Closed ideals of ring of integral Adeles

Question

If $\mathbb{A}_\mathbb{Z} = \mathbb{R} \times \hat{\mathbb{Z}}$, where $\hat{\mathbb{Z}} = \prod_{p \ prime} \mathbb{Z}_p$ denotes the profinite integers, is there anything that can be said about its closed ideals? In general there'd be no reason to suppose they would all be products of ideals in each component, but here's why something nice like that might happen:

• If J is a closed ideal of the profinite integers, then in particular, it is a closed subgroup and therefore of the form $\prod_{p \ prime} p^{n_p}\mathbb{Z}_p$ where each $n_p \in \mathbb{Z}_0^+ \cup \{+ \infty\}$ (this can be shown through Pontryagin duality, see Closed subgroups of $\hat{\mathbb{Z}}\cong\prod_{p}\mathbb{Z}_{p}$ as subsets?).

• The closed subgroups of the ring $\mathbb{Z}_p \times \mathbb{Z}_p$ won't necessarily have to be products of closed subgroups of the p-adic integers. This is related to the fact that $C_p \times C_p$ isn't isomorphic to $C_{p^2}$. So in a certain sense, what happens in the profinite integers turns out to be quite special (although admittedly, I don't even know if the closed ideals of say, $\mathbb{Z_2} \times \mathbb{Z_2}$, are of the mentioned form. If this happens to be the case, then this observation is meaningless with respect to this question.)

• All proper ideals of the real numbers are trivial.

With this in mind, are all the closed ideals of $\mathbb{A}_\mathbb{Z}$ "trivial" in the sense of being cartesian products of ideals?

Answer 1

An ideal in a product of two rings is always just a product of ideals on the factors (see here for instance). This isn't true for infinite products, but it does remain true if you restrict to closed ideals. To be precise:

Theorem: Let $(R_i)$ be a family of topological rings, let $R=\prod_i R_i$, and suppose $I\subseteq R$ is a closed ideal. Then $I=\prod I_i$ for some family $(I_i)$ of closed ideals $I_i\subseteq I$.

Proof: Identify $R_i$ with the subset of $R$ where all the coordinates except the $i$th are $0$. Let $I_i=I\cap R_i$; then $I_i$ is a closed ideal in $R_i$. Moreover, $I_i$ is also equal to the image of $I$ under the projection to $I_i$, since if $x\in I$ you can consider $e_ix\in I_i$ where $e_i\in R$ is the element with $i$th coordinate $1$ and all other coordinates $0$. So $I\subseteq \prod I_i$. For the reverse inclusion, note that $I_i\subseteq I$ for each $i$ and so $\bigoplus I_i\subseteq I$. But $\bigoplus I_i$ is dense in $\prod I_i$, so since $I$ is closed it contains all of $\prod I_i$.