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Let $K/F$ be a function field. First some notation:

$D_K$ is the group of divisors of $K$ (i.e. the free abelian group generated by the primes of $K$).

For $x \in K^{\times}$, $(x) = \sum_P ord_P(x)P $

$L(D) = \{x \in K^{\times} \,:\, (x) + D \geq 0\} \cup \{0\}$ for $D \in D_K$

$l(D) = \dim_F(L(D))$

Here is the problem:

Suppose $K/F$ is of genus $g\geq 2$, and $P$ a prime of degree $1$. For all integers $k$ we have $l(kP) \leq l((k+1)P)$. If we restrict $k$ to the range $0\leq k\leq 2g - 2$ show there are exactly $g$ values of $k$ where $l(kP) = l((k+1)P)$.

I am able to show that $l((k+1)P) - l(kP) \leq 1$ for $k \geq 0$ (this seems like it should help, but maybe not). I have also observed that by Riemann-Roch, $l(kP) = l((k+1)P)$ (with $0\leq k \leq 2g-2$) if and only if $l(C - kP) = l(C - (k+1)P) + 1$ where $C$ is a divisor in the canonical class $\mathcal{C}$.

Edit:

I'm not sure where to go from here. I tried to gain some intuition by looking at the case when $k = 0$. Of course $l(0P) = l(0) = 1$. But $l(P)$ can be either $1$ or $2$, and I don't see why $l(P)$ has to be $1$.

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1 Answer 1

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Notice that $l(0P) = 1$ and $l((2g-1)P) = \deg((2g - 1)P) - g + 1 = g$. Denote $n_0 = |K_0|$, where $K_0 = \{k \,:\,0\leq k \leq 2g-2 ; l(kP) = l((k+1)P) \}$ and let $K_1 = ([0,2g-2]\cap \mathbb{Z})\setminus K_0$. Then since for each $0\leq k\leq 2g-2$ we have $l((k+1)P)- l(kP) \leq 1$, for each $k \in K_1$, we must add $1$ to $l(0) = 1$ in order to reach $l((2g-1)P) = g$ (having a real hard time expressing this idea). Since $|K_1| = 2g-1 - n_0$, we get that $g = 2g-1 - n_0 + 1$ and rearranging terms we arrive at $n_0 = g$.

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