# What is known about the Brauer group of an arithmetic surface?

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.

• You should read Grothendieck's exposes in "Dix Exposes ..." – Jason Starr Aug 20 '14 at 17:58
• $Br(\mathbb{P}^1_\mathbb{Z}) = Br(\mathbb{Z}) = 0$. 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. – Daniel Loughran Aug 20 '14 at 18:49
• 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? – DonD Aug 21 '14 at 14:15

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.