Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose that $X$ and $Y$ are smooth complex algebraic varieties, and that $f:X\rightarrow Y$ is an etale morphism in the sense that $d_xf:T_xX\rightarrow T_{f(x)}Y$ is an isomorphism for all $x\in X$. Must $f$ be (at least locally) an open immersion? Certainly, difficulties can arise if the differential fails to be an isomorphism at a single point. Consider the map $\mathbb{A}^1\rightarrow\mathbb{A}^1$, $z\mapsto z^2$.

share|improve this question
3  
You can only say $f$ is a local biholomorphism around $x$. It is not, in general, a local isomorphism of algebraic varieties around $x$: just take a covering of curves of different genus minus the ramification points. Or, even simplier, take your example $\mathbb{A}^1\setminus \{ 0 \} \to \mathbb{A}^1\setminus \{ 0 \}$ $z\mapsto z^2$. The inverse function theorem however holds with respect to the étale topology for étale morphisms. –  Qfwfq May 25 '13 at 23:17
    
What about something of the form: "every smooth map is locally split over an étale cover"? I vaguely remember reading this somewhere, can't be sure it's true. –  David Roberts May 25 '13 at 23:43
4  
If $X=Y=\mathbb{A}^n$, your question is the famous Jacobian conjecture. –  Jérémy Blanc May 26 '13 at 6:52
2  
In some trivial but profound sense every etale map is locally an isomorphism. Let "locally" mean "locally in the etale topology". –  Tom Goodwillie May 26 '13 at 20:55

1 Answer 1

up vote 1 down vote accepted

Maybe this "answer" is too far away from schemes to be what is desired. But here it goes anyway.

I believe in general that for complex affine varieties $X,Y$, that a morphism $f:X\to Y$ is étale iff it is a local analytic isomorphism in the analytic topology. When $X,Y$ are smooth, it is enough to just check that the tangent spaces are isomorphic at every point.

For example, if a finite group $\Gamma$ acts on a connected and normal variety $X$ algebraically, then the mapping $X\to X//\Gamma$ is étale if and only if $\Gamma$ acts freely (i.e. stabilizers are trivial).

One can find projections like that who are not homeomorphic to their image; a necessary requirement for open immersions. For example, consider the étale map $\mathrm{SL}_2(\mathbb{C})\to \mathrm{SL}_2(\mathbb{C})/\mathbb{Z}_2\cong \mathrm{SO}(3,\mathbb{C})$.

Conversely however, open immersions are always étale.

EDIT: I added some detail to the general statement about finite quotients, and replaced my original example since it was not correct. In particular, the tangent map to $\mathbb{C}^* \to \mathbb{C}^*//\mathbb{Z}_2 \cong \mathbb{C}$ (where $\mathbb{Z}_2$ acts by $z\mapsto 1/z$) is not an isomorphism at $\pm 1$, and so it is not étale at those points (it is at all other points though).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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