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Let $R$ be a valuation ring, with fraction field $K$ and residue field $k$. Denote by $\Gamma=K^{\times}/R^{\times}$ the valuation group (assumed nontrivial).
The valuation $v:K^{\times}\to\Gamma$ decomposes $K^{\times}$ as a disjoint union of nonempty open subsets, indexed by $\Gamma$. Each of these is homeomorphic to $R^{\times}$, which is in turn (using the reduction map to $k$) a disjoint union of nonempty open subsets, indexed by $k^{\times}$.
We conclude that any basis for the topology of $K$ must have cardinality at least $\kappa:=\max(\mathrm{Card}\,\Gamma, \mathrm{Card}\,k)$.

Question: does there exist a basis of open subets of $K$ with cardinality $\kappa$?

(1) It is true if $v$ is discrete, i.e. $\Gamma\cong\mathbb{Z}$. Proof: take a set $S\subset R$ of representatives of $k$, and a uniformizing parameter $\pi$. Let $X\subset K$ be the set of finite sums $\sum_i s_i\pi^{n_i}$ ($s_i\in S$, $n_i\in\mathbb{Z}$). Then the balls centered on $X$ form a basis.
(2) I am especially intereseted in the case $\kappa=\omega$. Explicitly: if $\Gamma$ and $k$ are countable, does it follow that $K$ is second-countable (or, equivalently, separable)?

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up vote 2 down vote accepted

I now think that the answer is no. Let $K$ be the field of Hahn series for the group $\mathbb{Q}$ over a countable field $k$, in the variable $t$ (so $\kappa=\omega$). For any sequence $a=(a_i)$ of elements of $k$, and any increasing sequence $\gamma=(\gamma_i)$ of rational numbers between $0$ and $1$, we have an element $x_{a,\gamma}=\sum_{i\in\omega}a_it^{\gamma_i}$ in $K$. The distance between different such elements is less than $1$ (in the valuative sense), so if we take an open ball of radius $2$ around each we get a disjoint union of uncountably many balls.

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If the distance is less than 1, then a ball of radius 2 around one of them will contain all the others. In particular the balls will not be disjoint. What am I missing? –  Lior Bary-Soroker May 15 '12 at 19:34
I'm talking about valuative radius, so bigger is smaller... e.g. $t^2$ is closer to $0$ than $t$ –  Moshe May 15 '12 at 19:43
Very nice, thanks! –  Laurent Moret-Bailly May 16 '12 at 12:29
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