# 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.

• $\mathbb{A}_{K}$ is a direct limit of rings so Spec() will be an inverse limit of $Spec(\mathbb{A}_{S})$ You may find a better answer in the paper bu Brian Conrad: Some notes on topologizing the adelic points of schemes, unifying the viewpoints of Grothendieck and Weil. at Stanford (do a google search). – DBS Jul 9 '13 at 7:22

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:

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

2. 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:

1. 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 $$Spec(\mathbb Z_p) \hookrightarrow Spec(\hat{\mathbb Z})$$.

2. 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.

3. 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.

4. 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.