Ostrowski's theorem classifies all absolute values on a number field $K$.


  • More generally, can one classify all Hausdorff topologies on $K$ making $K$ into a topological field?

  • In particular, is every Hausdorff topology on a number field $K$ making $K$ into a topological field induced by an absolute value?

It would already be interesting to understand this when $K= \mathbb Q$. On the other hand, I'd be interested to understand this question for more general fields and rings as well. For "large" fields / rings, I imagine one might need to consider valuations in more general value groups as well. But I don't know a generally-accepted definition of "archimedean valuation" not over $\mathbb R$, so I'm not quite sure how to formulate a potentially-correct statement saying that "every topology comes from a generalized absolute value" in this context.

  • $\begingroup$ One generalisation of an archimedean valuation could be to take the order topology on any real closed field (and perhaps demand that the induced uniform structure is complete, to avoid things like $\mathbb R \cap \bar{\mathbb Q}$). I don't know if this actually happens in other examples, though. $\endgroup$ – R. van Dobben de Bruyn Sep 20 '18 at 21:41
  • $\begingroup$ @YCor I suppose you're right. Although this feels like a technicality -- it would be pretty natural to consider a "semi-absolute value" to be like a valuation but not required to satisfy $|x| = 0 \Rightarrow x = 0$, and ask whether every topological ring structure is induced by one of these. But certainly the main interest in the question lies in considering just the Hausdorff topologies. $\endgroup$ – Tim Campion Sep 20 '18 at 21:45
  • $\begingroup$ @R.vanDobbendeBruyn yeah, that seems pretty natural. I also speculated about another approach here. $\endgroup$ – Tim Campion Sep 20 '18 at 21:46
  • 2
    $\begingroup$ @TimCampion on a field, there's a single non-Hausdorff topology making it a topological ring: the indiscrete one. Indeed, being non-Hausdorff implies (for an arbitrary group) that the closure of $\{0\}$ is a nonzero subgroup, and as a topological ring, this closure has to be an ideal. Being a field, it is the whole field. $\endgroup$ – YCor Sep 20 '18 at 21:49
  • 1
    $\begingroup$ @EmilJeřábek: Ah, I was unaware of that. Do you have a reference? $\endgroup$ – R. van Dobben de Bruyn Sep 20 '18 at 21:52

Thanks to YCor's examples in the comments, I decided this question was worth a deep dive. It turns out on the one hand that

There are lots of exotic (Hausdorff) field topologies on $\mathbb Q$.

But on the other hand, it turns out that

Every locally compact (Hausdorff) ring topology on a field is induced by an absolute value.

See Theorem 16.3 In Warner's Topological Rings. Shanks and Warner also showed that every locally bounded (Hausdorff) ring topology on $\mathbb Q$ comes from an absolute value. Here a topological ring $R$ is locally bounded if there is a neighborhood $B$ of 0 which is bounded in the sense that for every neighborhood $U$ of 0 there is a neighborhood $V$ of 0 such that $VB \subseteq U$. A partial extension to global fields was given by Nichols and Cohen (not in alphabetical order).

It seems there's been work on constructing exotic topologies on more general rings and fields, but I didn't come across positive results constraining these more general topologies under reasonable conditions.

  • $\begingroup$ I think that there's ambiguity on the meaning of "induced by an absolute value". Whether you require multiplicativity, or only submultiplicativity, this will not include, or include, the topology, say, induced by inclusion into $\mathbf{Q}_2\times\mathbf{Q}_3$, that is, defined by the norm $\|\cdot\|_2+\|\cdot\|_3$. $\endgroup$ – YCor Sep 21 '18 at 6:43
  • 1
    $\begingroup$ Full reference info: Mutylin, A. F. An example of a nontrivial topologization of the field of rational numbers. Complete locally bounded fields. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 30 1966 873–890. (MR0195856) English translation: Amer. Math. Soc. Series 2, Vol. 73: Fourteen papers on algebra, topology, algebraic and differential geometry. American Mathematical Society, Providence, R.I. 1968 iv+260 pp. (MR0231674) $\endgroup$ – YCor Sep 21 '18 at 6:52
  • $\begingroup$ It seems that Mutlylin's exotic constructions have the property that, for each of them, some sequence of primes tends to 0. Indeed, this is in far contrast with all the topologies induced by some set of $p$-adic absolute values. $\endgroup$ – YCor Sep 21 '18 at 6:58
  • 2
    $\begingroup$ Would someone be willing to write the gist of the Mutylin's construction? The relevant pages of the google book are unavailable to me. $\endgroup$ – Somatic Custard Sep 27 '18 at 1:35

The following relevant classification result, due to Kowalsky and Dürbaum [2], appears in Appendix B of Engler and Prestel [1].

Let $(K,\tau)$ be a topological field. Then $\tau$ is called a V-topology if for every neighbourhood $W\ni0$, there exists a neighbourhood $U\ni0$ such that $(K\smallsetminus W)(K\smallsetminus W)\subseteq K\smallsetminus U$ (that is, for any $x,y\in K$, if $xy\in U$, then $x\in W$ or $y\in W$).

Theorem: The V-topologies on a given field are exactly the topologies induced by valuations (with arbitrary value groups) or by archimedean absolute values.

Note that nonarchimedean absolute values are also covered, being special cases of valuations (with, confusingly enough, value groups that are archimedean, i.e., rank 1).

In the special case of global fields (including number fields), all valuations have rank 1, i.e., they are equivalent to nonarchimedean absolute values (e.g., see Thm. 2.1.4 and Cor. 3.2.5 in [1]). Thus:

Corollary: If $K$ is a global field, the topologies on $K$ induced by absolute values are exactly the V-topologies.


[1] Antonio J. Engler and Alexander Prestel, Valued fields, Springer, 2005.

[2] Hans-Joachim Kowalsky and Hansjürgen Dürbaum, Arithmetische Kennzeichnung von Körpertopologien, Journal für die reine und angewandte Mathematik 191 (1953), 135­–152.

  • 1
    $\begingroup$ Thanks, this is really interesting! Somehow it seems more straightforward to me to reformulate the V-topology condition as follows: for $A \subseteq K$, if $0 \in \overline {A\cdot A}$ then $0 \in \overline{A}$. $\endgroup$ – Tim Campion Sep 27 '18 at 6:37

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.