Definition: Say that a domain $D$ with field of fractions $K$ satisfies property $(\rho)$ if countable intersections of nonzero ideals are nonzero, or equivalently if $D[[x]]_{D \setminus 0} = K[[x]]$.
I'd like to get a better understanding of how strong property $(\rho)$ is. I might even lamely conjecture that any domain with property $(\rho)$ is a field.
EDIT: A similar, weaker, property in the literature is that of being anti-archimedean, i.e. $\bigcap a^nD \not=0$ for all $a \in D$, or equivalently, $D[[x]]_{D \setminus 0}$ is local (has one maximal ideal).
Question: Do there exist non-field domains satisfying property $(\rho)$?
Of course any proper overring of such a ring will have this property too, so we could find a non-field valuation overring of $D$ having the property $(\rho)$ (if there weren't any, then the integral closure of $D$ would be $K$, hence $D = K$.)
So we've essentially reduced the question to valuation rings. However, I haven't been able to construct a nontrivial example with the desired property or able to see why such examples couldn't exist.
The following observation shows that any ring with property $(\rho)$ will be extremely large.
A ring $D$ with property $(\rho)$ is such that every proper saturated multiplicative set in $D$ is disjoint from uncountably many prime ideals. Consequently, every ideal of a valuation ring with property $(\rho)$ must contain uncountably many prime ideals.
Suppose $S$ is a saturated multiplicative set disjoint from only countably many primes. Then the localization $S^{-1}D$ has countably many primes, and by property $(\rho)$, there would be some $0 \not= a$ in the intersection of its primes, and thus $K = S^{-1}D[1/a]$. But then picking $0 \not= d \in \bigcap a^n$ (using property $(\rho)$ again) and examining $1/d \in S^{-1}D[1/a]$ shows that $a$ is a unit in $S^{-1}D$. Hence $K = S^{-1}D$ and $S = D \setminus 0$. $\square$.