What is Spec of the Adeles? Let $K$ be a global field and $A_K$ the ring of adeles. 

What are the prime ideals of $A_K$?

I have been told that a full proof of this is quite subtle, but have been unable to find a reference for this result.
 A: Let me do the case of the integral adeles $\mathbb A = \mathbb A_\mathbb Q$ -- I'm not feeling quite up to the task of a general number field. I think the following is about as explicit as one can be about the points of $Spec(\mathbb A)$, but perhaps there's more to be said about the topology.
Let $\Pi = \{2,3,5,\dots\}$ be the set of integral primes. Recalling that $\mathbb A = \mathbb Q \otimes \hat{\mathbb Z} \times \mathbb R$ where $\hat{\mathbb Z} = \prod_{p \in \Pi}\mathbb Z_p$, the following gets us most of the way to our goal:
Theorem: A radical ideal $I \in Spec(\hat{\mathbb Z})$ is specified by two pieces of data:


*

*a filter $\mathcal F$ on the set of primes.

*a subset $F \subseteq (\mathbb N \cup \{\infty\})^\Pi$ such that for $f,g: \Pi \to \mathbb N \cup \{\infty\}$,


*

*If $f \lesssim_\mathcal F g$ and $f \in F$, then $g \in F$.

*If $f,g \in F$, then $\min(f,g) \in F$.
Here, $f \lesssim_\mathcal F g$ means that there is a constant $C> 0$ such that $\{p \in \Pi \mid f(p) \leq C g(p)\} \in \mathcal F$.
Explicitly, the ideal $I = I(\mathcal F, F)$ corresponding to this data is:
$$I(\mathcal F, F) = \{x \in \hat{\mathbb Z} \mid \{p \mid (v(x))_p \in F\} \in \mathcal F\}$$
where $v(x): \Pi \to \mathbb N \cup \{\infty\}$ sends $p \in \Pi$ to the $p$-adic valuation $v_p(x_p)$ of $x_p$. We have $I(\mathcal F, F) \subseteq I(\mathcal G, G)$ if and only if $\mathcal F \subseteq \mathcal G$ and $F \subseteq G$.
The ideal $I(\mathcal F, F)$ is prime if and only if $\mathcal F \in \beta(\Pi)$ is an ultrafilter.
This gives a complete description of the points of $Spec(\hat{\mathbb Z})$ and a basis for its topology. To get $Spec(\mathbb A)$, throw out the points $I(\uparrow \{p\}, F)$ for $p \in \Pi$, where $F$ consists of those functions $f: \Pi \to \mathbb N\cup \{\infty\}$ with $f(p) \geq 1$ (to localize at $\mathbb Q$), and add a point for $Spec(\mathbb R)$.

Notes:


*

*If $\mathcal F = \uparrow\{p\}$ is a principal ultrafilter at $p \in \Pi$, then there are exactly two points of the form $I(\uparrow\{p\}, F)$; these are the two points in the image of the inclusion $Spec(\mathbb Z_p) \hookrightarrow Spec(\hat{\mathbb Z})$.

*If $\mathcal F$ is a nonprincipal ultrafilter, then the points of the form $I(\mathcal F, F)$ are exactly those in the image of the map $Spec(\hat{\mathbb Z}/\mathcal F) \to Spec(\hat{\mathbb Z})$ where $\hat{\mathbb Z}/\mathcal F = \prod_{p \in \Pi} \mathbb Z_p / \mathcal F$ is the ultraproduct.

*If $\mathcal F$ is a nonprincipal ultrafilter, then after modding out by $\mathcal F$, the functions $f: \Pi \to \mathbb N \cup \{\infty\}$ form a complete dense linear order of cardinality continuum. With a little more work (relating this to the set of functions $\Pi \to \mathbb R_{>0}$ and using the fact that any complete dense linearly ordered abelian group is isomorphic to $\mathbb R$), we see that the collection of points $I(\mathcal F, F) \in Spec(\hat{\mathbb Z})$ with the same ultrafilter part are homeomorphic to the half-open interval $(1,\infty]$ with the open-lower-interval topology.

*Thus $Spec(\hat{\mathbb Z})$ consists of


*

*a copy of $\beta(\Pi)$, the space of ultrafilters on the primes $\Pi$, corresponding to points $I(\mathcal F, F)$ where $F$ contains only functions which are constant at $\infty$ for some $T \in \mathcal F$ (quotienting $\hat{\mathbb Z}$ by one of these ideals is exactly taking the ultraproduct $\prod_{p \in \Pi} \mathbb Z_p / \mathcal F$).

*for each isolated point of $\beta(\Pi)$, an additional point connected to it (from (2) above). These points are discarded in $Spec(\mathbb A)$.

*for each non-isolated point of $\beta(\Pi)$, an interval emanating from it (from (4) above) in the open-lower-interval topology.
However, the topology is more complicated when one looks at more than one nonprincipal ultrafilter at a time.

Sketch of Proof:
If $I$ is an ideal, let $\mathcal F(I) = \{S \subseteq \Pi \mid z_S \in I\}$ where $(z_S)_p = \begin{cases} 0 & p \in S \\ 1 & p \not \in S \end{cases}$. It's not hard to see that $\mathcal F(I)$ is a filter, and an ultrafilter if $I$ is prime. Moreover, the ideal $(z_S \mid S \in \mathcal F(I))$ is contained in $I$. Then set $F(I) = \{f: \Pi \to \mathbb N \cup \{\infty\} \mid \exists x \in I,\, v(x) = f\}$. It's not hard to see that $F(I)$ satisfies the properties above, and that $I$ is generated by $\{(p^{f(p)})_{p \in \Pi} \mid f \in F(I)\}$. Conversely, it's easy to check that $I(\mathcal F, F)$ is a radical ideal, prime if and only if $\mathcal F$ is an ultrafilter.
