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Let $F$ be a function field with one variable with total constant field $k$, and let $X$ be the set of all places of $F$. How do I show that the genus of $k(T)$ is $0$ without using Riemann-Roch? Is there a way to do it a la establishing the surjectivity of $k[T] \to k((1/T))/k[[1/T]]$?

EDIT: My definition of genus is as follows. Let $S$ be a nonempty finite subset of $X$. Then the genus of $F$ is equal to the dimension over $k$ of the cokernel of $$O_S \to \bigoplus_{v \in S} F_v/O_v.$$

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    $\begingroup$ What is your definition of the genus of a function field over $k$? $\endgroup$
    – KConrad
    Jul 10, 2015 at 20:51
  • $\begingroup$ I recommend that you look up "Lagrange interpolation". That works best over an algebraically closed field, but surjectivity of your "principal parts homomorphism" can be checked after base change. $\endgroup$ Jul 10, 2015 at 23:25

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