5
$\begingroup$

Let $X$ be an arithmetic surface over $\mathbb{Z}$, that is we have $\pi: X\rightarrow Spec(\mathbb{Z})$, $X$ is integral, two-dimensional and regular and $\pi$ is projective and flat.

What is known about the Brauer group of $X$?

Is there a relation between the Brauer group of $X$ and the Brauer group of the generic fiber $X_{\eta}$? Or is there a relation with the special fibers?

For example can one find this group in the simplest example: what is $Br(\mathbb{P}^1_{\mathbb{Z}})$? How about if the generic fiber is an elliptic curve?

Any other interesting facts about $Br(X)$ or hints to literature is welcome.

$\endgroup$
  • 2
    $\begingroup$ You should read Grothendieck's exposes in "Dix Exposes ..." $\endgroup$ – Jason Starr Aug 20 '14 at 17:58
  • 2
    $\begingroup$ $Br(\mathbb{P}^1_\mathbb{Z}) = Br(\mathbb{Z}) = 0$. This latter vanishing follows from class field theory, as by reciprocity there is no non-trivial element of $Br(\mathbb{Q})$ which is unramified at all prime numbers. $\endgroup$ – Daniel Loughran Aug 20 '14 at 18:49
  • $\begingroup$ Ok, i see that $Br(\mathbb{Z})=0$. But how do i proof the first equality? I only know $Br(\mathbb{P}^1_k)=Br(k)$ for a field $k$. Is the same true for arbitrary rings? $\endgroup$ – DonD Aug 21 '14 at 14:15
7
$\begingroup$

As Jason says you should read "Dix Exposes". Part 3 of "Le Groupe de Brauer III" uses a result of Artin to show that $R^i\pi_*G_m =0$ for $i\geq2$ for surfaces such as you are interested in. The issue, of course, is dealing with $p$ torsion when the residue field of a valuation ring has characteristic $p$. So the Leray spectral sequence reduces to a long exact sequence involving cohomology of $\pi_*G_m$ and $R^1\pi_*G_m$. This immediately shows that $Br(P^1_{ Z})=0$. In general you need to use Neron models to carry out a full calculation since $\pi$ cannot be smooth over ${ Z}$ if the genus is positive. So I don't think there is a simple answer. In the case of a number field $K$ you might be reduced to calculating $H^1({ O_K}, A)$ where $A$ is an abelian scheme over the ring of integers $O_K$ if $\pi$ is smooth.

| cite | improve this answer | |
$\endgroup$

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.