# group of units in a topological ring

I am looking at some notes on Adeles and Ideles by Pete Clark here, and puzzling over exercise 6.9 (page 6), that if the group of units $$U$$ in a topological ring is an open subset, then multiplicative inversion on $$U$$ is continuous. I am supposing the idea is that projection from the locus of $$x y = 1$$ onto $$R^\times$$ is an open map (whence the result would follow), but I'm just not seeing it -- perhaps I'm being dense.

Pete notes that this implies that a (Hausdorff) topological ring whose underlying ring is a field (I guess this is what he means by "topological field") has this property. What is additionally puzzling me is that most of sources I've looked at (e.g. Wikipedia), when defining topological field, explicitly stipulate that inversion should be continuous, but if Pete were correct then that would be redundant (since a topological ring that is a field must be either indiscrete or is Hausdorff, due to the fact that the closure of $$\{0\}$$ is an ideal, and inversion is trivially continuous for the indiscrete case).

Edit: Looking over the monograph Topological Fields by Seth Warner, who is expert in this area, one gets a strong suspicion from how the theorems are formulated that inversion need not be continuous even if the group of units is open, contrary to the assertion in Pete's exercise. (Sample statement, pp. 109-110: "If $$\mathcal{T}$$ is a Hausdorff ring topology on a commutative ring $$A$$ with identity for which $$A^\times$$ is open, then of all the ring topologies on $$A$$ weaker than $$\mathcal{T}$$ for which inversion is continuous there is a strongest $$\mathcal{S}$$, $$\mathcal{S}$$ is Hausdorff, and $$A^\times$$ is open for $$\mathcal{S}$$.") But frustratingly, no examples of this (where $$\mathcal{S}$$ is strictly weaker than $$\mathcal{T}$$) are given! Perhaps Google Books cut me off before that point was reached.

Anyway, I would be much obliged if someone would give an example of a field with a Hausdorff ring topology but where inversion is not continuous.

• Ah well, I found a very simple solution. But there's still some time left to claim your prize, before I go ahead and post my solution. Commented Aug 19, 2015 at 15:09
• Another example is the topological ring of the adeles of $\Bbb Q$ (see here). Commented Sep 14, 2017 at 19:14
• @Watson The example you're referring to is well-known (the idele topology is strictly finer than the subspace topology coming from the adeles), but it doesn't fit the text of the question because the set of ideles is not open in the space of adeles (any basic open has lots of zero divisors). The question at the end asks actually for an example of a field with a ring topology but not a field topology. Commented Sep 14, 2017 at 20:02

(I've never offered a bounty before; I hope it's not bad etiquette to post a solution now, even though technically we're in a "grace period". Since there has been no activity, I'm guessing it'll be alright.)

There is a counterexample, suggested in Warner's book (linked in the question), p. 113, involving a ring topology on the rational numbers $\mathbb{Q}$. Take basic neighborhoods of the origin to be ideals of $\mathbb{Z}$, i.e., subsets of $\mathbb{Q}$ of the form $n\mathbb{Z}$ where $n \neq 0$ is an integer. By taking lcm's, the collection of such sets is closed under finite intersections. Thus a basis of the topology consists of sets of the form $q + n\mathbb{Z}$ where $q \in \mathbb{Q}$. That this gives a ring topology (i.e., a topology for which subtraction and multiplication on $\mathbb{Q}$ are continuous) may be boiled down to three easily checked claims (cf. Warner, theorems 11.2, 11.3, and 11.4, pp. 78-79):

• For each ideal $I$ there is an ideal $J$ such that $J + J \subseteq I$ and $-J = J$ (obviously $J = I$ will do);

• For each ideal $I$ and $r \in \mathbb{Q}$, there is an ideal $J$ such that $rJ \subseteq I$ (if $r = p/q$ for integers $p, q$, then $J = qI$ will do);

• For each ideal $I$ there is an ideal $J$ such that $J \cdot J \subseteq I$ (again $J = I$ will do).

Also notice that $\{0\}$ is closed in this topology, because if $q \neq 0$ and we pick $n \in \mathbb{Z}$ with $|n| > |q|$, then the open neighborhood $q + n\mathbb{Z}$ of $q$ doesn't contain $0$.

Note reciprocation = multiplicative inversion is not continuous w.r.t. this topology. Indeed, for the neighborhood $1 + 2\mathbb{Z}$ of $1$, its inverse image under reciprocation is the set $\{\frac1{1 + 2n}: n \in \mathbb{Z}\}$, and no basic open neighborhood $1 + m\mathbb{Z}$ of $1$ is small enough to fit inside this set. Thus, we have exhibited a counterexample.

The link provided to Pete's notes is broken, but usually, these rings (Adeles) are built to be locally compact (that's kind of the idea behind the restricted product). It is known for example that the inverse operation is continuous for locally compact division rings (fields) On locally compact fields, Y.Otobe.

So Im geussing Pete exescise assumes local compactiss.

Generally, I think it is interesting to note that (under some mild conditions) any commutative locally compact ring is a restricted product of local fields and valuation rings Locally Compact Rings , Goldman and Sah.

• Pete told me privately that the exercise was indeed mistaken, and fixed the mistake with a credit to me. So he wasn't assuming local compactness, although that's a reasonable assumption to make in this context. Commented Nov 23, 2021 at 14:44
• Meanwhile, Pete's updated notes can be found here: alpha.math.uga.edu/~pete/8410FULL.pdf, where the emended exercise appears on page 100 out of 157. Commented Nov 23, 2021 at 15:01
• @user43208 Thanks! Commented Dec 4, 2021 at 14:45