In a lot of the literature I read, especially stuff dealing with automorphic forms and representations, adelic (double) coset spaces are pretty ubiquitous. Frequently people seem to work only with finite adeles, and I have a few questions about some basic facts which seem implicit in the literature, but for which I can not find a reference (perhaps because they are self-evident to the authors). I should add that I believe I know the answer to (some of) these questions, but I'm not super confident about them, and am hoping that an expert could either confirm or refute my beliefs.

To set notation for the questions, let $K$ be a number field with integers $\mathscr{O}_K$, $\mathbf{A}_K$ its adele ring, and $\mathbf{A}_K^\times$ the unit group of $\mathbf{A}_K$ with its restricted direct product topology (so the idele group of $K$). The ring of finite adeles $\mathbf{A}_{K,f}$ can be defined as the restricted direct product $\prod_{v\nmid\infty}^\prime K_v$ of the non-Archimedean completions of $K$ with respect to their integer rings $\mathscr{O}_v$. We also have the group of finite ideles (I am not sure if this is standard terminology or not) $\mathbf{A}_{K,f}^\times$, which is the restricted direct product $\prod_{v\nmid\infty}K_v^\times$ with respect to the unit groups $\mathscr{O}_v^\times$. There is a natural continuous, surjective homomorphism $\mathbf{A}_K^\times\rightarrow\mathbf{A}_{K,f}^\times$ whose kernel is the Archimedean component $\prod_{v\mid\infty}K_v^\times$ embedded in $\mathbf{A}_K^\times$ with $1$'s at the non-Archimedean places. My first question:

(1) Is the continuous group isomorphism $\mathbf{A}_K^\times/\prod_{v\mid\infty}K_v^\times\cong\mathbf{A}_{K,f}^\times$ an isomorphism of topological groups, i.e., bicontinuous?

I believe the answer to (1) is "yes" because it seems to me that the natural map $\mathbf{A}_K^\times\rightarrow\mathbf{A}_{K,f}^\times$ takes the standard basic open subgroups around the identity of the source to the standard basic open subgroups around the identity of the target (I'm referring to the $S$-idele groups where $S$ is finite and contains the Archimedean primes).

Assuming I'm right about (1), and that the isomorphism of (1) is compatible with the diagonal image of $K^\times$ in both sides of the isomorphism, my impression is that it should (via the reciprocity law) set up a bijection between closed subgroups of $\mathbf{A}_{K,f}^\times/K^\times$ and abelian extensions of $K$ which are unramified at the Archimedean primes, and so, if I'm only interested in such extensions, I can restrict attention to the finite ideles. I worry a bit, however, that the fact that $K^\times$ is not usually discrete in $\mathbf{A}_{K,f}^\times$ could cause problems. I'm really mainly interested in the case where $K$ is totally imaginary though, and in this case $K^\times$ is discrete in the finite ideles, so I think everything works.

(2) Is my impression about abelian extensions unramified at the infinite primes correct, at least when $K$ is totally imaginary?

My last question is purely algebraic. The notation $\hat{K}^\times$ comes up frequently in the papers I read, and it is generally defined as $(K\otimes_\mathbf{Z}\hat{\mathbf{Z}})^\times$. This can be identified with $(K\otimes_\mathbf{Q}(\mathbf{Q}\otimes_\mathbf{Z}\hat{\mathbf{Z}}))^\times$. I can verify that the natural ring map $\mathbf{Q}\otimes_\mathbf{Z}\hat{\mathbf{Z}}\rightarrow\prod_p\mathbf{Q}_p$ induces an isomorphism onto $\mathbf{A}_{\mathbf{Q},f}$ (injectivity holds by thinking of tensoring with $\mathbf{Q}$ as localization). So when I tensor with $K$, I get an injection $\hat{K}\rightarrow\prod_p(K\otimes_\mathbf{Q}\mathbf{Q}_p)\cong\prod_p\prod_{v\mid p}K_v$, where the second isomorphism is standard and canonical. It's clear to me that the image of this map lands in $\mathbf{A}_{K,f}$. My final question is:

(3) Is the natural injective map $\hat{K}\rightarrow\mathbf{A}_{K,f}$ an isomorphism of rings, i.e., surjective?

I think the answer is "yes," and I can actually prove that the natural map $K\otimes_{\mathscr{O}_K}\prod_{v\nmid\infty}\mathscr{O}_v\rightarrow\mathbf{A}_{K,f}$ is an isomorphism (this is the analogue of the isomorphism $\mathbf{Q}\otimes_\mathbf{Z}\hat{\mathbf{Z}}\cong\mathbf{A}_{\mathbf{Q},f}$), so if I knew that $\prod_v\mathscr{O}_v$ could be identified with $\mathscr{O}_K\otimes_{\mathbf{Z}}\hat{\mathbf{Z}}$ via the natural map, I'd be done. By "the natural map" I mean the composite of the isomorphism $\mathscr{O}_K\otimes_\mathbf{Z}\hat{\mathbf{Z}}\rightarrow\prod_p(\mathscr{O}_K\otimes_\mathbf{Z}\mathbf{Z}_p)$ (the fact that this is an isomorphism can be verified by choosing a $\mathbf{Z}$-basis for $\mathscr{O}_K$) and the map $\prod_p(\mathscr{O}_K\otimes_\mathbf{Z}\mathbf{Z}_p)\rightarrow\prod_p\prod_{v\mid p}\mathscr{O}_v$. So I guess it all boils down to whether or not the map $\mathscr{O}_K\otimes_\mathbf{Z}\mathbf{Z}_p\rightarrow\prod_{v\mid p}\mathscr{O}_v$ is an isomorphism. This I'm not sure about.


I learnt a lot about this stuff from Gross's Algebraic Modular Forms paper and the references therein; he explains a lot of things very clearly and gives a handy guide to the 1960's literature.

(1) Yes, this morphism is a topological isomorphism and the proof is immediate (just write down a basis of open sets). The more general statement is that restricted direct product is associative, so if you have a set of topological spaces $\{X_i\}_{i \in I}$, each endowed with a choice of an open subset $\{Y_i\}_{i \in I}$, and $I = \bigsqcup_{j \in J} I_j$ is a partition of the index set $I$, then there is a topological isomorphism $$ \sideset{}{'}\prod_{i \in I} X_i = \sideset{}{'}\prod_{j \in J}\left(\sideset{}{'}\prod_{i \in I_j} X_i\right), $$ where the restricted direct product means "elements of the product that are in the distinguished subset for all but finitely many indices". This formulation is now so hyper-general that it must be either false or completely trivial, and fortunately it's the latter.

In particular taking $I$ to be all primes of $K$, $J = \{0, 1\}$, $I_1$ = finite primes, $I_2$ = infinite primes, then we see that $$ \mathbb{A}_K^\times = \mathbb{A}_{K,f}^\times \times \prod_{v \mid \infty} K_v^\times $$ as topological groups.

(2) This is almost right but one thing you say is a little wide of the mark. The image of $K^*$ in the finite ideles is discrete only when K is $\mathbb{Q}$ or imaginary quadratic, since the intersection of the image with the compact set $\widehat{O}_K^\times$ is the global unit group, which is finite only in these two cases. But you can just take its closure and everything works.

(3) The result you want about local unit groups is true, and standard (see e.g. Exercise 4 in $\S$ II.8 of Neukirch's Algebraic Number Theory). So you can treat $\widehat K$ as just another name for $\mathbb{A}_{K, f}$.

  • $\begingroup$ Thanks David! This is very helpful. $\endgroup$ – Keenan Kidwell Nov 9 '12 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.