Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a somewhat strange question perhaps, but I was wondering the following:

Suppose $X$ and $Y$ are schemes (lets say Noetherian, and integral as well if you'd like). Suppose that there are maps $f: Y \rightarrow X, g: X \rightarrow Y$ which realize Y as a finite etale cover of X, and vice versa. Does it follow that X and Y have to be isomorphic?

I would be happy with any result along these lines. That is, if it's not always true, are there nice situations where it is? Knowing about the case where $X, Y$ are assumed affine would be just fine too.

share|improve this question
    
Not every etale cover with a section is trivial. –  Martin Brandenburg Feb 14 '13 at 16:21

2 Answers 2

up vote 5 down vote accepted

No. For instance, there are non-isomorphic elliptic curves that satisfy this relationship (which is called "isogeny" in that case). I'm not sure of any non-drastic simplifying assumptions that make this true.

share|improve this answer
    
Thanks for clearing that up. I suppose then that this is still false even if $X$ and $Y$ are anabelian (say hyperbolic curves over a number field)? I guess that a morphism between etale fundamental groups with a section won't necessarily be an isomorphism... –  KristianJS Feb 14 '13 at 18:26

This is just a comment, but I don't have enough reputation yet.

It's true for compact hyperbolic curves (=smooth projective curves of genus at least two) for trivial reasons. In fact, if $X\to Y$ is a finite etale morphism of compact hyperbolic curves of degree $d$ and $Y\to X$ is a finite etale morphism of degree $e$ we obtain that $$2g_X -2 = (2g_Y-2)d = (2g_X-2)dd^\prime.$$ This implies that $dd^\prime =1$ and thus $d=d^\prime = 1$.

More generally, the above approach shows the following. Suppose that $f:X\to Y$ is finite etale and $g:Y\to X$ is finite etale, where $X$ and $Y$ are smooth projective varieties. Then, the "Riemann-Hurwitz formula" implies that $$K_X = f^\ast K_Y = f^\ast g^\ast K_X = (gf)^\ast K_X.$$ Maybe this implies that $K_X$ is torsion, but I'm not sure. Of course, I'm assuming $\deg g, \deg f>1$.

It might be more natural to just consider the isomorphism $(gf)^\ast \Omega^1_X \to \Omega^1_X$. You can compute the Euler characteristic of $X$ and $Y$ by taking the $n$-th Chern class ($n=\dim X=\dim Y$), and then you see that the Euler characteristic of $X$ (and $Y$) is zero (if the above surjective map is an isomorphism).

share|improve this answer
    
This looks very promising. I will have to study this comment more closely. Quick question: so there is no assumption on the groundfield here at all? –  KristianJS Feb 14 '13 at 19:14
    
It is true that $f^*\Omega_Y\to \Omega_X$ is an isomorphism when $f$ is étale (this is iff). –  user18119 Feb 14 '13 at 21:43
    
@QiL'8 Thanks. I wasn't completely sure. I'll edit the question accordingly. –  Ari Feb 15 '13 at 10:38
    
Okay, this looks nice, but I have some concern about assumptions on the groundfield. Doesn't the application of Hurwitz assume the groundfield is algebraically closed? Is it valid more generally, or is it true that if the degree is 1 after basechanging to the algebraic closure, then it must have been 1 over the original field as well? –  KristianJS Feb 15 '13 at 12:09
    
"Hurwitz" is valid over any groundfield, if you use the right definitions of degree. To answer your question, if $X\to Y$ is of degree $d$, then the base change $X_{\overline k} \to Y_{\overline k}$ is also of degree $d$. The degree being here the generic degree. (You might have been thinking about the degree of a divisor $D$ on a curve $X$. This also doesn't change when you go to $X_{\overline k}$, but in the definition of the degree $\deg_X D$ you take into account the residue degrees of the points. That's something you don't need to do over an algebraically closed field of course.) –  Ari Feb 17 '13 at 13:31

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.