# Topological fraction rings and fields

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 $(p,q)\to (q,p)$).
• The case $R=k[[x,y]]$, $k$ a field and $S^{-1}R$ is its fraction field, would be a good start ($R$ is not a neighborhood of 0 in its fraction field, if I'm correct). – YCor Mar 9 '17 at 6:29
• Do you want S closed? – David Roberts Mar 9 '17 at 6:53
• @DavidRoberts Well, I have no idea of the effect of $S$ closed. If you have some element of discussion, I would be happy to receive any ideas. – Duchamp Gérard H. E. Mar 9 '17 at 8:33
• @DavidRoberts [Do you want S closed?]---> Not necessarily. The facts I gave do not suppose $S$ closed. – Duchamp Gérard H. E. Mar 9 '17 at 23:54
• 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 $\mu(f^{-1} \infty) = 0$. – David Handelman Mar 20 '17 at 22:47

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 $\bar{\nu}$.