1
$\begingroup$

Is every surjective etale morphism from a connected separated scheme to $A^n_{\mathbb{C}}$ of finite type? Is it finite? We use Stacks project's definitions.

EDIT: From Jason Starr's answer, we learn that such a morphism indeed has to be of finite type, and since etale morphisms are locally quasi-finite, we infer that the morphism has to be quasi-finite.

Is it true that every surjective etale morphism from a connected separated scheme to $A^n_{\mathbb{C}}$ such that the cardinality of the fiber over a closed point is independent of the choice of the closed point is finite? I think that for $n=1$, this question should be answered positively by considering local affine coordinates for etale morphism and applying the fact that a univariate complex polynomial has a non-simple root iff its derivative has a common root with it. I am not sure about $n>1$ though.

$\endgroup$
4
  • 1
    $\begingroup$ @R.vanDobbendeBruyn but Remy, for the second example you will not have separatedness right? $\endgroup$
    – geometer
    Dec 26, 2018 at 11:31
  • 3
    $\begingroup$ @R. van Dobben de Bruyn: $z\mapsto z^2$ is not étale. One can take $z\mapsto z^3-3z$ from $\mathbb{A}^1\smallsetminus \{\pm 1\} $ to $\mathbb{A}^1$. $\endgroup$
    – abx
    Dec 26, 2018 at 11:36
  • $\begingroup$ Ok, it seems my comment was left in a rush, and neither part was addressed accurately. Now removed. $\endgroup$ Dec 26, 2018 at 23:48
  • $\begingroup$ You changed the question after you accepted an answer. Anyway, your addendum question is addressed here. $\endgroup$ Dec 28, 2018 at 13:53

1 Answer 1

4
$\begingroup$

The question is "really" about quasi-compactness, which is usually assumed as a hypothesis in versions of Zariski's Main Theorem. However, the other strong hypotheses of the OP imply quasi-compactness in this case. The key point is that an open immersion is quasi-compact if the target is Noetherian.

Lemma. Let $i:X\to Z$ be a separated morphism between irreducible schemes. If there exists a covering of $X$ by open affines $U$ such that each restriction $i|_U$ is an open immersion, then $i$ is an open immersion. If $Z$ is Noetherian, then $X$ is quasi-compact.

Proof. Up to replacing $Z$ by the open image of $i$, assume that $i$ is surjective. The goal is to prove that $i$ is an isomorphism. We construct the inverse isomorphism $i^{-1}:Z\to X$ by gluing. Let $U$ and $V$ be nonempty open subschemes of $X$. The cocycle condition for $i^{-1}$ is precisely the condition that $i^{-1}(i(U)\cap i(V))$ equals $U\cap V$.

Let $Y^o$ be a nonempty open affine subset of the open intersection $i(U)\cap i(V)$. Denote by $X^o$ the inverse image $i^{-1}(Y^o)$. Since $X$ is irreducible, the intersections of nonempty open subsets $U\cap X^o$ and $V^\cap X^o$ are dense. Denote these by $U^o$ and $V^o$. By construction, each of the following restrictions of $i$ is an isomorphism, $$i_U:U^o\to Y^o, \ \ i_V:V^o\to Y^o.$$ These isomorphisms agree on $U^o\cap V^o = (U\cap V)\cap X^o$.

Since $i$ is separated and since $Y^o$ is affine, the scheme $X^o$ is separated. Define $j$ to be the automorphism of $X^o$ whose restriction to $U^o$ equals $i_V^{-1}\circ i_U$ and whose restriction to $V^o$ equals $i_U^{-1}\circ i_V$. These glue since $i_U$ and $i_V$ agree on $U^o\circ V^o$. Moreover, $j$ equals the identity on $U^o\circ V^o$. Since $j$ and the identity agree on the dense open $U^o\circ V^o$, and since $X^o$ is separated, the morphism $j$ equals the identity. Thus, $U^o$ equals $V^o$. Since we can cover $i(U)\cap i(V)$ by such open affines, it follows that $i^{-1}(i(U)\cap i(V))$ equals $U\cap V$.

Finally, if $Y$ is Noetherian, then every open subset of $Y$ is quasi-compact. Thus, the scheme $X$ is quasi-compact. QED

Let $f:X\to Y$ be a locally finite type, separated morphism with finite fibers that is quasi-finite Zariski locally on $X$, and that is strongly dominant in the sense that the $f$-inverse image of every dense open subset of $Y$ is a dense open subset of $X$. Assume also that $X$ is normal and that $Y$ is quasi-compact, separated, excellent, integral, and normal.

Proposition.(Variant of Grothendieck's "Zariski Main Theorem") There exists a factorization of $f$ as the composition of a dense open immersion into a normal scheme, $i:X\hookrightarrow Z$, followed by a finite, strongly dominant morphism, $g:Z\to Y$. Moreover, this factorization is unique up to unique isomorphism.

Proof. Every irreducible component of $X$ dominates $Y$, i.e., every generic point of $X$ maps to the generic point of $Y$. By hypothesis, there are only finitely many preimages of the generic point of $Y$, i.e., $X$ has only finitely many irreducible components. Since $X$ is normal, these irreducible components are connected components. Without loss of generality, assume that $X$ is connected, i.e., $X$ has a unique generic point $\eta$.

Since $Y$ is excellent, the "integral closure" of $Y$ in the "function field" $\kappa(\eta)$ is a finite, strongly dominant morphism whose domain is normal, $$g:Z\to Y.$$ By the universal property of the normalization, there exists a unique morphism of schemes compatible with the specified morphisms to $Y$, $$i:X\to Z.$$ By Zariski's Main Theorem, working locally on $X$ with opens that are quasi-compact over $Y$, the morphism $i$ is an locally an open immersion. By the lemma, the morphism $i$ is an open immersion.QED

Now you can apply this when $Y$ is affine space. I recommend that you read Grothendieck's formulation of Zariski's Main Theorem in EGA.

$\endgroup$
4
  • $\begingroup$ Dr. Starr, could you please confirm my reasoning? Etale already means that $f$ is locally of finite type. Since X is assumed to be separated, $f$ is separated. Since $f$ is smooth and the base is smooth, $X$ is smooth over a field (so normal). We further assume that $f$ has finite fibers. A connected scheme smooth over a field is irreducible, so every non-empty open in X is dense and, in particular, $Y$ is dominant. A finite strongly dominant morphism is surjective(??), and since we assume that $f$ is already surjective, it follows from Proposition that $f$ itself is finite strongly dominant. $\endgroup$
    – geometer
    Dec 26, 2018 at 12:52
  • $\begingroup$ if my reasoning is correct, I think that the hypothesis that $f$ has finite fibers (which, since $f$ is etale, is equivalent to quasicompactness) is not an obvious consequence of other hypotheses. $\endgroup$
    – geometer
    Dec 26, 2018 at 12:53
  • $\begingroup$ I also struggle to see how this answer is consistent with abx's comment. The cardinality of a fiber over a closed point for a surjective finite etale morphism between integral smooth schemes over $\mathbb{C}$ should not jump, right? $\endgroup$
    – geometer
    Dec 26, 2018 at 13:03
  • 1
    $\begingroup$ @geometer. I recommend that you think about these things for yourself and re-read my post. Every etale cover of a normal scheme is also normal. Hence, its connected components are irreducible. If X is irreducible, then there is a unique generic point, and that is all that is used in the post. $\endgroup$ Dec 26, 2018 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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