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 $s_{frac}$

  • $((a,p),(b,q))\to (aq+bp,pq)$
  • $((a,p),(b,q))\to (ab,pq)$
  • $(a,p)\to (-a,p)$

    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 $(p,q)\to (q,p)$).
  • $\endgroup$
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      $\begingroup$ Do you want S closed? $\endgroup$ Mar 9, 2017 at 6:53
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      $\begingroup$ It might make the localisation have a slightly nicer topology... $\endgroup$ Mar 9, 2017 at 19:34
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      $\begingroup$ @DavidRoberts [Do you want S closed?]---> Not necessarily. The facts I gave do not suppose $S$ closed. $\endgroup$ Mar 9, 2017 at 23:54
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      $\begingroup$ For more on rings of quotients of C(X), there is a nice little monograph by Gilman, Jerison [the elder], and Lambek, mid-1960s, I think. Yes, $f^{-1}0$ having empty interior is exactly the condition for $f$ to be a non-zerodivisor (better expression than 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 $\mu(f^{-1} \infty) = 0$. $\endgroup$ Mar 20, 2017 at 22:47
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      $\begingroup$ oops, it's Fine, Gilman, and Lambek; downloadable at math.mcgill.ca/barr/papers/fgl.pdf $\endgroup$ Mar 21, 2017 at 4:08

    1 Answer 1


    A partial answer to (Q1) with a class of examples for (Q2) (please double-check me or ask me details if needed).

    We note $$ i_{R}^S\ :\ R\to S^{-1}R $$ the canonical map, defined by $i_{R}^S(x):=s_{frac}((x,1))$ ($i_{R}^S$ is into iff $S$ contains no annihilator).

    First remark that, due to the fact that $i_{R}^S$ solves a universal problem, if there exists some topological ring $T$ and an arrow (continuous ring morphism $f\ :\ R\to T$ with $f(S)\subset T^{\times}$) such that $f^{-1}(\mathcal{T}_T)=\mathcal{T}_R$ (the inverse image of the topology of $T$ is the exactly the given topology of $R$), then $$ (i_{R}^S)^{-1}(\mathcal{T}_{S^{-1}R})=\mathcal{T}_R\ . $$
    For (Q2), if the topology of $R$ is given by a valuation $\nu$ (in the general sense of wikipedia and Bourbaki, i.e. a mapping $\nu\ :\ R\to \Gamma\sqcup \{\infty\}$ where $\Gamma$ is some totally ordered abelian group, this includes Malcev Neumann series on $\Gamma$) then one can check that $\nu$, extended to $R\times S$ by $\nu(a,s)=\nu(a)+(-\nu(s))$, passes to quotient as an extension ($R$ has no zero divisor) of $\nu$ (call it $\bar{\nu}$) and that the topology on $S^{-1}R$ is given by $\bar{\nu}$.


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