What is the fraction field of $R[[x]]$, the power series over some integral domain? 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.
 A: 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 $\frac {1}{i!}\in \mathbb Z[\frac {1}{r}]$ 
A: I just want to share the following result of Philip Sheldon
(Trans. AMS Vol. 159, 1971):
Let $R\subset S$ be two subrings of the rational numbers.
Then the transcendence degree of the field extension
$\mathrm{Frac} (R[[x]])\subset\mathrm{Frac}(S[[x]])$
is infinite.
A: Suppose that $R$ is a UFD. Let $F(X) = X^{-n} \sum_{k\ge0} r_k X^k \in K((X))$, with $r_0 \neq 0$. Write
$$ r_k = \frac{p_k}{q_0\cdot q_1 \cdots q_k}, \quad (k \geq 0),$$
with $q_k$ prime with $p_k$ (they are unique up to units).
Suppose that $F \in \mathrm{Frac}(R[[X]])$. Then you can find $A(X) = \sum_{k\ge0} a_n X^k$ and $B(X) = \sum_{k\ge0}b_n X^k$ in $R[[X]]$ such that:
$$ \sum_{k\ge0}r_k X^k = (\sum_{k\ge0}a_n X^k)(\sum_{k\ge0}b_n X^k)^{-1}.$$
You can suppose that $b_0 \neq 0$. This imply that for all $k$:
$$a_n = \sum_{p+q=k} r_p b_q.$$
Multiply this equality by $q_0\cdot q_1 \cdots q_k$, then you deduce that $q_k$ divides $b_0$ (for all $k$).
So a necessary condition for $F(X)$ to be in $\mathrm{Frac}(R[[X]])$ is the following:
$$(*) \ \ \bigcap_{k \geq 0} q_kR \neq \{0\}, $$
i.e. $\gcd(q_0,q_1,\dots) < \infty$. This explains why $\exp(X)$ is not in $\mathrm{Frac}(R[[X]])$ as proved in your link. I expect that $(*)$ is also sufficient, but I am not sure.
A: Here's my try:
$Frac(D)[[x]]=\{\alpha\in Frac(D)((x)) \mid \exists d\in D[[x]]^*, d\alpha\in D[[x]]\}$
This just goes about identifying things whose "denominators can be cleared". The fact that not everything in $Frac(D)((x))$ cannot be shifted this way is because $D[[x]]$ is not dense in $Frac(D)((x))$.
For example, you have $\frac{1}{2}x^{-5}+\frac{1}{3}x^{-4}+\frac{1}{2}x^{-3}+\frac{1}{3}x^{-2}+\dots$ which can be shifted into $\mathbb{Z}[[x]]$ with $6x^5$, but I think there will be impossible to shift $\Sigma_{i=1}^\infty \frac{1}{2}^i x^i$ into $\mathbb{Z}[[x]]$.
