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:

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

$\displaystyle K\hookrightarrow \AA _K$ (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

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

both algebraically and topologically. Under this isomorphism,

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

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

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

with respect to the integers

$\displaystyle \O _v\cdot \omega_1 \oplus \cdots \oplus \O _v\cdot \omega_n.$ (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

$\displaystyle 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

$\displaystyle \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

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

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

$\displaystyle \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

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

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

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

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

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

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

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

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

$\displaystyle \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

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

such that

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


$\displaystyle r_p = 0$   almost all $p$$\displaystyle .$

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

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

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:

$\displaystyle \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

$\displaystyle \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

$\displaystyle \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$.)

William Stein 2004-05-06