Usbrth

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 $operatornameFrac(R)((x))$, the Laurent series over the fraction field of $R$, as seen here. Given that, I'm at a loss of other candidates for what $operatornameFrac(R[[x]])$ can be.

The required fraction field $K$ of our ring $R[[x]]$ consists of fractions $phi(x)=frac f(x)g(x)$ with $f(x), 0neq g(x)in R[[x]]$ .

If $F=Frac(R)$ we obviously have $Ksubset F((x))$ but the following analysis will show that we don't have equality in general.

Write $g(x)=x^k(r-xgamma(x))=x^kr(1-frac xrgamma(x))$ with $kgeq 0$ and $0neq rin R$ .

Then $frac 1g(x)=x^-ksum frac x^nr^n(gamma (x))^n$ and we see that $phi(x)=sum_i=m ^infty c_ix^i$ where $min mathbb Z$ depends on $phi$ and each $c_i$ is of the form $c_i=frac rho_ir^nu^i$ with $rho_iin R$ and $nu_iin mathbb N$.

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 $rin R$ (depending on $phi$) such that all $c_iin R[frac 1r]$.

For example it is clear that for $R=mathbb Z $ the series $e^x=sum frac x^ii!notin K$ since it is impossible to find $rin mathbb Z$ such that all $frac 1i!in mathbb Z[frac 1r]$

I just want to share the following result of Philip Sheldon
(Trans. AMS Vol. 159, 1971):

Let $Rsubset S$ be two subrings of the rational numbers.
Then the transcendence degree of the field extension

$mathrmFrac (R[[x]])subsetmathrmFrac(S[[x]])$

is infinite.

Suppose that $R$ is a UFD. Let $F(X) = X^-n sum_kge0 r_k X^k in K((X))$, with $r_0 neq 0$. Write
$$ r_k = fracp_kq_0cdot q_1 cdots q_k, quad (k geq 0),$$
with $q_k$ prime with $p_k$ (they are unique up to units).

Suppose that $F in mathrmFrac(R[[X]])$. Then you can find $A(X) = sum_kge0 a_n X^k$ and $B(X) = sum_kge0b_n X^k$ in $R[[X]]$ such that:
$$ sum_kge0r_k X^k = (sum_kge0a_n X^k)(sum_kge0b_n X^k)^-1.$$
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 $q_0cdot q_1 cdots q_k$, then you deduce that $q_k$ divides $b_0$ (for all $k$).

So a necessary condition for $F(X)$ to be in $mathrmFrac(R[[X]])$ 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 $mathrmFrac(R[[X]])$ as proved in your link. I expect that $(*)$ is also sufficient, but I am not sure.

Here's my try:

$Frac(D)[[x]]=alphain Frac(D)((x)) mid exists din D[[x]]^*, dalphain D[[x]]$

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 $frac12x^-5+frac13x^-4+frac12x^-3+frac13x^-2+dots$ which can be shifted into $mathbbZ[[x]]$ with $6x^5$, but I think there will be impossible to shift $Sigma_i=1^infty frac12^i x^i$ into $mathbbZ[[x]]$.