The purity of Brauer group states that for a smooth (quasi-)projective variety $X$ over a field $k$, removing a closed subscheme $Z \subset X$ of codimension at least $2$ ensures that the restriction map $H^2(X,\mathbb{G}\_m) \rightarrow H^2(X-Z,\mathbb{G}\_m)$ is an isomorphism. This means that the result cannot be extended to curves. Does anyone know of a simple counterexample, preferably when the smooth projective curve has genus at least $2$, if it matters?



Counterexample to purity of Brauer group for curves
The results of [this answer](https://mathoverflow.net/a/380825/82179) carry over (mutatis mutandis) to curves over finite fields. This shows that if $C$ is a (smooth, geometrically integral) curve over $\mathbf F\_q$ with function field $K$, then there is an exact sequence
$$0 \to \operatorname{Br}(C) \to \operatorname{Br}(K) \to \bigoplus\_{v \in C^{\text{cl}}} \mathbf Q/\mathbf Z,$$
obtained by taking the residues in $\operatorname{Br}(K\_v) \cong \mathbf Q/\mathbf Z$ for each completion $K\_v = \operatorname{Frac}\big(\widehat{\mathcal O\_{C,v}}\big)$ at closed points $v \in C^{\text{cl}}$. (That is, Brauer classes on $C$ are Brauer classes on $K$ that are unramified at all closed points of $C$.)


Thus if $\bar C$ is the smooth compactification of $C$, as soon as there exists a Brauer class on $C$ that is ramified at a point $v \in \bar C \setminus C$, then the restriction $\operatorname{Br}(\bar C) \to \operatorname{Br}(C)$ is not an isomorphism (it is always injective since $\operatorname{Br}(\bar C) \to \operatorname{Br}(K)$ is injective).


But by class field theory we can actually compute $\operatorname{Br}(C)$: by the [other post](https://mathoverflow.net/a/380825/82179) it sits in an exact sequence
$$0 \to \operatorname{Br}(C) \to \bigoplus\_{v \in \bar C \setminus C} \operatorname{Br}(K\_v) \to \mathbf Q/\mathbf Z \to 0.$$
If $\lvert \bar C \setminus C\rvert \geq 2$, we can produce classes whose invariants sum to $0$ in $\mathbf Q/\mathbf Z$, which therefore come from $\operatorname{Br}(C)$. Applying the same sequence to $\bar C$ shows $\operatorname{Br}(\bar C) = 0$. $\square$


Note: a direct computation of triviality of $\operatorname{Br}(\bar C)$ (not using class field theory) is given in [Brauer3, Rmq. 2.5(b)]. Presumably those texts contain many more useful computations and examples; for instance I expect that the passage from a curve to a dense open might be discussed somewhere as well.


**Remark.** On the other hand, if $C$ is a curve over an *algebraically closed* field $k$, then $\operatorname{Br}(C) \hookrightarrow \operatorname{Br}(K) = 0$ by Tsen's theorem, so $\operatorname{Br}(C) = 0$. So the Brauer classes really have to come from the nontriviality of $H^2(\kappa(v),\mathbf Z) \cong H^1(\kappa(v),\mathbf Q/\mathbf Z)$ for closed points $v \in \bar C \setminus C$, i.e. we need the residue fields $\kappa(v)$ not to be algebraically closed.




---


**References.**


[Brauer3] A. Grothendieck, *Le groupe de Brauer. III: Exemples et compléments.* Dix Exposés sur la Cohomologie des Schémas, Adv. Stud. Pure Math. **3**, 88-188 (1968). [ZBL0198.25901](https://zbmath.org/?q=an:0198.25901).


