Idle thoughts about module topology and adeles

Stream: maths

[Kevin Buzzard]

If $K$ is a number field then a cheap definition of the finite adeles of $K$ is $K\otimes_{\Z}{\widehat{\Z}}$ and a cheap definition of the infinite adeles of $K$ is $K\otimes_{\mathbb{Q}}\R$. Writing it like this, it's really clear that the finite adeles of $K$ are some kind of "nonarchimedean completion" of $K$ and the infinite adeles of $K$ are some kind of "archimedean completion" of $K$, so the adeles of $K$, defined as the product of the finite and infinite adeles, are some kind of "universal" completion of $K$.

But actually we're missing something here, because the adeles are a topological ring and the topology is really important. Where is the topology coming from? Well the infinite adeles of $K$ are a finite-dimensional real vector space and it's of course equipped with the usual topology that a finite-dimensional real vector space has, which is the $\R$-module topology in the sense of docs#moduleTopology . [The module topology is a canonical topology on any module over a topological ring; it's the finest topology making addition and scalar multiplication continuous (you can also use negation but this follows from the fact that it's multiplication by $-1$). ]

So the infinite adeles have a canonical topology and it's the right thing. Can we do the same hack with the finite adeles?

My first reaction was "unlikely", because most of the interesting results I've proved about the module topology are for finite modules, and $K$ is not a finite $\Z$-module so you wouldn't expect the finite adeles to be a finite $\widehat{\Z}$-module and indeed they are not.

But idly messing around today, I realised that actually the finite adeles of $\mathbb{Q}$ could well have the $\widehat{\Z}$-topology. This is because $\widehat{\Z}$ is an open subring of the finite adeles with the usual topology and, although I haven't formalised it yet, I like the look of the following argument:

Theorem: if $R$ is an open subring of a topological ring $A$, then $A$ has the $R$-module topology.

Proof (not checked carefully): Let $\tau$ denote $A$'s topology and let $m$ denote the $R$-module topology on $A$. The module topology is defined as an Inf, meaning that the identity map from $(A,m)$ to $(A,\tau)$ is continuous. Furthermore the $R$-module topology on $R$ equals $R$'s topology (this is docs#TopologicalSemiring.toIsModuleTopology ) and all $R$-linear morphisms between modules with the module topology are continuous (this is docs#IsModuleTopology.continuous_of_linearMap ) so the inclusion $R\to (A,m)$ is continuous. I think this must be enough -- it surely shows that the neighbourhood filters of $0$ on $(A,m)$ and $(A,\tau)$ are equal, and so by some nonsense which Patrick did for the perfectoid project the two topologies are equal (you can use addition to translate the filter around to any point; note that this doesn't work for semirings because you need subtraction).

If that's right, then just like we topologised $K\otimes_{\mathbb{Q}}\R$ with the $\R$-module topology, we could topologise $K\otimes_{\Z}\widehat{\Z}$ with the $\widehat{\Z}$-module topology. Does this give the usual topology? I don't know but it might not be hard. The proof above doesn't apply because $\widehat{\Z}$ is not open in the finite adeles of $K$ if $K\not=\mathbb{Q}$, but on the other hand I do know that the finite adeles of $K$ are finite over the finite adeles of $\mathbb{Q}$ (because $K$ is finite over $\mathbb{Q}$) and that the finite adeles of $K$ do have the (finite adeles of $\mathbb{Q}$)-module topology, so if there's some kind of transitivity lemma there which I haven't thought carefully about then maybe we win. Alternatively one could go via the profinite completion of the integers of $K$ which might be easier; I don't know if it's easier for the open subring to be finite over a subring or whether it's easier to be finite over something with an open subring. Is it true in general that if $A\subseteq B\subseteq C$ are topological rings (with all topologies on subsets being induced) and $C$ has the $B$-module topology and $B$ has the $A$-module topology then $C$ has the $A$-module topology? Or if $A\subseteq B$ are a top ring and a subring, and if $B$ has the $A$-module topology, then do the $A$-module and $B$-module topologies on a $B$-module coincide?

[Scott Carnahan]

Regarding the finiteness problem, would it help to construct a "smaller finite adeles" ring, replacing $K$ with $\mathcal{O}_K$? The product with infinite adeles still gives you the usual adeles ring.

[Scott Carnahan]

Never mind, I see you basically wrote that near the end.

[Kevin Buzzard]

Yeah this might be the key.