I have a question similar to 74335.
Let $R$ be an integral domain. Is there a nice description of the fraction field of the power series $R[[x]]$?
I know that this field can be a proper subfield of $\operatorname{Frac}(R)((x))$, the Laurent series over the fraction field of $R$, as seen here. Given that, I'm at a loss of other candidates for what $\operatorname{Frac}(R[[x]])$ can be.
Best Answer
The required fraction field $K$ of our ring $R[[x]]$ consists of fractions $\phi(x)=\frac {f(x)}{g(x)}$ with $f(x), 0\neq g(x)\in R[[x]]$ .
If $F=Frac(R)$ we obviously have $K\subset F((x))$ but the following analysis will show that we don't have equality in general.
Write $g(x)=x^k(r-x\gamma(x))=x^kr(1-\frac {x}{r}\gamma(x))$ with $k\geq 0$ and $0\neq r\in R$ .
Then $\frac {1}{g(x)}=x^{-k}\sum \frac {x^n}{r^n}(\gamma (x))^n$ and we see that $\phi(x)=\sum_{i=m }^{\infty } c_ix^i$ where $m\in \mathbb Z$ depends on $\phi$ and each $c_i$ is of the form $c_i=\frac {\rho_i}{r^{\nu^i}}$ with $\rho_i\in R$ and $\nu_i\in \mathbb N$.
In other words in the investigated field $K$ every element is a power series $\phi(x)=\sum_{i=m }^{\infty } c_ix^i$ but these satisfy the strong requirement that there exists an element $r\in R$ (depending on $\phi$) such that all $c_i\in R[\frac {1}{r}]$.
For example it is clear that for $R=\mathbb Z $ the series $e^x=\sum \frac {x^i}{i!}\notin K$ since it is impossible to find $r\in \mathbb Z$ such that all $\frac {1}{i!}\in \mathbb Z[\frac {1}{r}]$