8
$\begingroup$

This question is the two-dimensional analogue of Etale coverings of certain open subschemes in Spec O_K

There I asked if one could characterize in a way certain covers of $\textrm{Spec} \ O_K$. As Cam Mcleman answered, this is basically done by the Galois group of the maximal extension unramified outside $D$. A covering of $U$ is of the form $O_L[\frac{1}{D}]$, where $L$ is any extension of $K$.

Here I would like to ask the same question, only now for $X=\mathbf{P}^1_{\mathbf{Z}}$.

Let $D$ be a normal crossings divisor on $\mathbf{P}^1_{\mathbf{Z}}$ and let $U$ be the complement of its support.

Q1. Is there an "equivalence of categories" as Georges Elencwajg mentions in his answer for the analytic case. (See above link.) Basically, is there an arithmetic Grauert-Remmert theorem?

Q2. What is known about the etale fundamental group in this case? Is it "finitely generated"? Has anybody studied the maximal pro-p-quotients of these groups?

Q3. The analytic analogue would be to consider the same question for $\mathbf{P}^1_{\mathbf{C}} \times \mathbf{P}^1_{\mathbf{C}}$.

Q4 Lars (see above link) mentions a result for $\mathbf{P}_{\mathbf{Q}}^1$. Is there something similar for $\mathbf{P}^2_{\mathbf{Q}}$?

$\endgroup$
2
  • $\begingroup$ I've actually been thinking about similar things. If I have more time later I'll write some more as an answer. Just to nitpick, though, the analytic analogue isn't $\mathbb{P}^2_{\mathbb{C}}$ but $\mathbb{P}^1_{\mathbb{P}^1_{\mathbb{C}}}=\mathbb{P}^1_{\mathbb{C}}\times\mathbb{P}^1_{\mathbb{C}}$ (these are birational but not isomorphic). $\endgroup$ – James D. Taylor Jan 20 '11 at 1:37
  • $\begingroup$ Also, it turns out that covers of the arithmetic line don't follow the same behavior one would expect from a Grauert-Remmert type result. $\endgroup$ – James D. Taylor Jan 20 '11 at 1:55
2
$\begingroup$

Regarding Q3: For any scheme $X$ of finite type over $\mathbb{C}$ the Riemann-Existence Theorem (See SGA1 XII.5) says that the category of finite étale coverings of $X$ is equivalent to the category of finite covering spaces of the associated analytic space $X^{an}$. This implies that the finite quotients of the topological fundametal group of $X^{an}$ are the same as the finite quotients of the étale fundamental group, and one obtains that the étale fundamental group of $X$ is isomorphic to the profinite completion of the topological fundamental group of $X^{an}$.

Q4: The same short exact sequence that I mentioned in your other question is still valid.

As in your other question, I cannot say anything about the situation over $\mathbb{Z}$ :)

$\endgroup$
4
$\begingroup$

This is a question which is too long to put in a comment box: What exactly do you mean by a simple normal crossings divisor in $\mathbb P^1_{\mathbb Z}$?

Let me recall that an irreducible divisor in $\mathbb P^1_{\mathbb Z}$ consists either of a Galois conjugacy class of $\overline{\mathbb Q}$ points of $\mathbb P^1$ (closed up to make a divisor in $\mathbb P^1_{\mathbb Z}$ (these are the horizontal divisors), or else one of the closed fibres $\mathbb P^1_{\mathbb F_p}$ of the map $\mathbb P^1_{\mathbb Z} \to \mathbb Z$ (these are the vertical divosors).

The vertical divisors are mutually disjoint, while the horizontal divisors meet each vertical divisor in a point (if the horizontal divisor is the point $\alpha \in \overline{\mathbb Q} \cup \{\infty\}$, and the vertical divisor is $\mathbb P^1_{\mathbb F_p}$, then they intersect in the point $\overline{\alpha}$ obtained by specializing $\alpha$ into char. $p$). Two horizontal divisors, say corresponding to the points $\alpha, \beta \in \overline{\mathbb Q}\cup \infty$, may or may not intersect; they meet in a point lying over the point $p \in $ Spec $\mathbb Z$ if and only if $\alpha$ and $\beta$ have the same specialization to char. $p$, and one can define a multiplicity of intersection, which reflects the precise power of $p$ modulo which they are congruent.

I'm not exactly sure in this arithmetic context what the s.n.c. condition is.

$\endgroup$
1
  • $\begingroup$ I think I mixed up the terminology. Sorry. It should just be a divisor with normal crossings. $\endgroup$ – Ariyan Javanpeykar Jan 30 '11 at 9:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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