 Let’s step away from $\zeta$-function stuff for a bit, and talk about something different. 1 In an earlier post, I mentioned these local fields like $\Q_p$ that are useful for studing things “one prime at a time” 2 (whatever that means). Corresponding to this local fields, one also has global objects (e.g. $\Q$) from which they arise, but in some sense, these global objects (i.e. global fields) don’t have all the information of the local objects readily available 3. Because of this, it may be nice to consider different global objects that combine all the local ones in a more straightforward manner.

# Definitions

Let $K$ be a global field. Technically, this means that $K$ is a number field or a function field of a curve over $\F_q$ (i.e. $K/\F_q(t)$ is finite), but the example I’ll have in mind if of $K$ as a number field. At some point I may explicitly say to let $K$ be a number field to simplify things, but know that most (all?) of what I do can be done for a general global field.

We want to construct the adele ring of $K$ which morally is just the topological ring

where $v$ ranges over all places of $K$, and $K_v$ is the completion of $K$ at $v$. However, this product is stupid-big (so maybe not the easiest thing to work), and doesn’t reflect some of the nice finiteness properties of global fields (e.g. the valuation of $x\in K$ is zero for almost all places). Because of this, we’ll replace it with a so-called restricted product.

Fix some set $I$ of indices, and some locally compact Hausdorff groups (or rings or fields) $G_i$ with compact, open (hence closed (!)) subgroups $H_i\le G_i$. The restricted (direct) product of the $G_i$ with respect to the $H_i$ is $$\prodp_{i\in I}(G_i,H_i):=\bracks{(g_i)\in\prod_{i\in I}G_i:g_i\in H_i\text{ for all but finitely many }i}.$$ We make this a topological group, but not by giving it the subspace topology it inherits from the direct product of the $G_i$. Instead, a neighborhood base of the identity consists of sets of the form $\prod N_i$ where $N_i$ is a neighborhood of $1\in G_i$ for all $i$, and $N_i=H_i$ for all but finitely many $i$.

Given this, we define the finite adele ring of $K$ (or ring of finite adeles) to be the (topological) ring

where the notation $v\nmid\infty$ means we’re ranging only over finite (i.e. non-archimedean) places, and $\ints v\subset K_v$ is the ring of integers (i.e. elements of norm at most 1). The adele ring of $K$ is obtained from the finite adeles by throwing in the infinite places. In other words, it is

Since there are only finitely many infinite places 4, if we let $\ints v=K_v$ for $v\mid\infty$, we could have just defined 5

Now that we know what the adele ring is, a few remarks about why this restricted direct product is nicer than the ordinary direct product. First, topologically speaking, $\prod_vK_v$ is not locally compact 6 essentially because the product topology requires open sets to be entire spaces in all but finitely many factors. However,$\dots$

Any restricted product $$\prodp_{i\in I}(G_i,H_i)$$ is locally compact.
Let $S\subset I$ be finite, and consider the subgroup $$G_S=\prod_{v\in S}G_v\by\prod_{v\not\in s}H_v.$$ Finite products of locally compact spaces are locally compact (and arbitrary products of compact spaces are compact), so $G_S$ is locally compact in the product topology. However, since $S$ is finite, the product topology on $G_S$ coincides with the subspace topology it inherits from the restricted direct product, so $G_S$ is locally compact there as well. Since every element of the restricted product belongs to a set of this form, it is locally compact.

This means that $\A_K$ is the product of finitely many locally compact spaces, so it is itself locally compact. In this way, it is not as stupid big as the ordinary direct product even though it looks massive at a glance. It’s also worth noting that $K\into\A_K$ as a discrete subgroup via the diagonal (algebraic) embedding

since $x$ has zero valuation at all but finitely many places 7. Elements of the image of this embedding are sometimes referred to as “principal adeles.” The embedding $K\into\A_K$ should be thought of as an analouge of $\Z\into\R$ (e.g it’s discrete 8 and we’ll later see that $\A_K/K$ is compact). The unit group $\units\A_K$ of the adele group is also important.

Prove that $$\units\A_K=\prod_{v\mid\infty}\units K_v\by\prodp_{v\nmid\infty}(\units K_v,\units{\ints v})$$ as groups. The right hand side is denoted $\I_K$, and is called the idele group of $K$. It's topology as a restricted direct product is different (stronger) than its topology as a subgroup of $\A_K$.

We similarly have a diagonal embedding $\units K\into\I_K$. Finally, we can extend the absolute value $\nabs_v$ on $K_v$ to $\A_K$ via $\abs{x}_v=\abs{x_v}_v$ for $x\in\A_K$, and then we can combine these to define a global absolute value 9

on $\A_K$ which converges since $\abs{x_v}_v\le1$ for all but finitely many $v$.

# Basic Properties

Fix a global field $K$. Let $\A_\omega=\prod_{v\mid\infty}K_v\by\prod_{v\nmid\infty}\ints v$. Then, $$\A_K=K+\A_\omega\text{, and }K\cap\A_\omega=\ints K$$ where $\ints K$ is the ring of integers of $K$ (elements with absolute value $\le1$ at all places).
Here, $K$ is embedded diagonally into $\A_K$. We want to show that given any $x\in\A_K$, there's some $\mu\in K$ such that each component of the difference $x-\mu$ is a local integer. Let $\mfp\subset\ints K$ be prime, and write $\mfp\cap F=(p)$ where $F\subseteq K$ is $\Z$ or $\F_q[t]$. Then, multiplying by $p$ will reduce the $\mfp$-adic absolute value of $x$. Since $\abs x_v\le1$ for all but finitely many places, this means that there's some $m\in F$ such that $mx$ is integral at all finite primes. Let $\{\mfp_1,\dots,\mfp_r\}$ be the set of primes of $K$ (i.e. prime ideals of $\ints K$) that divide (i.e. contain) $m$, and let $n_1,\dots,n_r$ be naturals such that $\mfp_j^{n_j}\nmid(m)$ (i.e. $n_j>v_{\mfp_j}(m)$), then the Chinese remainder theorem let's us find some $\lambda\in\ints K$ such that $$\lambda\equiv mx_j\pmod{\mfp_j^{n_j}},$$ where $x_j$ is the component of the adele $x$ corresponding to $\mfp_j$. Let $\mu=\lambda/m$, and note that $x-\mu=\inv m(mx-\lambda)$ is integral at each of the primes $\mfp_j$ (since we chose $n_j$ large). At other primes, its absolute value is the same as $mx-\lambda$'s so it's integral everywhere. Hence, we win.

One can also proove

Fix a finite place $v_0$ on a global field $K$, and let $\A_{K,v_0}=\prodp_{v\neq v_0}(K_v,\ints v)$ where $v$ ranges over all finite places. Then, $K$ is dense in $\A_{K,v_0}$.

but doing so requires saying the words “Haar measure,” and I’d rather not get into that, so I’ll skip the proof of this fact 11. If you recall from before, $K$ is discrete in $\A_K$, and so certainly not dense. This results says that if we remove a single (finite) place from $\A_K$, then $K$ goes from being discrete to being dense!

For an arbitrary Dedekind domain $A$, the weak approximation theorem says that for any finite set of primes $\mfp_i$ along with integers $e_i\ge0$, there exists some $a\in A$ such that $v_{\mfp_i}(a)=e_i$ for all $i$. Here, $v_{\mfp_i}$ is the $\mfp_i$-adic valuation. Sincr rings of integers of global fields are Dedekind domains, you may want to take a minute to think about how strong approximation compares/relates to weak approximation.

We’ll next take a look at how $\A_E$ is related to $\A_K$ when $E/K$ is a(n) (finite) 12 extension of global fields.

Fix some (finite, separable) extensions $E/K$, and fix a place $v$ on $K$. Let

where $w\mid v$ means that $w$ is a place on $E$ which restricts to $v$ on $K$. We can use to notation to build $\A_E$ from $K$ as the following lemma illustrates.

Let $E/K$ be a finite, spearable extension. Then, $$\A_E\cong\prod_{v\mid\infty}E_v\by\prodp_{v\nmid\infty}(E_v,\wt{\ints v}),$$ where $v$ ranges over places of $K$.
It is clear that $$\prod_{v\mid\infty}E_v\cong\prod_{w\mid\infty}E_w,$$ where the left product ranges over (archimedean) places of $K$ and the right ranges over (archimedean) places of $E$ just because the LHS expands out into exactly the RHS. Hence, to prove the claim, it suffices to show that the finite adeles on both sides agree. First note that the natural map $$\mapdesc f{\prodp_{v\mid\infty}(E_v,\wt{\ints v})}{\A_{E,\mrm{fin}}}{((x_w)_{w\mid v})_v}{(x_w)_w}$$ is well-defined because both the RHS and the LHS allow only finitely many non-integral elements, and it is visible bijective. Furthermore, because open sets in $E_v$ are products of open sets in $E_w$ for $w\mid v$, and because $\wt{\ints v}$ literally is the product of $\ints w$ for $w\mid v$, this map is clearly continuous and open. Hence, $f$ is a homeomorphism. It is even easy to see that $f$ is a ring map, so $f$ really is an isomorphism of topological rings.

The above isn’t the only way to get from $K$ (or $\A_K$) to $\A_E$, however.13

Let $E/K$ be a finite, spearable extension. Then, $\A_E\cong\A_K\otimes E$ where this time the isomorphism is only algebraic.
Like before, it suffices to only give an isomorphism between the finite adeles on both sides. That is, we only need to show that $\A_{E,\mrm{fin}}\cong\A_{K,\mrm{fin}}\otimes E$. Let $e_1,\dots,e_n$ (here, $n=[E:K]$) be an integral (i.e. $e_i\in\ints E$) basis for $E/K$. Hence, it is also a basis for $E_v/K_v$ for all places $v$ on $K$. Tensor products behave as you would expect with restricted direct products, so $$\A_{K,\mrm{fin}}\otimes E\cong\prodp_{v\nmid\infty}(K_v,\ints v)\cong\prodp_{v\nmid\infty}(K_v\otimes E,\ints v\otimes E)\cong\prodp_{v\nmid\infty}(E_v,\ints ve_1\oplus\dots\oplus\ints ve_n).$$ Finally, $\ints ve_1\oplus\cdots\oplus\ints ve_n\cong\wt{\ints v}$ for allmost all $v$, so the above is also isomorphic to $\prodp_{v\nmid\infty}(E_v,\wt{\ints v})\cong\A_E$.
We can easily upgrade the algebraic isomorphism above to a topological one by giving $\A_K\otimes E$ the topology induced from this isomorphism.
Let $K/E$ be a degree $n$, separable extension of global fields. As (additive) topological groups, we have $$\A_E\cong\prod_{i=1}^n\A_K.$$

Moving on, we claimed that the diagonal embedding $K\into\A_K$ turns $K$ into a discrete subgroup of $\A_K$. Let’s actually prove this in the case that $K$ is a number field. We’ll leave the function field case as an exercise. 14

Let $K$ be a number field. The image of the diagonal embedding $K\into\A_K$ is discrete.
Let $\sigma_1,\dots,\sigma_n:K\to\C$ be the $n=[K:\Q]$ (algebraic) embeddings of $K$ into $\C$. Take $x\mapsto\abs{\sigma_i(x)}$ to be the representatives for the $n$ archimedean places of $K$. Note that this means that $$\prod_{v\mid\infty}\abs x_v=\abs{\knorm(x)}$$ for all $x\in K$. Let $$U=\prod_{v\mid\infty}B_v(0,2^{-1})\by\prod_{v\nmid\infty}\ints v,$$ where $B_v(0,2^{-n})=\bracks{x\in K_v:\abs x_v<2^{-1}}$. Note that $U\subset\A_K$ is open, and consider any $x\in U\cap K$. Then, $x\in\ints v$ for all finite places $v$, so $x\in\ints K$ and hence $\knorm(x)\in\ints{\Q}=\Z$. Furthermore, $x\in U$ implies $$\abs{\knorm(x)}=\prod_{v\mid\infty}\abs x_v\le\prod_{v\mid\infty}2^{-1}=2^{-n}<1.$$ Since $\knorm(x)\in\Z$, this implies that $\knorm(x)=0$ so $x=0$. Thus, $U\cap K=\{0\}$ and the claim follows.

With that proven, let’s complete the second part of the $Z\into\R$ analogy by showing that $K$ is cocompact in $\A_K$. Note that, if $K$ is a number field, then $\A_K\cong(\A_{\Q})^n$ as topological groups where $n=[K:\Q]$. Hence, $(\A_K/K)\cong(\A_{\Q}/\Q)^n$ is compact iff $\A_{\Q}/\Q$ is. Thus, we can restrict our attention for the next proof (we could have use the same trick with discreteness if we wanted).

Let $K$ be a number field. Then, $\A_K/K$ is compact.
By the discussion above the theorem statement, it suffices to prove this for $K=\Q$, so that's what we'll do. Consider the compact set $$U=\prod_{p<\infty}\Z_p\by\sqbracks{-\frac12,\frac12}\subseteq\A_{\Q}.$$ We will show that the map quotient map $\A_{\Q}\to\A_{\Q}/\Q$ restricted to $U$ is surjective (i.e. for all $\alpha\in\A_{\Q}$, there's some $\beta\in\Q$ s.t. $\alpha-\beta\in U$), which will prove the result. Fix any $\alpha=(\alpha_p)_p\in\A_{\Q}$, and let $p$ be a prime for which $\abs{\alpha_p}>1$, so $\alpha_p=z_pp^{-k}$ for some $z_p\in\Z_p$ and $k>0$. Fix an integer $z_p'$ such that $z_p'\equiv z_p\pmod{p^k}$ and let $r_p=z_p'p^{-k}\in\Z$. Then, $$\abs{\alpha_p-r_p}_p=\abs{\frac{z_p-z_p'}{p^k}}_p=\abs{\frac{dp^k}{p^k}}_p=\abs d_p\le1,$$ for some $d\in\Z_p$. Furthermore, for any prime $q\neq p$, we have $$\abs{\alpha_q-r_p}_q\le\max(\abs{\alpha_q}_q,\abs{r_p}_q)\le\max(\abs{\alpha_q}_q,1),$$ so $\abs{\alpha_q-r_p}_q\le1$ if $\abs{\alpha_q}_q\le1$. This means that we can replace $\alpha$ by $\alpha-r_p$ to reduce the number of places it's nonintegral at by 1. After finitely many such replacements, we can assume that $\alpha$ is integral at all finite places. To finish, we observe that there exists some $s\in\Z$ such that $\alpha_\infty-s\in[-1/2,1/2]$, and hence $(\alpha-s)\mapsto\alpha+\Q\in\A_{\Q}/\Q$. Since $\alpha$ was arbitrary (really, sufficiently arbitrary since we added the assumption that it's integral at all finite places), we win.

To wrap up this section, we give a proof of the Artin product formula which says that

for all $a\in\units K$. We’ll continue our trend of only proving things for number fields, so assume $K$ is one of those. Note that

so it suffices to prove this for $K=\Q$. Since $\nabs:\A_K\to\R_{\ge0}$ is multiplicative, we can further simplify to the case that $a=p$ is prime. Here, there are only two nonunit absolute values, $\abs p_p=\frac1p$ and $\abs p_\infty=p$. Thus, we win.

# Class Groups

At this point we know a thing or two about adeles, but maybe we don’t know what they’re good for. One of the classic reasons for studying adeles is to give a more memorable proof of things like the finiteness of the class group of a number field. A love the geometry of numbers as much as the next guy 15, but the topologist in me refuses to believe in any finiteness prove that doesn’t end with “This space is both compact and discrete and hence finite.”

It turns out that there are many “class groups” one can define using adeles. We won’t bother looking at all (most?) of them. For the remainder of this section, fix a number field $K$.

Recall that $\units K\into\I_K$ embeds as a discrete subgroup. The idele class group of $K$ is the quotient $C_K:=\I_K/\units K$.

It turns out that $C_K$ is not necessarily compact, but the so-called norm-one idele class group of $K$ is.

Let $\I_K^1:=\ker(\nabs_{\A_K})$. The norm-one idele class group of $K$ is the quotient $C_K^1=\I_K^1/\units K$.
For a number field $K$, $C_K^1$ is compact.
Omitted. The proof technique is similar to the one used to show that $\A_K/K$ is compact in that you have an explicit compact subset of $\I_K^1$ which surjects onto $C_K^1$.

$\newcommand{\tints}{\wh{\mathscr O}_{#1}}$ We next relate these to the traditional class group $\Cl_K$ of $K$. Let $J_K$ denote its group of fraction ideals (finitely generated $\ints K$-submodules of $K$), so $\Cl_K=J_K/\units K$. Recall that there is a one-to-one correspondence $v\mapsto\mfp_v$ between the finite places of $K$ and the nonzero prime ideals of $\ints K$. Let $C_{K,\mrm{fin}}=\I_{K,\mrm{fin}}/\units K$. Finally, let $\tints K=\prod_{v\nmid\infty}\ints v$ and $\units{\tints K}=\prod_{v\nmid\infty}\units{\ints v}$.

\begin{align*} J_K\cong\I_{K,\mrm{fin}}/\units{\tints K}&&\Cl_K\cong C_{K,\mrm{fin}}/\units{\tints K}\units K \end{align*}
First, consider the map $$\mapdesc\phi{\I_{K,\mrm{fin}}}{J_K}{(\alpha_v)_v}{\prod_{v\nmid\infty}\mfp_v^{v(\alpha_v)}}$$ which visibly has kernel $\units{\tints K}$ and is visibly surjective. Hence, we get the first isomorphism. Now, note that, for $\alpha\in\units K$, we have $$\phi(\alpha)=\phi(\alpha,\alpha,\dots)=\prod_{v\nmid\infty}\mfp_v^{v(\alpha)}=\alpha\ints K$$ so $\phi$ descends to a map $\psi:C_{K,\mrm{fin}}\to\Cl_K,\alpha\units K\mapsto\phi(\alpha)\units K$. We claim that $\ker(\psi)=\units{\tints K}\units K$. It is clear that $\units{\tints K}\units K\subseteq\ker\psi$, so we focus on the revese direction. Pick some $\xi\units K\in C_{K,\mrm{fin}}$ with $\psi(\xi\units K)=\ints K\units K$, so $\prod_v\mfp_v^{v(\xi_v')}=\ints K$ for some representative $\xi'\in\xi\units K$. Hence, $\xi'\in\tints K$, and so $\xi\units K=\xi'\units K\in\tints K\units K$. This gives the second isomorphism.
It is a general fact about topological groups that, for $H\le G$, the quotient map $G\to G/H$ is open and that the quotient space $G/H$ is discrete iff $H$ is an open subgroup. In the above theorem, $\units{\tints K}$ is visibly an open subgroup of $\I_{K,\mrm{fin}}$, and so it descends to an open subgroup $\units{\tints K}\units K$ of $C_{K,\mrm{fin}}$. This means that $\Cl_K$ is discrete with the topology given by the above isomorphism.
The class group $\Cl_K$ of a number field is finite.
Consider the map $$\mapdesc f{\I_K^1}{J_K}{(\alpha_v)_v}{\prod_{v<\infty}\mfp_v^{v(\alpha_v)}}.$$ This map is visibly surjective, and similarly to last time, descends to a map $C_K^1\to\Cl_K$. Thus, $\Cl_K$ is the continuous image of a compact set, and hence compact. However, we saw in the previous remark that it was discrete, so it must be finite.

Another classic use of adeles is proving Dirichlet’s theorem about the rank of the unit group of the ring of integers of a number field. It’s also worth mentioning that both this and the finiteness of the class group have generalizations to $S$-integers which can be proven with adelic methods. However, instead of covering these, I think I will stop here.

1. Secretly, we’re not stepping that far away from it. One does Fourier analysis on R to prove nice things (i.e. existence of a functional equation) about the Riemann zeta function. Analagously, one does Fourier analysis on these adelic rings to prove nice things about more general L-functions. We won’t really touch on this here, but it’s lurking in the background.

2. Better put, “one place at a time”

3. Of course, given e.g. Q, you can complete it at various places to obtain all the Q_p’s you could want. However, if you want to study e.g. Q_2 and Q_5 at the same time, then you can’t complete Q because this will kill valuable information, but Q itself is somehow not the best place to work to understand Q_2 and Q_5 simultaneously.

4. Exercise: prove this (hint: infinite places on number fields come from embeddings into C)

5. If we did this, we would need to amend our definition of restricted products to not require all H_i to be compact. Instead, we’d only require all but finitely many H_i are compact. I’ll leave it up to the reader to figure out how to modify arguments for this slightly more general definition.

6. You want this to be able to do analysis-y type stuff. Locally compact topolgical groups have Haar measures which let you do Fourier Analysis (I guess in this setting it’s typically called harmonic analysis) on them (maybe just when they’re abelian).

7. Exercise: prove this (hint: factor (x))

8. Exercise: prove this (hint: suffices to find a neighborhood around 1 containing no other principal adele) (hint2: It’s possible you want to hold off on proving this until you see the product formula)

9.  technically, for this to make sense we need to choose a representative of each place on K. Just choose the normal ones (e.g. for v finite, choose p =1/q where p is a uniformizer and q is the size of the residue field)

10. Quotes because I won’t give all the details for most (any?) of the things here

11. I don’t think I’ll need it for anything. If I do and it bothers you that I haven’t proved it, you can find a proof in these notes

12. What’s the correct way to notate that the two choices are “a finite extension” and “an extension”?

13. TODO: double check that {e_i} gives a basis for E_v/K_v for all v and not just all but finitely many v (proof should be salvagable in either case)

14. hint: for F_q(t), the t-adic absolute value (I think this is usually considered the infinite place) should play the role of the archimedean places in the number field proof. I could be wrong about this; I really haven’t spent much time with function fields.

15. sarcasm: I’m not a fan of it