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?](https://mathoverflow.net/questions/226660/is-k-x-y-a-topological-field)
and so, it seems that it is quite a confusing question to topologize a fractional field?



Any "natural" topology on fractional field of a topological ring?
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. 


