Linked to this question and as a sequel to my answer of it.

Let $R$ be a topological (commutative, unital) ring and set $S$ be a submonoid of $(R,\times,1_R)$.

Let
$$
s_{frac}\ :\ R\times S\to S^{-1}R
$$
be the canonical surjection. We endow $R\times S$ with the product topology and $S^{-1}R$ with the quotient topology.

Then, due to the continuity of the following formulas
and their compatibility with $s_{frac}$

one has that $S^{-1}R$ is automatically a topological ring and that the arrow $s_{frac}$ solves the classical universal problem in the category of topological rings.

Q1) Is it known a way to compute the neighbourhoods of zero in $S^{-1}R$ and/or in $Frac(R)$ more explicit than the images, through $s_{frac}$, of the neighbourhoods of zero saturated by the fraction ring equivalence ?

(i.e. $s(s^{-1}(s(U\times (V\cap S)))$ where $U$ is a neighbourhood of zero in $R$ and $V$ an open set.)

Q2) Are there nice examples where the computation is not trivial, explicit (and good looking) ?

Q3) In case $R$ has no zero divisor and with $S=R\setminus \{0\}$, in TG.III.6 Exercice 27, Bourbaki gives a sufficient condition so that the topology induced by that of $S^{-1}R=Frac(R)$ on $R$ is the given topology (that $s_{frac}$ be open). Are there other examples ?

Some facts:

- If $R$ is Hausdorff and, if $S$ contains no zero divisor, then $S^{-1}R$ is Hausdorff.
- In the case when $R$ is a topological (commutative) integral domain, setting $S=R\setminus \{0\}$ $$ s_{frac}\ :\ R\times R'\to Frac(R) $$ one sees that the topological ring $Frac(R)$ is automatically a topological field (due to the continuity and compatibility of $(p,q)\to (q,p)$).

nonzero divisor,but I keep using the latter). Another example is $L^{\infty}(Y,\mu)$, whose classical ring of quotients (despite the algebra being non-separable in general) is its complete ring of quotients, and consists of measurable $f$ such that $\mu(f^{-1} \infty) = 0$. $\endgroup$ – David Handelman Mar 20 '17 at 22:4721more comments