topology on units in a topological ring

Question

Let $R$ be a topological ring and $R^\times$ the group of units. I read that $R^\times$ with the subspace topology from R does not make the multiplicative inverse map continuous. (Apparently one has to additionally require the map $R^\times \to R \times R$ (given by $x \mapsto (x, x^{-1})$ is continuous). What is a simple example illustrating this point (i.e. where giving the subspace topology makes the inverse not continuous)?

For example for $R=\mathbb{Z}_p$ (p-adic numbers) I might be mistaken but I think the open neighborhoods of $1$ in $R^\times$ are $1+p^n\mathbb{Z}_p$, which is the subspace topology from $\mathbb{Z}_p$

Answer 1

That example isn’t obvious to come up with in the first place, but it’s somewhat important.

Take $A$ the ring of finite adeles of $\mathbb{Q}$, ie $A$ is the subring of $\prod_p{\mathbb{Q}_p}$ made with the families $(z_p)_p$ such that $z_p \in \mathbb{Z}_p$ for all but finitely many $p$.

However, $A$ does not have the topology induced by the product topology. Instead, we endow $A$ with the *restricted product topology, ie a basis of open subset is given by $\prod_{s \in S}{U_s} \times \prod_{p \in P}{\mathbb{Z}_p}$, where $S$ is any finite set of prime numbers, and $U_s \subset \mathbb{Q}_s$ is an open subset for each $s \in S$.

Now, $(z_p)_p \in A^{\times}$ iff for each $p$, $z_p \neq 0$ and for all but finitely many $p$, $z_p \in \mathbb{Z}_p^{\times}$. Let $i:A^{\times} \rightarrow A^{\times}$ be the inversion. Let $U=i^{-1}\left(\prod_p{\mathbb{Z}_p}\right)$.

If $U$ is an open subset of $A^{\times}$, then there is a finite set $S$ of primes, open subsets $1 \in U_s \subset \mathbb{Q}_s$ for each $s\in S$ such that $V=A^{\times} \cap \prod_{p \notin S}{\mathbb{Z}_p} \times \prod_{s \in S}{U_s} \subset U$. But consider the element $\zeta=(z_p)_p$ where $z_p=p$ for some prime $p \notin S$, and $z_q=1$ for every other prime. Then $\zeta \in V$ but $i(\zeta)_p=p^{-1} \notin \mathbb{Z}_p$ so that $\zeta \notin U$.

So $U$ isn’t open, thus $i$ isn’t continuous.

Answer 2

Let $R$ be an integral domain, not a field. Then it is easy to see that the set of all non-zero ideals form a neighborhood base of $0$ for a ring topology on its field of quotients. A straightforward calculation shows that inversion is continuous, if and only if the Jacobson radical (= intersection of all maximal ideals) does not equal $\{0\}$. Hence in the above mentioned example of p-adic numbers inversion is continuous, but for $\mathbb{Z}[X]$ it isn't.