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See here. Crossposted from math.stackexchange since there's no good answer despite $>$ 20 upvotes.

Let $k$ be a field of characteristic $\neq 2$, and consider the quadratic extension $F$ of $k(T)$ generated by $\sqrt{T^3 + 1}$.

I'm asking the question: what is the cokernel of $O_S \to F_\infty/O_\infty$ and its dimension?

The progress I have so far: let $S = \{\infty\} \subset X$ ($X$ is the set of all places of $F$) is the unique place of $F$ that $\text{ord}_\infty(T) < 0$. Note that $O_S = k[T, \sqrt{T^3 + 1}]$. Denote $T$ by $x$ and denote $\sqrt{T^3 + 1}$ by $y$ (so $y^2 =x^3 + 1$). Note also that the integral closure of $k[1/T]$ in $F$ is $k[1/x,\, y/x^2]$ and we have$$\left({y\over{x^2}}\right)^2 = \left({1\over{x}}\right)^4 + {1\over{x}}.$$But I am not sure what to do from here on out.

EDIT: Maybe Lagrange interpolation might be helpful?

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  • $\begingroup$ The cokernel has dimension one and is generated by the image of a element of $F_{\infty}$ with a simple pole. This follows from Riemann-Roch or can be proved by hand by simplifying the steps of a proof of Riemann-Roch. Not MO, though. $\endgroup$
    – Felipe Voloch
    Commented Aug 4, 2015 at 13:25

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