# What are unramified morphisms like?

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.

• 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 Nov 16, 2010 at 5:10

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.)

• 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. Nov 16, 2010 at 4:02
• 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. Nov 16, 2010 at 4:17
• 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. Nov 16, 2010 at 4:46
• @BCnrd: a slightly more general hypothesis is "geometrically unibranch". See EGA IV-4, (18.10.1). Nov 17, 2010 at 16:32
• We have the normalization map $\mathbb{A}^1 \to C$ by $a\mapsto (a^2, a^3)$ of the cuspidal cubic as well as a map $C \to \mathbb{A}^1$ given by projection onto either factor. If I'm not mistaken, these give us a finite surjective map that is unramified (resp. smooth) but not smooth (resp. unramified). And the singularity is also geometrically unibranch. Oct 1, 2015 at 2:18

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

• 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. Nov 16, 2010 at 3:11
• Matt, thanks. I was going to say pretty much the same, but you said it better. 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. Nov 16, 2010 at 4:00