Let $k$ be a number field and $\mathbf{A}_k$ the adele ring of $k$. What can be said about the Krull dimension of $\mathbf{A}_k$?

More generally, I do not know if something can be said about the Krull dimension of an infinite product of rings: is it possible to bound it above by the supremum of the dimensions of the rings?

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    This is just random thoughts. Note that $\mathbf{A}_k$ is not an infinite direct product itself. Namely, for a finite subset $S$ of the finite places of $k$ let us denote $$\mathbf{A}_{k,S}:=\left\{(x_v):x_v\in \mathcal{O}_{k_v}\text{ for }v\notin S\right\}\cong \prod_{v\in S}k_v\times \prod_{v\notin s}\mathcal{O}_{k_v}$$ then $$\mathbf{A}_k=\varinjlim_S \mathbf{A}_{k,S}$$ as topological rings. In particular, we deduce that $$\mathrm{Spec}(\mathbf{A}_k)=\varprojlim \mathrm{Spec}(\mathbf{A}_{k,s})$$ Now the transition maps are open embeddings (I think?) and the pices are all – Alex Youcis Dec 12 '19 at 15:32
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    infinite dimensional (e.g. see Theorem 3.3 of this https://www.jstor.org/stable/2154134?seq=5#metadata_info_tab_contents) and since the inverse limit is non-empty I think one can show that it's infinite-dimensional, but this is just idle speculation. – Alex Youcis Dec 12 '19 at 15:33
  • I would also like to point out that while this is an interesting idle curiosity I've never seen the spectrum of the adele ring be important. Is there a reason you ask? – Alex Youcis Dec 12 '19 at 15:50
  • @AlexYoucis: Your argument works. First, if $S\subseteq S'$ then $$\mathrm{Spec}(\mathbf{A}_{k,S})=\left(\coprod_{v\in S}\mathrm{Spec}(k_v)\right)\coprod\left(\coprod_{v\in S'-S}\mathrm{Spec}(\mathcal{O}_v)\right)\coprod\mathrm{Spec}(\prod_{v\not\in S'}\mathcal{O}_v)$$ and $$\mathrm{Spec}(\mathbf{A}_{k,S'})=\left(\coprod_{v\in S}\mathrm{Spec}(k_v)\right)\coprod\left(\coprod_{v\in S'-S}\mathrm{Spec}(k_v)\right)\coprod\mathrm{Spec}(\prod_{v\not\in S'}\mathcal{O}_v)$$ so that the transition map $\mathrm{Spec}(\mathbf{A}_{k,S'})\rightarrow \mathrm{Spec}(\mathbf{A}_{k,S})$ is an open immersion... – Gaussian Dec 12 '19 at 20:58
  • ... since it is given by the disjoint union of the identity on the components indexed by some $v\in S$, of the open immersions $\mathrm{Spec}(k_v)\rightarrow\mathrm{Spec}(\mathcal{O}_v)$ for $v\in S'-S$ and the identity on $\mathrm{Spec}(\prod_{v\not\in S'}\mathcal{O}_v)$. – Gaussian Dec 12 '19 at 20:59
  • @AlexYoucis: sorry, I did not mean that the whole argument works: just the fact that the transition maps are open immersions is true. I could further say that the inverse limit you wrote holds in the category of topological spaces (cf Proposition 8.2.9 in EGA IV, Troisième Partie), which means that the topological space of $\mathrm{Spec}(\mathbf{A}_k)$ is the inverse limit of the inverse system given by the $\mathrm{Spec}(\mathbf{A}_{k,S})$. – Gaussian Dec 12 '19 at 21:33

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