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I'm wondering if finite unramified morphism between reduced schemes decomposes as closed immersions and etale morphisms. Suppose I have a morphism between reduced schemes which is finite, surjective and unramified, is it necessarily etale? I think this is certainly true if both source and target are curves, but I'm not sure about higher dimensional examples. Thanks

EDIT: to avoid trivial example let's assume the source and target are connected. What I'm wondering is precisely when one can deduce flatness from these conditions.

edited Nov 16 2010 at 3:57

asked Nov 16 2010 at 2:31

Yuhao Huang
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Dear Yuhao, See the edit to my answer, as well as Brian's attendant comments, for an answer to your edited-in question. Regards, Matt – Emerton Nov 16 2010 at 5:10

Emerton · Accepted Answer · 2010-11-16 04:33:20Z UTC

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Finite, surjective, and unramified does not imply etale. E.g. suppose that $Y$ is a proper closed subscheme of $X$, and we consider the map $X \coprod Y \to X$ defined as the disjoint union of the identity on $X$, and the given closed immersion $Y \to X$ on $Y$.

Then this map is finite, unramified, and surjective, but not etale. (See Sandor's answer for the missing condition, which is flatness!)

Added: A more interesting example is given by letting $X$ be a nodal cubic, letting $\tilde{X}$ be the normalization, and considering the natural map $\tilde{X} \to X.$ This map is not flat and certainly not etale, but it is unramified. (Formally, each branch through the node maps by a closed immersion into the nodal curve.)

edited Nov 16 2010 at 4:33

answered Nov 16 2010 at 3:00

Emerton
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See my edit: what if I add the condition on connectedness? I'm wondering if there is anything can imply the flat locus is both open and closed. Thank you. – Yuhao Huang Nov 16 2010 at 4:02

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I have been wondering about this for a bit of time (with reduced connected noetherian schemes), but am not yet satisfied. In the connected normal (implies irreducible) noetherian case flatness is easy to verify by passing to strict henselizations (which are domains, by normality) -- this argument already appears early in the Freitag-Kiehl book, for example -- and my feeling is that without normality there should be counterexamples, even with irreducibility. No success yet in finding one. Need some interesting non-normal singularities. – BCnrd Nov 16 2010 at 4:17

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Oh drat, I was being an idiot; somehow I fooled myself into thinking the normalization of the nodal cubic is not unramified. But it is, as Emerton properly explains. OK, so that's the end of this one: normality is the "right" hypothesis for flatness to be automatic. – BCnrd Nov 16 2010 at 4:46

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@BCnrd: a slightly more general hypothesis is "geometrically unibranch". See EGA IV-4, (18.10.1). – Laurent Moret-Bailly Nov 17 2010 at 16:32

Sándor Kovács · Answer 2 · 2010-11-16 02:59:00Z UTC

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You also need flat. See Hartshorne, Ex. III.10.3.

answered Nov 16 2010 at 2:59

Sándor Kovács
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Dear Harry, But since smooth and finite implies etale, this undercuts the need for assuming unramified at all. The point of Sandor's answer is that the conditions of flatness and being unramified are orthogonal to one another; neither impies or precludes the other --- they are just different conditions. And when you have both, you have an etale morphism. – Emerton Nov 16 2010 at 3:11

Matt, thanks. I was going to say pretty much the same, but you said it better. – Sándor Kovács Nov 16 2010 at 3:19

What I'm wondering is precisely whether these conditions somehow implies flatness, because the only examples of finite unramified morphisms I've seen are closed immersions and finite etale maps. It will be cool if you can show me some non-trivial example. – Yuhao Huang Nov 16 2010 at 4:00

What are unramified morphisms like?

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