41
$\begingroup$

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.

$\endgroup$
  • 11
    $\begingroup$ $\mathbb{A}_{K}$ 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). $\endgroup$ – DBS Jul 9 '13 at 7:22
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.