The Adele Ring

Let $K$ be a global field. For each normalization $\left\vert \cdot \right\vert _v$ of $K$, let $K_v$ denote the completion of $K$. If $\left\vert \cdot \right\vert _v$ is non-archimedean, let $\O _v$ denote the ring of integers of $K_v$.

Definition 20.3.1 (Adele Ring)   The $\AA _K$ of $K$ is the topological ring whose underlying topological space is the restricted topological product of the $K_v$ with respect to the $\O _v$, and where addition and multiplication are defined componentwise:

$$(\mathbf{x}\mathbf{y})_v = \mathbf{x}_v \mathbf{y}_v \qquad (\mathbf{x}+ \mathbf{y})_v = \mathbf{x}_v + \mathbf{y}_v \mathbf{x}, \mathbf{y}\in\AA _K.$$ (20.4)

It is readily verified that (i) this definition makes sense, i.e., if $\mathbf{x}, \mathbf{y}\in \AA _K$, then $\mathbf{x}\mathbf{y}$ and $\mathbf{x}+\mathbf{y}$, whose components are given by (20.3.1), are also in $\AA _K$, and (ii) that addition and multiplication are continuous in the $\AA _K$-topology, so $\AA _K$ is a topological ring, as asserted. Also, Lemma 20.2.4 implies that $\AA _K$ is locally compact because the $K_v$ are locally compact (Corollary 17.1.6), and the $\O _v$ are compact (Theorem 17.1.4).

There is a natural continuous ring inclusion

$$K\hookrightarrow \AA _K$$ (20.5)

that sends $x\in K$ to the adele every one of whose components is $x$. This is an adele because $x\in \O _v$ for almost all $v$, by Lemma 20.1.2. The map is injective because each map $K\to K_v$ is an inclusion.

Definition 20.3.2 (Principal Adeles)   The image of (20.3.2) is the ring of .

It will cause no trouble to identify $K$ with the principal adeles, so we shall speak of $K$ as a subring of $\AA _K$.

Formation of the adeles is compatibility with base change, in the following sense.

Lemma 20.3.3   Suppose $L$ is a finite (separable) extension of the global field $K$. Then

$$\AA _K \otimes _K L \cong \AA _L$$ (20.6)

both algebraically and topologically. Under this isomorphism,

$$L\cong K\otimes _K L \subset \AA _K \otimes _K L$$

maps isomorphically onto $L\subset \AA _L$.

Proof. Let $\omega_1,\ldots, \omega_n$ be a basis for $L/K$ and let $v$ run through the normalized valuations on $K$. The left hand side of (20.3.3), with the tensor product topology, is the restricted product of the tensor products

$$K_v \otimes _K L \cong K_v \cdot\omega_1 \oplus \cdots \oplus K_v\cdot \omega_n$$

with respect to the integers

$$\O _v\cdot \omega_1 \oplus \cdots \oplus \O _v\cdot \omega_n.$$ (20.7)

(An element of the left hand side is a finite linear combination $\sum \mathbf{x}_i \otimes a_i$ of adeles $\mathbf{x}_i \in \AA _K$ and coefficients $a_i \in L$, and there is a natural isomorphism from the ring of such formal sums to the restricted product of the $K_v\otimes _K L$.)

We proved before (Theorem 19.1.8) that

$$K_v \otimes _K L \cong L_{w_1} \oplus \cdots \oplus L_{w_g},$$

where $w_1,\ldots, w_g$ are the normalizations of the extensions of $v$ to $L$. Furthermore, as we proved using discriminants (see Lemma 20.1.6), the above identification identifies (20.3.4) with

$$\O _{L_{w_1}} \oplus \cdots \oplus \O _{L_{w_g}},$$

for almost all $v$. Thus the left hand side of (20.3.3) is the restricted product of the $L_{w_1} \oplus \cdots \oplus L_{w_g}$ with respect to the $\O _{L_{w_1}} \oplus \cdots \oplus \O _{L_{w_g}}$. But this is canonically isomorphic to the restricted product of all completions $L_w$ with respect to $\O _w$, which is the right hand side of (20.3.3). This establishes an isomorphism between the two sides of (20.3.3) as topological spaces. The map is also a ring homomorphism, so the two sides are algebraically isomorphic, as claimed. $\qedsymbol$

Corollary 20.3.4   Let $\AA _K^+$ denote the topological group obtained from the additive structure on $\AA _K$. Suppose $L$ is a finite seperable extension of $K$. Then

$$\AA _L^+ = \AA _K^+ \oplus \cdots \oplus \AA _K^+, \qquad ([L:K]$$    summands$$).$$

In this isomorphism the additive group $L^+\subset \AA _L^+$ of the principal adeles is mapped isomorphically onto $K^+\oplus \cdots \oplus K^+$.

Proof. For any nonzero $\omega \in L$, the subgroup $\omega\cdot \AA _K^+$ of $\AA _L^+$ is isomorphic as a topological group to $\AA _K^+$ (the isomorphism is multiplication by $1/\omega$). By Lemma 20.3.3, we have isomorphisms

$$\AA _L^+ = \AA _K^+ \otimes _K L \cong \omega_1\cdot \AA _K^+ \o... ... \oplus \omega_n \cdot \AA _K^+ \cong \AA _K^+ \oplus \cdots \oplus \AA _K^+.$$

If $a\in L$, write $a=\sum b_i \omega_i$, with $b_i \in K$. Then $a$ maps via the above map to

$$x = (\omega_1\cdot \{b_1\},\ldots, \omega_n \cdot \{b_n\}),$$

where $\{b_i\}$ denotes the principal adele defined by $b_i$. Under the final map, $x$ maps to the tuple

$$(b_1,\ldots, b_n) \in K\oplus \cdots \oplus K \subset \AA _K^+ \oplus \cdots \oplus \AA _K^+.$$

The dimensions of $L$ and of $K\oplus \cdots \oplus K$ over $K$ are the same, so this proves the final claim of the corollary. $\qedsymbol$

Theorem 20.3.5   The global field $K$ is discrete in $\AA _K$ and the quotient $\AA _K^+/K^+$ of additive groups is compact in the quotient topology.

At this point Cassels remarks

``It is impossible to conceive of any other uniquely defined topology on $K$. This metamathematical reason is more persuasive than the argument that follows!''

Proof. Corollary 20.3.4, with $K$ for $L$ and $\mathbf{Q}$ or $\mathbf{F}(t)$ for $K$, shows that it is enough to verify the theorem for $\mathbf{Q}$ or $\mathbf{F}(t)$, and we shall do it here for $\mathbf{Q}$.

To show that $\mathbf{Q}^+$ is discrete in $\AA _\mathbf{Q}^+$ it is enough, because of the group structure, to find an open set $U$ that contains $0 \in \AA _\mathbf{Q}^+$, but which contains no other elements of $\mathbf{Q}^+$. (If $\alpha\in\mathbf{Q}^+$, then $U+\alpha$ is an open subset of $\AA _\mathbf{Q}^+$ whose intersection with $\mathbf{Q}^+$ is $\{\alpha\}$.) We take for $U$ the set of $\mathbf{x}=\{x_v\}_v \in \AA _\mathbf{Q}^+$ with

$$\left\vert x_\infty\right\vert _\infty < 1$$   and$$\qquad \left\vert x_p\right\vert _p \leq 1$$   (all $p$)$$,$$

where $\left\vert \cdot \right\vert _p$ and $\left\vert \cdot \right\vert _\infty$ are respectively the $p$-adic and the usual archimedean absolute values on  $\mathbf{Q}$. If $b\in \mathbf{Q}\cap U$, then in the first place $b\in\mathbf{Z}$ because $\left\vert b\right\vert _p\leq 1$ for all $p$, and then $b=0$ because $\left\vert b\right\vert _\infty<1$. This proves that $K^+$ is discrete in $\AA _\mathbf{Q}^+$. (If we leave out one valuation, as we will see later (Theorem 20.4.4), this theorem is false--what goes wrong with the proof just given?)

Next we prove that the quotient $\AA _\mathbf{Q}^+/\mathbf{Q}^+$ is compact. Let $W\subset \AA _\mathbf{Q}^+$ consist of the $\mathbf{x}=\{x_v\}_v \in \AA _\mathbf{Q}^+$ with

$$\left\vert x_\infty\right\vert _\infty \leq \frac{1}{2}$$   and$$\qquad \left\vert x_p\right\vert _p \leq 1$$   for all primes $p$$$.$$

We show that every adele $\mathbf{y}=\{y_v\}_v$ is of the form

$$\mathbf{y}= a + \mathbf{x}, \qquad a\in \mathbf{Q}, \quad \mathbf{x}\in W,$$

which will imply that the compact set $W$ maps surjectively onto $\AA _\mathbf{Q}^+/\mathbf{Q}^+$. Fix an adele $\mathbf{y}=\{y_v\}\in\AA _\mathbf{Q}^+$. Since $\mathbf{y}$ is an adele, for each prime $p$ we can find a rational number

$$r_p = \frac{z_p}{p^{n_p}}$$   with$$\quad z_p \in \mathbf{Z}$$   and$$\quad n_p \in \mathbf{Z}_{\geq 0}$$

such that

$$\left\vert y_p - r_p\right\vert _p \leq 1,$$

and

$$r_p = 0$$   almost all $p$$$.$$

More precisely, for the finitely many $p$ such that

$$y_p = \sum_{n\geq -\left\vert s\right\vert} a_np^n \not\in\mathbf{Z}_p,$$

choose $r_p$ to be a rational number that is the value of an appropriate truncation of the $p$-adic expansion of $y_p$, and when $y_p\in \mathbf{Z}_p$ just choose $r_p = 0$. Hence $r=\sum_{p} r_p\in\mathbf{Q}$ is well defined. The $r_q$ for $q\neq p$ do not mess up the inequality $\left\vert y_p - r\right\vert _p \leq 1$ since the valuation $\left\vert \cdot \right\vert _p$ is non-archimedean and the $r_q$ do not have any $p$ in their denominator:

$$\left\vert y_p - r\right\vert _p = \left\vert y_p - r_p - \sum_{... ...vert _p, \left\vert\sum_{q\neq p} r_q\right\vert _p\right) \leq \max(1,1) = 1.$$

Now choose $s\in\mathbf{Z}$ such that

$$\left\vert b_\infty - r-s\right\vert \leq \frac{1}{2}.$$

Then $a=r+s$ and $\mathbf{x}= \mathbf{y}- a$ do what is required, since $\mathbf{y}- a = \mathbf{y}- r - s$ has the desired property (since $s\in\mathbf{Z}$ and the $p$-adic valuations are non-archimedean).

Hence the continuous map $W\to \AA _\mathbf{Q}^+/\mathbf{Q}^+$ induced by the quotient map $\AA _\mathbf{Q}^+ \to \AA _\mathbf{Q}^+/\mathbf{Q}^+$ is surjective. But $W$ is compact (being the topological product of the compact spaces $\left\vert x_\infty\right\vert _\infty\leq 1/2$ and the $\mathbf{Z}_p$ for all $p$), hence $\AA _\mathbf{Q}^+/\mathbf{Q}^+$ is also compact. $\qedsymbol$

Corollary 20.3.6   There is a subset $W$ of $\AA _K$ defined by inequalities of the type $\left\vert x_v\right\vert _v \leq \delta_v$, where $\delta_v=1$ for almost all $v$, such that every $\mathbf{y}\in\AA _K$ can be put in the form

$$\mathbf{y}= a + \mathbf{x}, \qquad a\in K, \quad \mathbf{x}\in W,$$

i.e., $\AA _K = K + W$.

Proof. We constructed such a set for $K=\mathbf{Q}$ when proving Theorem 20.3.5. For general $K$ the $W$ coming from the proof determines compenent-wise a subset of $\AA _K^+\cong \AA _\mathbf{Q}^+\oplus \cdots \oplus \AA _\mathbf{Q}^+$ that is a subset of a set with the properties claimed by the corollary. $\qedsymbol$

As already remarked, $\AA _K^+$ is a locally compact group, so it has an invariant Haar measure. In fact one choice of this Haar measure is the product of the Haar measures on the $K_v$, in the sense of Definition 20.2.5.

Corollary 20.3.7   The quotient $\AA _K^+/K^+$ has finite measure in the quotient measure induced by the Haar measure on $\AA _K^+$.

Remark 20.3.8   This statement is independent of the particular choice of the multiplicative constant in the Haar measure on $\AA _K^+$. We do not here go into the question of finding the measure $\AA _K^+/K^+$ in terms of our explicitly given Haar measure. (See Tate's thesis, [Cp86, Chapter XV].)

Proof. This can be reduced similarly to the case of $\mathbf{Q}$ or $\mathbf{F}(t)$ which is immediate, e.g., the $W$ defined above has measure $1$ for our Haar measure.

Alternatively, finite measure follows from compactness. To see this, cover $\AA _K/K^+$ with the translates of $U$, where $U$ is a nonempty open set with finite measure. The existence of a finite subcover implies finite measure. $\qedsymbol$

Remark 20.3.9   We give an alternative proof of the product formula $\prod\left\vert a\right\vert _v=1$ for nonzero $a\in K$. We have seen that if $x_v\in K_v$, then multiplication by $x_v$ magnifies the Haar measure in $K_v^+$ by a factor of $\left\vert x_v\right\vert _v$. Hence if $\mathbf{x}=\{x_v\}\in\AA _K$, then multiplication by $\mathbf{x}$ magnifies the Haar measure in $\AA _K^+$ by $\prod \left\vert x_v\right\vert _v$. But now multiplication by $a\in K$ takes $K^+\subset \AA _K^+$ into $K^+$, so gives a well-defined bijection of $\AA _K^+/K^+$ onto $\AA _K^+/K^+$ which magnifies the measure by the factor $\prod\left\vert a\right\vert _v$. Hence $\prod\left\vert a\right\vert _v=1$ Corollary 20.3.7. (The point is that if $\mu$ is the measure of $\AA _K^+/K^+$, then $\mu = \prod\left\vert a\right\vert _v \cdot \mu$, so because $\mu$ is finite we must have $\prod\left\vert a\right\vert _v=1$.)