Htykuut

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.

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 $f: Pi to mathbb Ncup {infty}$ 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.