I've asked some questions on Math Stackexchange regarding areas around my research but I received very little success with responses, so I thought I might try to share some of my other problems here instead.

The (cohomological) Brauer group of a field $k$ is given by $\mathrm{Br}(k) = H^2(k,\bar{k}^*)$. Here we let $k$ be a number field. Consider a smooth geometrically integral curve $C$ and the Hochschild-Serre spectral sequence $$H^p(k,H^q_{\mathrm{et}}(C_\bar{k},\mathbb{G}_m)) \implies H^{p+q}_{\mathrm{et}}(C,\mathbb{G}_m).$$ When $C$ is proper (hence projective), we obtain the map $\alpha:\mathrm{Br}(k) \rightarrow \mathrm{Br}_1(C)$ as part of the spectral sequence in low degree terms. This map enables us to obtain the isomorphism $$\frac{\mathrm{Br}_1(C)}{\mathrm{im}(\alpha)} \cong H^1(k, \mathrm{Pic}(C_{\bar{k}})).$$ However, when $C$ is not assumed to be proper, we instead obtain the map $$\alpha':H^2(k,\bar{k}[C]^*) \rightarrow \mathrm{Br}_1(C).$$ Question 1. Clearly, the inclusion $\bar{k}^* \rightarrow \bar{k}[C]^*$ induces the map $\mathrm{Br}(k) \rightarrow H^2(k,\bar{k}[C]^*)$. How do we interpret this map? Are there any explicit descriptions or known properties that can help us understand it?

To see why things wouldn't be as straightforward, the isomorphism that we will obtain involving $\alpha'$ is $$\frac{\mathrm{Br}_1(C)}{\mathrm{im}(\alpha')} \cong \mathrm{ker}(H^1(k,\mathrm{Pic}(C_\bar{k})) \rightarrow H^3(k,\bar{k}[C]^*)).$$

Let us denote the map above whose kernel we are interested in by $\gamma$.

Question 2. Once again, are there any ways to interpret $\gamma$? Or, more specifically, how does one understand an element of $\mathrm{ker}(\gamma)$?

  • 2
    $\begingroup$ Just a remark: if $C_\bar{k}=C'\smallsetminus S$, with $C'$ smooth projective and $S$ finite, there is an exact sequence $1\rightarrow \bar{k}^*\rightarrow \bar{k}[C]^*\rightarrow \mathbb{Z}^S\rightarrow 0$. This might help describing $H^2(k, \bar{k}[C]^*)$. $\endgroup$
    – abx
    Feb 27, 2021 at 8:07
  • $\begingroup$ @abx thanks, this is actually related to a question I just posted. If we set $C' = C_{1,\bar{k}}$, would $C' \backslash C_{\bar{k}}$ be finite? Also, from the exact sequence you provided, it seems like $\mathrm{Br}(k) \rightarrow H^2(k,\bar{k}[C]^*)$ is injective. $\endgroup$
    – oleout
    Feb 27, 2021 at 8:17
  • 2
    $\begingroup$ 1) It is finite, see my comment to your new question. $\ $ 2) I don't see why, there is the image of $H^1(k,\mathbb{Z}^S)$ in the kernel. $\endgroup$
    – abx
    Feb 27, 2021 at 8:25
  • 1
    $\begingroup$ Yes, that's right, with the natural action of the Galois group on $S$. $\endgroup$
    – abx
    Feb 27, 2021 at 11:26
  • 1
    $\begingroup$ The map $\overline{k}[C]^* \to \overline{k}^*$ obtained by evaluating at the fixed $k$-point is Galois-equivariant and splits the natural map $\overline{k}^*\to \overline{k}[C]^* $. $\endgroup$
    – Will Sawin
    Jul 23, 2021 at 12:43


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