# 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.

• When I say Laurent series I only allow finitely many non-zero negative coefficients. – miforbes May 2 '12 at 20:07
• Can you be more specific about what you would like in your description beyond the description given by the construction of $Frac(R[[x]])$? That construction is already pretty concrete. – rschwieb May 2 '12 at 20:13
• @rschwieb: A more concrete description would be something like: the fraction field is the laurent/power series with coefficients of form X''. – miforbes May 2 '12 at 21:56

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 $\frac {1}{i!}\in \mathbb Z[\frac {1}{r}]$

• If I'm not mistaken, $K[[x]][1/x]$ is just $K((x))$, the Laurent series over $K$. But by the link I provided in the question, the fraction field of $\mathbb{Z}[[x]]$ does not contain the power series for $exp(x)$, but that power series is contained in the power series over $\mathbb{Q}$. So I'm not sure your answer is correct. – miforbes May 2 '12 at 21:59
• @miforbes, Georges said that $K[[x]][1/x]$ is the fraction field of $R[[x]]$, where $K = \operatorname{Frac}(R)$. So since $\mathbb{Q} = \operatorname{Frac}(\mathbb{Z})$, the fraction field of $\mathbb{Z}[[x]]$ is $\mathbb{Q}[[x]][1/x] = \mathbb{Q}((x))$, which does indeed contain the power series for $\exp$, as you noted. – Antonio Vargas May 2 '12 at 22:11
• Now that I read user10676's answer, I'm less convinced my comment above is correct. Is it true that $K[[x]][1/x] = K((x))$? If so, does it not contradict the link the OP posted? – Antonio Vargas May 2 '12 at 23:27
• Yes, $K[[x]][1/x]=K((x))$, because every (nonzero) Laurent series is $x^i p(x)$ where $i$ is an integer and $p(x)$ is a power series with nonzero constant term. If you have adjoined an inverse for $x$, this is always possible. Since any power series with nonzero constant is a unit in $K[[x]]$, and you can "shift" to that unit with the unit $x^i$, everything is a unit. – rschwieb May 3 '12 at 0:28
• Robert Gilmer has shown (Proc. AMS Vol.18, No. 6 (1967)) that $F((X))$ is the field of fractions of $R[[x]]$ if and only if $K[[x]]=(R\setminus 0)^{−1}R[[x]]$. This shows some similarity to Georges necessary requirement for elements of $F((x))$ to be in $K$. The latter by the way directly implies that $K\neq F((x))$ if $R$ is a 1-dimensional valuation ring. Since these rings are "large" compared to $F$ this seems to indicate that the equality $K=F((x))$ holds only infrequently. – Hagen Knaf May 3 '12 at 16:06

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.

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.

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 $\mathbb{Z}[[x]]$.