exist, we define a morphism $f: X\to A$ to be associated to the moprhism of $k$-algebras $k[t_1,\ldots, t_n]\to O(X)$, $t_i\mapsto f_i$. Then by hypothesis on the $df_i$'s, the canonical morphism $f^{\star}\Omega_{A/k}\to \Omega_{X/k}$ is an isomorphism. Therefore $f$ is étale.  
Remark: it is not enough to suppose that $\Omega_{X/k}$ is free to insure that $X$ is étale over $A$. For example, if $X$ is an elliptic curve $E$ minus the origin, then $\Omega_{X/k}$ is free (because $\Omega_{E/k}$ is !), but at least in characteristic $0$, $X$ is not étale over $\mathbb A^1_k$ (if so, it would be finite and étale over $\mathbb A^1$ by considering the extension $E\to \mathbb P^1_k$, but $\mathbb A^1_k$ is simply connected in characteristic $0$). I don't known whether this can happen in positive characteristic. Note that this is already an example (in characteristic 0) of a smooth affine curve which is not étale over $\mathbb A^1$. 
Now we construct in any characteristic a smooth affine curve $X$ such that $\Omega_{X/k}$ is not free. By the above, $X$ will not be étale over ${\mathbb A^1}$. Fix a projective smooth connected curve $C$ over $k$ of genus $g>1$. Suppose that for any affine open subset $X$ of $C$, $\Omega_{X/k}$ is free. We want to find a contradiction. 


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*We first reduce to the case $k$ is algebraically closed (for simplicity, this is actually not necessary). I claim that over the algebraic closure $\bar{k}$ of $k$, $\Omega_{X'/\bar{k}}$ is free for any affine open subset $X'$ of $C_{\bar{k}}$. Indeed, $X'$ is defined over a finite extension $K/k$, and the projection $C_{K}\to C$ induces an étale morphism from $X'$ to an affine open subset $X$ of $C$, thus $\Omega_{X'/K}\simeq \Omega_{X/k}\otimes O_{X'}$ is free. 

*Now we suppose that $k$ is algebraically closed. For any closed point $x\in C$, $\Omega_{X/k}$, where $X=C \setminus {x}$, is free. So $\Omega_{C/k}|_X$ is trivial. This implies that the canonical divisor $K_C$ on $C$ is trivial on $X$, hence linearly equivalent to $(2g-2)[x]$. Fix a point $x_0\in C(k)$. Then the immersion $C\to {\rm Jac}(C)$, $x\mapsto [x-x_0]$, maps $C$ into the $(2g-2)$-torsion part of ${\rm Jac}(C)(k)$ which is finite. Contradicton. 
Final remark: if you have a separated scheme $X$ which is étale (or more generally quasi-finite) over any affine scheme $A$, then it has to be quasi-affine. This is because by Zariski's Main Theorem, $X$ is an open subscheme of a scheme $\overline{X}$ which is finite over $A$, hence $\overline{X}$ is affine. 
A: This paper by Kedlaya might be what you want, since it contains some rearrangement of the words you used, but I can't really tell from the question.  If you want a proper F-scheme to have an etale map to affine space, it has to be a disjoint union of finite F-schemes, and the affine space has to be zero dimensional. 
