Let $F$ be a global function field with full constant field $\mathbb{F}\_q$. We fix a place $\infty$ and let $A$ be the ring of elements of $F$ regular away from $\infty$. We denote by $v\_\infty$ the normalized valuation associated to $\infty$. If $a\in A$ and $v\_\infty(a)\geq 0$ then $a$ has no poles. Therefore (see for example Corollary 1.1.20 in Stichtenoth's Algebraic Function Fields and Codes) $a$ is algebraic over $\mathbb{F}\_q$ and then $a\in \mathbb{F}\_q$ since it is the full constant field. We then have $v\_\infty(a) < 0$ for all $a\in A\setminus \mathbb{F}\_q$. My questions are: 


1. Is $D\_{F,\infty} := $Max$\{v\_\infty(a)/ a\in A, v\_\infty(a)<0\}$ equal to -1? i.e. is there an element of $A$ of valuation -1 at infinity?
2. If $D\_{F,\infty}$ is not -1, can it be computed explicitly, for instance in terms of the genus $g\_F$ of $F$ (I am just really making a guess here, thinking about Riemann-Roch and its corollaries)?
3. Is there any algorithm to compute $D\_{F,\infty}$?


Thank you.



On the maximal value of the valuation at infinity of elements in the ring of integers of a global function field
1. Typically, no. If $\infty$ has degree $1$ then $D\_{F,\infty}=-1$ if and only if $F$ is rational, since in this case $a$ would have degree $1$ so $F=\mathbb{F}\_q(a)$. When $\infty$ has degree $1$, $-D\_{F,\infty}$ is the first non-gap in the Weierstrass semigroup of $\infty$. Lots of literature on that. It can take any value between $2$ and $g+1$ and is "usually" $g+1$.
2. Riemann-Roch tells you that $-D\_{F,\infty} \le r$ if $r > g/ \deg \infty$ so $D\_{F,\infty}=-1$ if $\deg \infty > g$.
3. Yes, since you can compute Riemann-Roch spaces. So compute $\ell(r\infty)$ until it is $>1$.


