Let $F[[x]]$ be the ring of formal power series in the indeterminate x with coefficients in F. Show that the field of fractions of $F[[x]]$ is the ring $F((x))$ of formal Laurent series.
I've been going in circles on this one for what seems like ever now! If anyone could lend me some insight I would be exceedingly grateful! Thanks!
On
Hint:
A formal power series in $F[[x]]$ is invertible if and only if its constant term is not $0$ (more generally, a formal power series in $R[[x]]$, $R$ being a commutative ring, is invertible if and only if its constant term is a unit in $R$).
Let $p$ be the order of a formal power series. It can be written as $$\sum_{k\ge p}a_k x^k=a_px^p\Bigl(1+\frac{a_{p+1}}{a_p}x +\frac{a_{p+2}}{a_p}x^2+\dotsm\Bigr),$$ i.e. as the product of $\;a_px^p$ by an invertible power series. So in order to find the reciprocal of the formal power series, you just have to find the reciprocal of $x^p$.
$\sum_{k=-n}^\infty a_kx^k=\frac{\sum_{k=0}^\infty a_{k-n}x^k}{x^n}$, so every Laurent series can be written as a quotient of power series.
Conversely, given a fraction $\frac{F(x)}{G(x)}$, write $G(x)=a_nx^n(1+xH(x)),$ where $H(x)$ is a power series. Then $$ \frac1{G(x)}=\frac1{a_nx^n}\cdot \frac{1}{1+xH(x)}=\frac1{a_nx^n}\sum_{k=0}^\infty(-1)^kx^kH(x)^k $$ When $H(x)^k$ is expanded into powers of $x$, and like terms are collected, the above writes $\frac1{G(x)}$ as a Laurent series. Since Laurant series are closed under multiplication, $F(x)\cdot \frac1{G(x)}$ is also a Laurent series.