We can code the integers into the orders at zero of elements of $F_q(t)$ There is the following part in the paper that I am reading: 
We can code the integers into the orders at zero of elements of $F_q(t)$ (the field of rational functions in $t$ with coefficients in a finite field $F_q$ with $q=p^r$ elements, $p$ a prime) 
$+$ in $\mathbb{Z}$ corresponds to multiplication in $F_q(t)$, and the relation "there is an $s$ with $m=p^sn$" corresponds to "there are $x$ and $y$ so that $y$ has order $m$ and $x$ has order $n$ and $y$ is a $p^s$th power of $x$".
Could you explain to me the first sentence? 

We can code the integers into the orders at zero of elements of $F_q(t)$. 

 A: It seems to me that the "coding" here means simply to map an integer $n$ to the element $t^n\in\Bbb{F}_q(t)$. If $n>0$ then $t^n$ has a zero of order $n$ at $t=0$. OTOH if $n<0$ then $t^n$ has a pole of order $|n|$ at $t=0$.
The following is standard:
Order at zero (or at any element of the algebraic closure $\overline{\Bbb{F}_q}$) gives a valuation $\nu_0$ of the function field $\Bbb{F}_q(t)$. For any polynomial $f(t)\in\Bbb{F}_q[t]$ we declare that $\nu_0(f)=n$, if $t^n\mid f(t)$ but $t^{n+1}\nmid f(t)$. We extend this to fractions by defining
$$
\nu_0(\frac {f(t)}{g(t)})=\nu_0(f(t))-\nu_0(g(t)).
$$
Every element $f(t)/g(t)$ of the field of rational functions can uniquely be written in the form $t^n \tilde f(t)/\tilde g(t)$, where $n\in\Bbb{Z}$ and 
$\tilde f(0)\neq0,\tilde g(0)\neq0$.
Any irreducible polynomial $p(t)$ can be used in place of $t$ here to give you a valuation $\nu_p$. This gives us all the valuations of $\Bbb{F}_q(t)$ with the exception of the valuation at infinity:
$$
\nu_\infty(\frac {f(t)}{g(t)})=\deg g(t)-\deg f(t)
$$
that calculates the order of the zero/pole at $t=\infty$.
