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 $O\_S \to F\_\infty/O\_\infty$?



What is the cokernel of $O_S \to F_\infty/O_\infty$?
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.


