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.

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    $\begingroup$ @R.vanDobbendeBruyn but Remy, for the second example you will not have separatedness right? $\endgroup$
    – geometer
    Commented Dec 26, 2018 at 11:31
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    $\begingroup$ @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
    Commented 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. $\endgroup$ Commented Dec 26, 2018 at 23:48
  • $\begingroup$ You changed the question after you accepted an answer. Anyway, your addendum question is addressed here. $\endgroup$ Commented Dec 28, 2018 at 13:53

1 Answer 1


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.

  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented Dec 26, 2018 at 13:03
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    $\begingroup$ @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. $\endgroup$ Commented Dec 26, 2018 at 13:06

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