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I am looking for examples of smooth affine surfaces over algebraically closed fields with trivial $\ell$-torsion of the Brauer group.

Related questions: Schemes with trivial brauer group and Brauer group of projective space

The affine plane $\mathbf{A}^2_k$ should be one example. Edit: I am especially interested in cases where the surface is not rational.

Note that it is difficult to pass from projective surfaces to affine surfaces because of purity for Brauer groups: $0 \to \mathrm{Br}(S)(\ell) \to \mathrm{Br}(S - C)(\ell) \to H^1(C,\mathbf{Q}_\ell/\mathbf{Z}_\ell)$. Here, $H^1(C,\mathbf{Q}_\ell/\mathbf{Z}_\ell) = 0$ iff $C$ has genus $g(C) = 0$, is affine and $\bar{C} - C$ (the divisor at infinity) is a one-point set.

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I think the following should work, but I have not checked all the details.

Let $k$ be an algebraically closed field and let $\ell$ be coprime to the characteristic of $k$. Then the Grothendieck purity sequence implies that any non-trivial element of $\mathrm{Br}(k(x,y)) \{\ell\}$ must be ramified along at least two irreducible divisors in $\mathbb{P}^2$ (I think this can also be proved using the Faaddev reciprocity law, see Section 5 of the appendix to Chapter 2 of Serre's book on Galois cohomology, p114).

Hence, again by purtiy, it follows that if $f \in k[x,y]$ is irreducible and we take $U = \mathbb{P}^2 \setminus \{f = 0\}$, then we have $\mathrm{Br}(U) \{\ell\} = 0$.

This generalises the case of $\mathbb{A}^2$ which you already gave.

Edit about the non-rational case: If you want the $\ell$-primary part of the Brauer group to be non-trivial, then certainly you need the $\ell$-primary part of a smooth compactification $S$ to be non-trivial. By a result of Grothendieck, this can happen only if $b_1(S) = p_g(S) = 0$. For minimal surfaces, it seems like $S$ must be either an Enriques surface $(\ell > 2)$ or a surface of general type satisfying these cohomological restrictions. It seems possible that the complement of a hyperplane section of a surface of general type in $\mathbb{P}^3$ might give an example.

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    $\begingroup$ No, that's wrong. There are plenty of Brauer classes in $\mathrm{Br}(k(x,y))$ which are ramified along a smooth connected curve $C$ in $\mathbb{P}^2$, for instance those given by conic bundles. In view of the purity exact sequence, the curve $C$ should be rational, e.g. a conic, or (with a little more work) a cuspidal cubic. $\endgroup$ – abx Jun 29 '14 at 16:19
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    $\begingroup$ Yes, I see where my argument breaks down. I don't have time to try to salvage my answer at the moment, so I will just leave it as it is. $\endgroup$ – Daniel Loughran Jun 30 '14 at 7:56
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Here is an approach that should work. We know that $Br(\bf{A}_k ^2)=0$ if $k$ is an algebraically closed field. Now let $G$ be a finite group of order $n$ equipped with an action on $\bf{A}_k ^2$ such that $X$ is the quotient surface and $\bf{A}_k ^2\to X$ is flat. Now if $(\ell,n)=1$ then $Br(X)$ has no $\ell$ torsion since $Br(X)$ is split by a finite covering of order $n$ prime to $\ell$ and so must consist of torsion elements of order dividing $n$.

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    $\begingroup$ The OP asked for smooth examples, but it seems like in general these quotients will have singularities. $\endgroup$ – Daniel Loughran Jun 30 '14 at 15:11
  • $\begingroup$ If $Z \xrightarrow{f} X \xrightarrow{g} S$ are morphisms of schemes with $f$ fppf and $g \circ f$ smooth, then $g$ is smooth, so shouldn't $X$ in question be $k$-smooth because $\mathbb{A}_k^2 \rightarrow X$ is fppf, being a $G_X$-torsor? $\endgroup$ – Kestutis Cesnavicius Jun 30 '14 at 17:16
  • $\begingroup$ @DanielLoughran: If the characteristic is $p$, then there are actions on $\mathbb{A}^2_k$ of $p$-groups such that the quotient map is finite 'etale. For instance, translations have order $p$. $\endgroup$ – Jason Starr Jun 30 '14 at 19:48
  • $\begingroup$ My point was that if $k$ has characteristic zero, it seems that the Chevalley–Shephard–Todd theorem implies that the only smooth quotient which arises this way is $X = \mathbf{A}^2$ itself. $\endgroup$ – Daniel Loughran Jul 1 '14 at 7:47

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