2
$\begingroup$

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 $$O_S \to \bigoplus_{v \in S} F_v/O_v.$$

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ What is your definition of the genus of a function field over $k$? $\endgroup$
    – KConrad
    Commented 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. $\endgroup$
    – Jason Starr
    Commented Jul 10, 2015 at 23:25

0

Reset to default

You must log in to answer this question.