Let $\mathbb{F}_q$ be a finite field with characteristic $p$, let $\mathbb{F}_q[t]$ be the ring of polynomials with coefficients in $\mathbb{F}_q$, let $\mathbb{F}_q(t)$ be the quotient field of rational functions with coeffizients in $\mathbb{F}_q$ and let $\mathbb{F}_q((t))$ be the field of formal Laurent series with coefficients in $\mathbb{F}_q$.
In many text books on algebraic number theory we can read that $\mathbb{F}_q((t))$ is the completion of $\mathbb{F}_q(t)$, with respect to a certain valuation.
However I spend hours in the library looking this up in any reference I could find, but I couldn't come up with any single detailed proof. In addition many textbooks appear to be very cryptic when it comes to the definition of the appropriate valuation function. (Maybe this is because number theory is not my primary field of research, or maybe because most reference are pretty old)
Anyway, is someone willing to share a clear proof, thats shows how $\mathbb{F}_q((t))$ is the completion of $\mathbb{F}_q(t)$?
There is an absolute value on $\mathbb{F}_q[t]$ : $ | f| = 1$ if $f(0) \ne 0$, $|t^n f| = 2^{-n} |f|$.
$\mathbb{F}_q[[t]]$ (the ring of formal power series) is the completion of $\mathbb{F}_q[t]$ for $|.|$
$\mathbb{F}_q((t))$ is its field of fraction.
$\mathbb{F}_q(t)$ is the field of fractions of $\mathbb{F}_q[t]$. The absolute extends naturally with $|f/g| = |f|/|g|$.
Therefore $\mathbb{F}_q((t))$ is also the completion of $\mathbb{F}_q(t)$ for $|.|$
Otherwise, you can look at the ring of Laurent polynomials $\mathbb{F}_q[t,t^{-1}]$ and take the completion for $|.|$ to obtain the ring of formal Laurent series and show it is field, ie. $\mathbb{F}_q((t))$