Gerbes on the multiplicative group

Question

Let $k$ be an arbitrary field with absolute Galois group $\Gamma$. The group $\text{Hom}(\Gamma,\mathbb{Q}/\mathbb{Z})$ injects into $H^2(\mathbb{A}^1 \setminus \{ 0 \},\mathbb{G}_m)$, as one can see e.g. by computing $\text{Ext}^2(\mathbb{G}_m,\mathbb{G}_m) = H^2(\Gamma,\mathbb{Z})$. If such a gerbe coming from $\Gamma \to \mathbb{Q}/\mathbb{Z}$ extends to $\mathbb{A}^1$, is it necessarily trivial? It's not clear to me because there are some weird gerbes on $\mathbb{A}^1$ if $k$ is not perfect.

Answer 1

I believe the answer is "yes, such a gerbe is necessarily trivial".

We first note that for any field $K$ and open subscheme $U$ of $\mathbb{A}^{1}_{K}$, the inclusion $\operatorname{Br}(U) \subseteq \mathrm{H}_{et}^{2}(U,\mathbb{G}_{m}) $ is an isomorphism. Also the coboundary $\operatorname{Hom}(\Gamma,\mathbb{Q}/\mathbb{Z}) \simeq \mathrm{H}^{1}(\Gamma,\mathbb{Q}/\mathbb{Z}) \to \mathrm{H}^{2}(\Gamma,\mathbb{Z})$ is an isomorphism since $\mathrm{H}^{i}(\Gamma,\mathbb{Q}) = 0$ for $i \ge 1$.

The Leray spectral sequence for $\mathbb{A}^{1}_{k} \to \operatorname{Spec} k$ is \begin{align} \mathrm{E}_{2}^{p,q} = \mathrm{H}^{p}(\Gamma,\mathrm{H}_{et}^{q}(\mathbb{A}^{1}_{k^{\mathrm{sep}}},\mathbb{G}_{m})) \implies \mathrm{H}_{et}^{p+q}(\mathbb{A}^{1}_{k},\mathbb{G}_{m}) \end{align} with differentials $\mathrm{E}_{2}^{p,q} \to \mathrm{E}_{2}^{p+2,q-1}$. This and the analogous spectral sequence for $\mathbb{A}^{1}_{k} \setminus \{0\} \to \operatorname{Spec} k$ gives a commutative diagram $\require{AMScd}$ \begin{CD} 0 @>>> \operatorname{Br}(k) @>\xi_{1}>> \operatorname{Br}(\mathbb{A}^{1}_{k}) @>\xi_{2}>> \mathrm{H}^{0}(\Gamma,\operatorname{Br}(\mathbb{A}^{1}_{k^{\mathrm{sep}}})) \\ @. @V\rho_{1}VV @V\rho_{2}VV @V\rho_{3}VV \\ 0 @>>> \operatorname{Br}(k) \oplus \mathrm{H}^{2}(\Gamma,\mathbb{Z}) @>>\xi_{1}'> \operatorname{Br}(\mathbb{A}^{1}_{k} \setminus \{0\}) @>>\xi_{2}'> \mathrm{H}^{0}(\Gamma,\operatorname{Br}(\mathbb{A}^{1}_{k^{\mathrm{sep}}} \setminus \{0\})) \\ \end{CD}

where the $\rho_{i}$ are induced by restriction along the open immersion $\mathbb{A}^{1}_{k} \setminus \{0\} \subset \mathbb{A}^{1}_{k}$. Here both rows are exact because $\operatorname{Pic}(\mathbb{A}^{1}_{k^{\mathrm{sep}}}) = 0$ and $\operatorname{Pic}(\mathbb{A}^{1}_{k^{\mathrm{sep}}} \setminus \{0\}) = 0$ (so that $\mathrm{E}_{2}^{p,1} = 0$ for $p \ge 0$ for both spectral sequences).

With respect to the above commutative diagram, the question may be rephrased as follows:

Suppose $\alpha_{1}' \in \mathrm{H}^{2}(\Gamma,\mathbb{Z})$ is such that $\xi_{1}'(\alpha_{1}') = \rho_{2}(\alpha_{2})$ for some $\alpha_{2} \in \operatorname{Br}(\mathbb{A}^{1}_{k})$. Is $\xi_{1}'(\alpha_{1}')$ necessarily $0$?

For such $\alpha_{1}',\alpha_{2}$ we have $\rho_{3}(\xi_{2}(\alpha_{2})) = \xi_{2}'(\rho_{2}(\alpha_{2})) = \xi_{2}'(\xi_{1}'(\alpha_{1}')) = 0$. Since $\mathbb{A}_{k}^{1}$ and $\mathbb{A}_{k^{\mathrm{sep}}}^{1}$ are regular Noetherian (and also since $\mathrm{H}^{0}(\Gamma,-)$ is left exact), the restriction maps $\rho_{2}$ (and hence also $\rho_{1})$ and $\rho_{3}$ are injective. Thus $\xi_{2}(\alpha_{2}) = 0$, thus there exists some $\alpha_{1} \in \operatorname{Br}(k)$ such that $\xi_{1}(\alpha_{1}) = \alpha_{2}$; then $\xi_{1}'(\rho_{1}(\alpha_{1})) = \rho_{2}(\xi_{1}(\alpha_{1})) = \rho_{2}(\alpha_{2}) = \xi_{1}'(\alpha_{1}')$, but injectivity of $\xi_{1}'$ implies $\alpha_{1}' = \rho_{1}(\alpha_{1})$; thus both $\alpha_{1} = 0$ and $\alpha_{1}' = 0$ since $\alpha_{1}',\rho_{1}(\alpha_{1})$ are in different summands of $\operatorname{Br}(k) \oplus \mathrm{H}^{2}(\Gamma,\mathbb{Z})$; thus $\xi_{1}'(\alpha_{1}') = 0$ as well.