# Finiteness of surjective etale morphisms

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.

• @R.vanDobbendeBruyn but Remy, for the second example you will not have separatedness right? – geometer Dec 26 '18 at 11:31
• @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 $\mathbb{A}^1$. – abx Dec 26 '18 at 11:36
• Ok, it seems my comment was left in a rush, and neither part was addressed accurately. Now removed. – R. van Dobben de Bruyn Dec 26 '18 at 23:48
• You changed the question after you accepted an answer. Anyway, your addendum question is addressed here. – R. van Dobben de Bruyn Dec 28 '18 at 13:53

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.

• 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. – geometer Dec 26 '18 at 12:52
• 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. – geometer Dec 26 '18 at 12:53
• 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? – geometer Dec 26 '18 at 13:03
• @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. – Jason Starr Dec 26 '18 at 13:06