5
$\begingroup$

I am encountering the following situation which is similar to the Abhyankar's higher dimensional conjecture on étale fundamental groups, but with much stronger assumptions:

Let $S$ be a finitely generated subring of $\mathbb{C}$, let $X$ be a smooth affine variety over $S$, and let $G$ be a finite group such that the following holds. For any large enough prime $p$ and a base change $S\to k$ to an algebraically closed field of characteristic $p$, the variety $X_{k}$ admits a Galois covering with the Galois group $G$. Does this imply that $X_{\mathbb{C}}$ also admits a Galois covering with the group $G?$

I believe the answer is "yes" if $X$ is a curve, or is a complement of divisors with normal crossings in a projective space (by a result of Abhyankar).

Any suggestions or references would be greatly appreciated.

$\endgroup$
7
  • $\begingroup$ When you say "variety" over $S$, do you mean "scheme"? Also, is your Galois covering (finite) etale? If yes, for $d\geq 1$, let $M_d$ be the stack of finite etale covers of $X$ over $A$. This is a finite type stack over $A$. Like "everything" over $A$, it is generically flat over the function field $K(A)$ of $A$. Therefore, by spreading out (see Rydh's appendix in arxiv.org/abs/0904.0227), replacing Spec $A$ by a dense open if necessary, we may and do assume that $M_d$ is flat over $A$.... $\endgroup$ Commented Jun 25, 2017 at 20:17
  • $\begingroup$ ....In particular, any $k$-point of characteristic $p>0$ (with $k$ the residue field of a maximal ideal of $A$) lifts to characteristic zero (and thus to a $\mathbb C$-point). So, if you have a finite etale cover of $X_k$, it will "come" from a finite etale cover of $X$ (via specialization). $\endgroup$ Commented Jun 25, 2017 at 20:18
  • $\begingroup$ Yes, I do assume that Galois coverings are finite etale coverings, and variety means integral scheme of finite type. $\endgroup$
    – John Z.
    Commented Jun 25, 2017 at 20:43
  • $\begingroup$ @AriyanJavanpeykar Why is the stack of finite type? Here $X$ is affine, and if we are in characteristic $p$, we probably do not even have such a stack, and even if we did, the moduli problem would be unbounded... $\endgroup$ Commented Jun 25, 2017 at 20:44
  • 1
    $\begingroup$ @AriyanJavanpeykar The situation in characteristic $p$ is quite perplexing: any $f\in A$ gives rise to the Artin-Schreier $\mathbb{Z}/p$-covering $t^p - t = f$ of $X={\rm Spec} A$, and $f$ and $g$ yield the same covering iff $f-g = h^p - h$ for some $h\in A$. In particular, unlike in char. 0, one can easily have "families" of coverings, e.g. $t^p-t=xy$, as a family of coverings of $\mathbb{A}^1$ parametrized by $\mathbb{A}^1$. On the other hand, by the unique lifting property of etale maps, every such family is formally constant. This seems to mean that there is no reasonable moduli problem. $\endgroup$ Commented Jun 25, 2017 at 21:02

1 Answer 1

7
$\begingroup$

Here's a sketch of a possible way: Since $p$ is large enough, we can assume that $G$ is of order prime to $p$. If $X$ was projective, then the prime-to-$p$ completions of $\pi_1(X_{\mathbb{C}})$ and $\pi_1(X_{\bar k})$ would be isomorphic by the results of SGA1, and we would be done. To "reduce" to this case, we might assume that we have a smooth compactification $\bar X$ of $X$ such that $\bar X\setminus X$ is a divisor with simple normal crossings. In this case we can use Abhyankar's lemma to extend the covering of $X_{\bar k}$ to a tamely ramified covering of $\bar X_{\bar k}$. The magic of log geometry should allow one to lift this covering to characteristic zero exactly as in the case without ramification. In particular, a single $p$ (satisfying some good reduction hypotheses) should be enough.

EDIT. Let me add some details. First, a definition (the terminology is not standard).

Definition. (1) (cf. SGA1 Exp. XIII, 2.1) Let $\pi:X\to S$ be a smooth morphism of schemes, $D\subseteq X$ an effective Cartier divisor on $X$. We say that $D$ has normal crossings relative to $S$ if locally on $X$, there exists an etale $S$-morphism $g:X\to \mathbf{A}^n_S$ such that $D=g^*(D(x_1\cdot\ldots\cdot x_r))$ for some $r$. We will call $(X, D)$ a log smooth pair over $S$.

(2) We say that a finite morphism $f:Y\to X$ over $S$ is a tame cover of a log smooth pair $(X, D)$ if etale locally on $Y$, there exists a morphism $g:X\to \mathbf{A}^n_S$ as in (1), integers $e_1 ,\ldots, e_r$ invertible on $S$, and an isomorphism over $X$ between $Y$ and the pullback of $(x_1, \ldots, x_n)\mapsto (x_1^{e_1}, \ldots, x_r^{e_r}, x_{r+1}, \ldots, x_n):\mathbf{A}^n_S\to \mathbf{A}^n_S$ along $g$.

Then Abhyankar's lemma (see SGA1 Exp. XIII, Appendice I, Proposition 5.5) can be phrased as follows:

Lemma 1. Assume that $S={\rm Spec}\, k$, and let $(X, D)$ be a log smooth pair over $S$. Set $U=X\setminus D$. Then the restriction functor $$ \left( \text{tame covers of }(X,D)\right) \longrightarrow \binom{\text{finite etale covers of }U}{\text{tame along }D} $$ is an equivalence.

The "log magic" I referred to above is hidden in the proof of the following lemma:

Lemma 2. Let $S\to \widetilde{S}$ be a nilpotent closed immersion, let $(\widetilde X, \widetilde D)$ be a log smooth pair over $\widetilde S$, $(X, D)$ its base change to $S$. Then the restriction functor $$ \left(\text{tame covers of }(\widetilde X, \widetilde D)\right) \longrightarrow\left(\text{tame covers of }(X, D)\right)$$ is an equivalence.

This can be deduced from Kato's first article on log geometry "Logarithmic structures of Fontaine-Illusie", section 3. (Probably there is an easier way of seeing this directly.) The point is that "tame covers" are log etale if $X$ and $Y$ are given their natural log structures. If we replace $\Omega^1_{Y/X}$ with its variant with log poles, we get the zero sheaf, which is why we obtain a unique extension property for tame covers.

From this we obtain the result you need. More precisely, suppose that $R$ is a henselian dvr with residue field $k$ and that $(X, D)$ is a log smooth pair over $S={\rm Spec}\, R$ such that $X$ is proper over $S$. Let $U=X\setminus D$. If $V_k\to U_k$ is a connected $G$-torsor, where $G$ is of order invertible in $k$, then Lemma 1 provides an extension to a connected $G$-tame cover $Y_k$ of $(X_k, D_k)$. Applying Lemma 2 to all ${\rm Spec}\,k\to {\rm Spec}\, R/m^n$, we obtain a connected $G$-tame cover $\mathcal{Y}$ of the formal completion $(\mathcal{X}, \mathcal{D})$. By Grothendieck's existence theorem applied to $f_*\mathcal{O}_{\mathcal{Y}}$, this comes from a connected $G$-tame cover $Y$ of $(X, D)$, which stays connected on the geometric generic fiber $(X_{\bar K}, D_{\bar K})$. Restricting to $U_{\bar K}$, we obtain a desired $G$-cover of your affine scheme $U_{\bar K}$.

$\endgroup$
4
  • 2
    $\begingroup$ This is a good suggestion. I just want to point out that this line of reasoning is used in the appendix of Olsson-Lieblich arxiv.org/abs/math/0703139 and in the recent paper by Bhatt-Gabber-Olsson arxiv.org/pdf/1705.07303.pdf (see Lemma 2.8 and the proof of Theorem 4.1). Instead of the magic of log-geometry, they use the magic of root stacks (which is "similar" magic). $\endgroup$ Commented Jun 25, 2017 at 21:16
  • $\begingroup$ Unfortunately I am rather ignorant when it comes to log geometry, I do agree that an appropriate lifting result of tamely ramified covers to characteristic 0 would do the job. $\endgroup$
    – John Z.
    Commented Jun 25, 2017 at 21:55
  • 1
    $\begingroup$ Wild guess: For $G$ a finite group and $X$ a finite type scheme over $A$ (a finitely generated $\mathbb {Z}$-subring of $\mathbb{C}$) the stack of $G$-covers of $X$ of a "specific topological type" is a finite type algebraic stack over $\mathbb Z[1/\# G]$. If that's true, you should be able to use the spreading out comment above (instead of log-geometry) to conclude the desired statement. $\endgroup$ Commented Jun 25, 2017 at 22:23
  • 1
    $\begingroup$ Thank you for your excellent answer. Also the appendix of Olsson-Lieblich paper (Corollary A.12) mentioned above does have the lifting result that applies here. $\endgroup$
    – John Z.
    Commented Jun 26, 2017 at 16:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .