# How are formal Laurent series the completition of rational functions over finite fields

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 $\mathbb{F}_q(t)$?

• The formal power series – reuns Apr 27 '17 at 9:35
• It depends on what definition you choose. But anyway the trick is to take the completion before or after taking the fraction field, since the absolute value is multiplicative, the two are the same. – reuns Apr 27 '17 at 9:54

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. $\mathbb{F}_q((t))$

• Ok, but how do we show, that every Chauchy series in any of the completition converges there? Why is $\mathbb{F}_q[[t]]$ the completition of $\mathbb{F}_q[t]$? Because the only limits of series $f_n$ in $\mathbb{F}_q[t]$ that are not in $\mathbb{F}_q[t]$ are those where $|f_n|$ goes to infinity? Then again why are formal Laurent series the completion of fractions? I don't see this at all. What are the limits that don't exists in fractions? – Bobby Apr 27 '17 at 10:12
• @Bobby Show that $|.|$ is a metric with $d(f,g) = |f-g|$. By definition a sequence converges in the completion iff it is Cauchy for the metric. Then, use that the absolute value is multiplicative and play with the quotients $|f/g|$ in the field of fraction, and show that it doesn't change anything taking the completion before or after taking the fractions – reuns Apr 27 '17 at 10:17