What is the cokernel of $O_S \to F_\infty/O_\infty$? Let $k$ be a field of characteristic $\neq 2$ and consider the quadratic extension $F$ of $k(T)$ generated by $\sqrt{T^3 + 1}$. Let $X$ be the set of all places of $F$. Let $S = \{\infty\} \subset X$ where $\infty$ is the unique place of $F$ such that $\text{ord}_\infty(T) < 0$. Note that $O_S = k[T, \sqrt{T^3 + 1}]$.
What is the cokernel of $O_S \to F_\infty/O_\infty$?
 A: I realize now that this is the type of question that should be solved by the OP.  However, since my guess above is incorrect (which I should have realized before I clicked submit), I feel obliged to submit a correct answer.
The cokernel is a one-dimensional $k$-vector space, generated by the class of $u^{-1}$.  By Hensel's Lemma, there exists $v\in 1 + \mathfrak{m}_{\mathcal{O}_\infty}$ such that $v^2 = 1 + (1/T)^3$.  Thus, for $w=uv$, then $\mathcal{O}_\infty \cong k[[w]]$ and $w^2 = (1/T)$.  So, as in the comment above, $F_\infty/\mathcal{O}_\infty$ equals the isomorphic image of the $k$-subspace $kw^{-1}\oplus kw^{-2} \oplus kw^{-3} \oplus \dots$ of $K_\infty$.  In particular, the image of $T^r$ equals $w^{-2r}$.  Similarly, the image of $T^r\sqrt{T^3+1}$ equals $\pm w^{-2r-1}$ modulo lower order terms in $w^{-1}$.  Thus, by induction, the cokernel is generated as a $k$-vector space by the image of $w^{-1}$.
On the other hand, there is no $f\in \mathcal{O}_S$ that has image equal to $w^{-1}$.  If there were, then $f$ would be a rational function on the elliptic curve $y^2 = T^3+1$ that has a simple pole at $\infty$ and no other poles.  That means that $f$ would be an isomorphism of the elliptic curve with the projective line over $k$.  Thus, the image of $w^{-1}$ is nonzero in the cokernel.  Thus the image of $w^{-1}$ is a $k$-basis for the cokernel.
