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

Let $X \to Y$ be a smooth proper morphism. Let $y$ be a geometric point of $Y$. Is the kernel of the natural map of etale fundamental groups $\pi_1^{et}(X_y) \to \pi_1^{et} (X)$ abelian?

This is true for analytic fundamental groups, by the long exact sequence of homotopy groups.

It is not obviously true for etale fundamental groups of schemes over $\mathbb C$, since the profinite completion functor is not exact.

This article provides a version of the long exact sequence of etale homotopy, that specifically avoids saying this about $\pi_1$. Thus I suspect there is some kind of obvious counterexample to at least the most bold possible version of this question. If so, what is that counterexample? Is any special case of this statement true?

share|improve this question
    
Here is one attempt at a counterexample. First of all, there is a theorem of Fulton and Deligne (extending an earlier claim of Zariski), that for every plane curve with only ordinary double points, the "tame" fundamental group of the complement is Abelian. Now consider a pencil of degree $d$ plane curves whose general member is smooth, and consider the plane curve that is the union of all the singular members of this pencil. In characteristic 0, this will basically never be a curve with ordinary double points. However, in positive characteristic, it can be. (continued). –  Jason Starr Dec 5 '12 at 12:15
    
(continued.) Let $p>0$ be the characteristic of the algebraically closed ground field $k$. Let $d$ be $p^r$ for some positive integer $r$. Let $L_1,\dots, L_d, M$ be general linear polynomials on $\mathbb{P}^2$. Consider the pencil of degree $d$ homogeneous polynomials, $L_1\cdots L_d - t M^d$. The only singular members occur for $t=0$ and for $t=\infty$. Thus the union of all singular members is the plane curve $Z(L_1\cdots L_d\cdot M)$. Now blow up the base locus of this pencil, and remove the two singular fibers. This fibration seems to be a counterexample (I need to double-check). –  Jason Starr Dec 5 '12 at 12:24
    
@Will: Since Friedlander works with $\ell$-completions of the higher étale homotopy groups, do you also want to look only at the maximal prime-to-$p$ quotients of the étale fundamental group? If so, my counterexample below works. –  Jason Starr Dec 6 '12 at 13:17
    
I'm more interested in the full etale fundamental groups. But your example is pretty convincing that it won't be true regardless. –  Will Sawin Dec 6 '12 at 16:11
add comment

1 Answer

up vote 3 down vote accepted

Regarding the suggestion in my comment, there is a subtle point regarding my proposed pencil of curves; the Euler identity fails, so one has to check by hand that the singular members occur only for $t=0$ and $t=\infty$. In fact, this is not always the case. However, I checked this in one case.

Let $k$ be an algebraically closed field of characteristic $2$. Let $a,b\in k$ be elements such that none of $a,b,a+b$ equals $0$. Denote $\mathbb{P}^1_k$ by $\overline{Y}$, and let $[s,t]$ be homogeneous coordinates on $\mathbb{P}^1_k$. Let $[u,v,w]$ be homogeneous coordinates on $\mathbb{P}^2_k$. Let $\overline{X}$ be the Cartier divisor in $\mathbb{P}^1_k \times_k \mathbb{P}^2_k$ with defining equation $f(s,t;u,v,w) = suvw(u+v+w) - t(au+bv-(a+b)w)^4$. Consider the projection morphism $\overline{\pi}:\overline{X}\to \overline{Y}$. By direct computation, the only singular points of the morphism occur where $([s,t],[u,v,w])$ equals one of $$ ([1,0],[1,0,0]), ([1,0],[0,1,0]), ([1,0],[0,0,1]), $$ $$ ([1,0],[1,1,0]), ([1,0],[1,0,1]), ([1,0],[0,1,1]), ([0,1],[u,v,w]). $$ The tricky coordinates, of course, are $[u,v,w]=[1,1,1]$. However, the conditions on $a$ and $b$ guarantee this is contained in the member $[s,t]=[0,1]$ of the pencil. So this is a pencil to which my comments above apply. Define $Y\subset \overline{Y}$ to be the open complement of $\{[1,0],[0,1]\}$. Define $X\subset \overline{X}$ to be the inverse image of $Y$ under $\overline{\pi}$. Finally, define $\pi:X\to Y$ to be the restriction of $\overline{\pi}$. Then $\pi$ is a proper, smooth morphism.

To summarize the comments, by Deligne-Fulton the tame fundamental group of the open complement $U:= \mathbb{P}^2 \setminus Z(uvw(u+v+w)(au+bv-(a+b)w))$ is Abelian. This is also a dense open subset of $X$. Thus the induced map from the étale fundamental group of $U$ to the étale fundamental group of $X$ is surjective. In particular, the tame fundamental group of $X$ is also Abelian. The geometric generic fiber $X_{\overline{\eta}}$ of $\pi$ is a smooth, genus $3$ curve. Therefore its tame fundamental group is non-Abelian. So the kernel of the homomorphism of étale tame fundamental groups $\pi_1^t(X_{\overline{\eta}})\to \pi_1^t(X)$ is non-Abelian.

$\textbf{Edit}.$ I am a little worried now about the distinction between the tame fundamental group and the full étale fundamental group in this argument. Take any finite group $G$ of order prime to $p$. There is an injective homomorphism into a finite symmetric group $S_{p^r}$. Yet the maximal prime-to-$p$ quotient of $S_{p^r}$ is just $S_2$ (or trivial if $p$ equals $2$). So perhaps the kernel of the map of tame fundamental groups is much larger than the kernel of the map of full étale fundamental groups.

$\textbf{Second Edit}.$ I looked in the article of Friedlander that Will Sawin linked. Friedlander also needs to work with completions of the higher étale homotopy groups at primes different from the characteristic; in particular, this factors through the maximal prime-to-$p$ quotient. My example above also works if we look at the maximal prime-to-$p$ quotients of the étale fundamental group, or if we look at an $\ell$-completion for $\ell$ a prime different from $p$, since both of these factor through the tame fundamental group. So I feel that this example does explain why Friedlander did not state his theorem for $\pi_1$ as well as for $\pi_n, n\geq 2$.

share|improve this answer
add comment

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.