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By a famous theorem of Tate, we know that the Tate conjecture holds for a product of curves over a finite field.

But this implies that the Brauer group of a product of curves (over finite field) is finite. What do we know about this Brauer group apart from the fact that it's finite ? Does there exist some computation done, for example for $\mathbb{P}^1 \times \mathbb{P}^1$ (or any other simple example) ?

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    $\begingroup$ $X:=\mathbb{P}^1 \times \mathbb{P}^1$ is rationally connected over $\mathbb{C}$, so its Brauer group is isomorphic to the torsion subgroup of $H^3(X, \, \mathbb{Z})$, which is $0$. $\endgroup$
    – Francesco Polizzi
    Commented Sep 16, 2016 at 12:31
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    $\begingroup$ In fact, somewhere in Manin's book "Cubic Forms" it is proved that for any field $K$, the pullback map $\text{Br}(K)\to \text{Br}(\mathbb{P}^1_K)$ is an isomorphism. Thus, over any field $K$ (including finite fields), the pullback map $\text{Br}(K)\to \text{Br}(\mathbb{P}^1_K\times_K \mathbb{P}^1_K)$ is an isomorphism. Combined with Wedderburn's theorem (reproved by Chevalley via the study of C1 fields), for $K$ a finite field, the Brauer group of $\mathbb{P}^1_K\times_K \mathbb{P}^1_K$ is trivial. $\endgroup$
    – Jason Starr
    Commented Sep 16, 2016 at 12:35
  • $\begingroup$ @JasonStarr How does the fact that $Br(K) \to Br(\mathbb P^1_K)$ is an isomorphism imply that the pull-back map $Br(K) \to Br(\mathbb{P}^1_K \times \mathbb P^1_K)$ is an isomorphism? I would personally argue differently: $\mathbb P^1_K \times \mathbb P^1_K$ is birational to $\mathbb P^2_K$, so that $Br(\mathbb P^1\times \mathbb P^1_K) = Br(\mathbb P^2_K)$ (see Thm 42.5 in Manin). However, the latter group is $Br(K)$ (see Theorem 42.8 in Manin). $\endgroup$
    – Ariyan Javanpeykar
    Commented Sep 28, 2016 at 13:27
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    $\begingroup$ @AriyanJavanpeykar. Let $L/K$ be the function field of $\mathbb{P}^1_K$. Then $\text{Br}(L)\to \text{Br}(\mathbb{P}^1_L)$ is an isomorphism. By "purity theorems" (the easy half), the map $\text{Br}(\mathbb{P}^1_K\times \mathbb{P}^1_K) \to \text{Br}(\mathbb{P}^1_L)$ is an injection. Thus, every element of $\text{Br}(\mathbb{P}^1_K\times \mathbb{P}^1_K)$ is pulled back from $\text{Br}(L)$. Now restrict a Brauer class on $\mathbb{P}^1_K\times \mathbb{P}^1_K$ to $\{[0,1]\} \times \mathbb{P}^1_K$ to see that the class is actually pulled back from $\text{Br}(\mathbb{P}^1_K)$. $\endgroup$
    – Jason Starr
    Commented Sep 28, 2016 at 14:04

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Suppose that $k$ is a finite field with $q$ elements. Let $C_i$, $i = 1, 2$ be two smooth projective geometrically irreducible curves over $k$. Set $X = C_1 \times_{\text{Spec}(k)} C_2$. In principle we know how to compute the Brauer group of $X$ in terms of the action of Frobenii on the \'etale cohomology of $C_1$ and $C_2$.

Let's assume there is no nontrivial isogeny from the Jacobian of $C_1$ to the Jacobian of $C_2$, or equivalently that the characteristic polynomials $P_1, P_2$ of the Frobenius endomorphisms of these Jacobians are relatively prime. This means that the Picard group of $X$ is generated by pullbacks of invertible sheaves from $C_1$ and $C_2$. Thus for $n$ prime to $q$ the exact sequence $$ \text{Pic}(X) \to H^2_{\acute{e}t}(C_1 \times C_2, \mu_n) \to Br(X)[n] \to 0 $$ coming from the Kummer sequence and cohomology is (in principle) computable. Working through what is known about \'etale cohomology of curves over finite fields you'll get $$ \left( H^1_{\acute{e}t}(C_{1, \overline{k}}, \mathbf{Z}/n\mathbf{Z}) \otimes H^1_{\acute{e}t}(C_{2, \overline{k}}, \mathbf{Z}/n\mathbf{Z}) \otimes \mu_n \right)^{\text{Gal}(\overline{k}/k)} $$ for the $n$ torsion in the Brauer group of $X$. If there are nonzero isogenies between the Jacobians, then you'll just get a quotient of this. If $n$ is not coprime to $q$, then you'll have to look at papers of Milne to see what happens.

Now it is clear that you can certainly get lot's of Brauer classes. In fact, you can show from the above that if $P_1$ and $P_2$ have a factor in common modulo $\ell$, then there'll be an $\ell$-torsion class in $Br(X)$ for $\ell$ a prime not dividing $q$. And given a situation as above and an $\ell$, you can always find a finite extension $k'$ of $k$ such that this thing happens over $k'$. Enjoy!

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