Is the Brauer group of a surface an elliptic curve? Of course not.
But after reading a bit, some points make me believe it should be:
Let $S$ be a nice$^{\*}$ surface defined over $Spec\ \mathbb{Z}$.


*

*The Brauer group $Br(S\otimes \bar{\mathbb{Q}})$ is an abelian divisible group,

*It is also a $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ module,

*For good primes there are reductions $Br(S)\rightarrow Br(S\otimes \mathbb{F}_q)$,

*These $Br(S\otimes \mathbb{F}_q)$ are finite,

*There is a formal Brauer group $\hat{Br}(S)$ of dimension 1,

*The coefficients of $\hat{Br}$, in suitable natural coordinates, relate to $|Br(S\otimes \mathbb{F}_q)|$.

*There are some examples where the associated L-function comes from a modular form (of weight 3). I'm not sure if this is conjectured (let alone known) in general.


Since the Brauer group observes many characteristics of an abelian variety (all properties) of dimension 1 (properties 5 and 7 [weight isn't two, but it's the right space]), my vague question is: how far is it from actually being a variety?
There are some easy examples of $S$ with $|Br(S\otimes \mathbb{F}_q)|$ varying between $1$ and $4(q-4)$, as $q$ varies over the primes. This is a clear point of departure from elliptic curves and varieties in general.
Maybe there's a family of natural galois-module homomorphisms into certain abelian varieties defined over $\mathbb{Q}$, commuting with the reduction maps and restriction (or some other appropriate term) to formal groups?
What's going on with these Brauer groups?
$^\*$ say a K3 surface. Something that (1) is true for (so not a rational surface) and (4) is proven for.
 A: This is not a general answer to your question, but evidence of the intriguing connection between Brauer groups of surfaces and elliptic curves. Let $X$ be a K3 surface over the complex numbers $\mathbb{C}$. Then, the rank of $H^2(X,\mathbb{Z})$ is $22$, and the Hodge numbers are $h^{0,2}=h^{2,0}=1$ and $h^{1,1}=20$. If the Neron-Severi group of $X$ is as big as possible, namely if it has rank $20$, then the image of $H^2(X,\mathbb{Z})\rightarrow H^2(X,\mathcal{O}_X)$ has rank $2$ and so is a lattice. Therefore, the cokernel of this map is an elliptic curve. But, since $H^3(X,\mathbb{Z})=0$, there is an exact sequence $$H^2(X,\mathbb{Z})\rightarrow H^2(X,\mathcal{O}_X)\rightarrow H^2(X,\mathcal{O}_X^*)\rightarrow 0.$$ Therefore, $H^2(X,\mathcal{O}_X^*)$ is an elliptic curve $E$. The torsion of $H^2(X,\mathcal{O}_X^*)$ is therefore precisely the Brauer group of $X$, and these are precisely the torsion points of $E$.
One can produce similar examples using abelian surfaces. For instance, if $E$ is a CM elliptic curve, then $E\times E$ is an abelian surface such that $H^2(E\times E,\mathcal{O}^*)$ is isomorphic to $E$.
