Prove the field of fractions of $F[[x]]$ is the ring $F((x))$ of formal Laurent series.
$F[[x]]$ is contained in $F((x))$. So there's at least a ring homomorphism that is injective. Can also see it's injective because the kernel of such a mapping would be trivial because $0$ is the same in either. Not sure if showing they are isomorphic is the best way to do this.
$\displaystyle \sum_{n \ge N} a_nx^n \in F((x))$
Im not sure how I would define the mapping. maybe theres a better way
The field of fractions of an integral domain is the smallest field that the domain injects into. The homomorphism that sends a power series to itself is an injective homomorphism into $F((x))$, since every power series is also a Laurent series. If $F$ is the field of fractions of $F[[x]]$, then $f$ injects into $F((x))$, so we just need to check that the smallest subfield of $F((x))$ containing $F[[x]]$ is $F((x))$ itself. But this is clear: since $x^n$ is a power series for all $n\geq 0$, then any subfield containing all power series contains $x^{-n}$ as well. Thus, such a subfield contains all sums of power series and finitely many negative powers of $x$, which is exactly the field of Laurent series $F((x))$. Thus, $F((x))$ is indeed the field of fractions of $F[[x]]$.
