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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?

Gaussian
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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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Let's treat $K=\Bbb Q$ first. It's well-known that $\Bbb A_{\Bbb Q}\cong (\Bbb Q \otimes_{\Bbb Z}\widehat{\Bbb Z}) \times \Bbb R$. We can just focus on $\Bbb Q \otimes_{\Bbb Z}\widehat{\Bbb Z}$ for the dimension. I'll take for granted the fact that $\widehat{\Bbb Z}$ has infinite Krull dimension.

Consider the terminal morphism $\mathrm{Spec}(\widehat{\Bbb Z}) \to \mathrm{Spec}(\Bbb Z)$. We can decompose $\mathrm{Spec}(\widehat{\Bbb Z})$ as a set (not as a space) as the disjoint union of the fibers of this morphism:

$\mathrm{Spec}(\widehat{\Bbb Z})=\mathrm{Spec}(\Bbb Q \otimes \widehat{\Bbb Z})\sqcup \bigsqcup_p \mathrm{Spec}(\Bbb F_p \otimes_{\Bbb Z} \widehat{\Bbb Z})$

Now $\Bbb F_p \otimes_{\Bbb Z} \widehat{\Bbb Z}\cong \Bbb F_p$, so for each $p$, the fiber $\mathrm{Spec}(\Bbb F_p \otimes_{\Bbb Z} \widehat{\Bbb Z})$ is just a point, which is closed because $(p) \in \mathrm{Spec}(\Bbb Z)$ is closed. We can also think of $\Bbb Q \otimes_{\Bbb Z} \widehat{\Bbb Z}$ as the the localization $S^{-1}\widehat{\Bbb Z}$, where $S=\Bbb N$. This implies a canonical homeomorphism: $$\mathrm{Spec}(\Bbb Q \otimes_{\Bbb Z} \widehat{\Bbb Z})\cong\{\mathfrak{p} \in \mathrm{Spec}(\widehat{\Bbb Z}) \mid S \cap \mathfrak{p} = \varnothing\}$$

To say that $\widehat{\Bbb Z}$ has infinite Krull dimension is to say that for every $n$, there's a prime ideal in $\widehat{\Bbb Z}$ of height at least $n$. Fix $n \in \Bbb N$ and take a prime ideal $\mathfrak{q}$ of height at least $n+1$, so that we obtain a chain of prime ideals $\mathfrak{p}_0 \subsetneq \ldots \subsetneq \mathfrak{p}_n \subsetneq \mathfrak{q}$. Clearly it is impossible in this situation that $\mathfrak{p}_i$ is maximal for any $i$. But as we have seen before, the complement of $\mathrm{Spec}(\Bbb Q \otimes \widehat{\Bbb Z})$ inside $\mathrm{Spec}(\Bbb Z)$ consists of closed points.

From a standard lemma about prime ideals in localizations, this means that the chain $\mathfrak{p}_0 \subsetneq \ldots \subsetneq \mathfrak{p}_n$ gives rise to a chain $S^{-1}\mathfrak{p}_0 \subsetneq \ldots \subsetneq S^{-1}\mathfrak{p}_n$ of prime ideals in $S^{-1}\widehat{\Bbb Z}=\Bbb Q \otimes_{\Bbb Z} \widehat{\Bbb Z}$. As $n$ was arbitrary, this implies that $\Bbb Q\otimes_{\Bbb Z} \widehat{\Bbb Z}$ has infinite Krull dimension.

For a general number field $K$, note that $\Bbb Q\hookrightarrow K$ is an integral morphism and $\Bbb Z \to \widehat{\Bbb Z}$ is flat, so that $\Bbb Q \otimes_{\Bbb Z} \widehat{\Bbb Z} \hookrightarrow K \otimes_{\Bbb Z} \widehat{\Bbb Z}$ defines an integral extension of rings. Any integral extension of rings preserves the Krull dimension. $\Bbb A_K$ is a direct product of $K \otimes_{\Bbb Z}\widehat{\Bbb Z}$ with a finite number of fields. Thus $\Bbb A_K$ has infinite Krull dimension.

Lukas Heger
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