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Let $R$ be a topological integral domain. Let $K=\mathrm{Frac} R$. Is there any "natural" topology on $K$? Actually, since $K$ can be regarded as a quotient of $R\times R$ quotient some equivalence relation, so maybe we can equip $R\times R$ with the product topology, and that induces topology on $K$?

In particular, suppose $R$ has the complete $(p)$-adic topology, where $(p)$ is the ideal $pR$, then what does the induced topology on $K$ look like?

I did came across an old question, Is $k(\!(x,y)\!)$ a topological field? and so, it seems that it is quite a confusing question to topologize a fractional field?

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    $\begingroup$ You could check out Bourbaki, TG.III.6 Exercice 27 for some results on this. $\endgroup$ – Fred Rohrer Feb 28 '17 at 11:24
  • $\begingroup$ @FredRohrer Do you mean General topology, Chapter 3 (topological groups), section 6 (topological groups with operators.....)? There are only 26 exercises for that section....no Exercise 27.... $\endgroup$ – guestguest Feb 28 '17 at 11:44
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    $\begingroup$ @guestguest: I refer to the French edition from 1971 which contains said exercise. $\endgroup$ – Fred Rohrer Feb 28 '17 at 12:57
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    $\begingroup$ Not strange at all. The English translation is based on the 1966 edition. $\endgroup$ – Fred Rohrer Feb 28 '17 at 14:29
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    $\begingroup$ A difference between $A=k[[x]]$ and $A=k[[x,y]]$ ($k$ field) is that in the second case $A$ is not open in its fraction field. $\endgroup$ – YCor Mar 9 '17 at 6:26
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As said Fred Rohrer, exercise 27 exists in the French edition and seems to answer the question of the MO. Let $$ s_{frac}\ :\ R\times R'\to Frac(R) $$ ($R'=R\setminus \{0\}$) be the canonical surjection. The "quotient field equivalence" $\equiv_{frac}$ on $R\times R'$ is that given by $s_{frac}$ i.e. $$ (a,p)\equiv_{frac} (b,q) \Longleftrightarrow s_{frac}(a,p)=s_{frac}(b,q) \Longleftrightarrow aq=bp\ . $$ As a matter of fact, if the product topology on $R\times R'$ passes to quotient as a topology on $Frac(R)$ compatible with its field structure and inducing the given one on $R$, it is the finest of all topologies (a) compatible with the field structure (b) inducing the given topology on $R$. Bourbaki's exercise gives a sufficient condition ($\equiv_{frac}$ is an open equivalence relation on $R$) so that this occured (i.e. the quotient topology on $Frac(R)$ induces the given topology on $R$).

Update In the general case $S^{-1}R$ can also be constructed as the direct limit $\lim_\rightarrow R_t$ (thanks to David Handelman) where $$ R_t:=(t^{\mathbb{N}})^{-1}R\ ;\ t^{\mathbb{N}}:=\{t^n\}_{n\in \mathbb{N}} $$ the direct system being preordered by divisibility $$ s\prec t\Longleftrightarrow (\exists u\in S)(t=us) $$ one can check that the two constructions (by usual quotient or by direct limit) solve the same universal problem (algebraic and topological i.e. in the category of topological rings), their topology is therefore the same. The topology of $(t^{\mathbb{N}})^{-1}R$ could be easier to describe.

Hope this helps.

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  • $\begingroup$ So if $\mathbb Z$ is discrete, what topology do we get for $\mathbb Q$? Coarser than the usual topology? $\endgroup$ – Gerald Edgar Feb 28 '17 at 14:06
  • $\begingroup$ @GeralEdgar I might be wrong, but you get the quotient topology of $\mathbb{Z}\times (\mathbb{Z}\smallsetminus\{0\})$ by the action of $\mathbb{Z}\smallsetminus \{0\}$, which if I'm not mistaken is the discrete topology on $\mathbb{Q}$ $\endgroup$ – Denis Nardin Feb 28 '17 at 14:27
  • $\begingroup$ So "coarser" means what? Fewer open sets? $\endgroup$ – Gerald Edgar Feb 28 '17 at 14:31
  • $\begingroup$ @GeraldEdgar II didn't make the exercise yet (let me until thursday). They seem to say that this topology is extreme. Let me see. $\endgroup$ – Duchamp Gérard H. E. Feb 28 '17 at 15:17
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    $\begingroup$ @GeraldEdgar Since $\mathbb{Z}\times(\mathbb{Z}\smallsetminus\{0\})$ is discrete, its quotient by any equivalence relation is discrete. $\endgroup$ – Denis Nardin Mar 1 '17 at 17:07

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