Let $E$ be an elliptic curve over a number field $k$ and let $O$ be the distinguished $k$-rational point of $E$. Let $K$ be the function field of $E$. By Milne's Etale cohomology we have an exact sequence
$$0 \rightarrow \mathrm{Br}\,E \rightarrow \mathrm{Br}\, K \rightarrow \bigoplus_{x \in E^{(1)}}\mathrm{Hom}(k(x),\mathbb{Q}/\mathbb{Z}),$$
where $E^{(1)}$ denote the set of closed points of $E$, and $k(x)$ is the residue field of the closed point $x$. I am interested in the image of the residue map
$$\chi = \chi_O: \mathrm{Br}\, K \rightarrow \mathrm{Hom}(k(x),\mathbb{Q}/\mathbb{Z})$$
with respect to the $k$-point $O$. By Colliot-Thelene and Skorobogatov's The Brauer-Grothendieck group, Theorem 3.6.2,
(Witt) Let $R$ be a henselian discrete valuation ring with fraction field $K$ and perfect residue field $k$. Then there is a split exact sequence $$0 \rightarrow \mathrm{Br}\,k \rightarrow \mathrm{Br}\,K \rightarrow \mathrm{Hom}(k,\mathbb{Q}/\mathbb{Z}) \rightarrow 0.$$
I'm trying to fit my situation into the one stated in the theorem.
- Is the function field $K$ of $E$ the fraction field of some henselian DVR?
I did some searching and found out that for the (affine) elliptic curve given by the equation $f(X,Y) = 0$, its field of rational function is the fraction field of $k[X,Y]/(f)$, but I don't think this is a DVR.
- Does the residue field of $O$ happen to be $K$ itself?
I'm trying to show that the image of $\chi$ should be zero, but I'm not sure if this is even right.