Strong Approximation

We first prove a technical lemma and corollary, then use them to deduce the strong approximation theorem, which is an extreme generalization of the Chinese Remainder Theorem; it asserts that $K^+$ is dense in the analogue of the adeles with one valuation removed.

The proof of Lemma 18.4.1 below will use in a crucial way the normalized Haar measure on $\AA _K$ and the induced measure on the compact quotient $\AA _K^+/K^+$. Since I am not formally developing Haar measure on locally compact groups, and since I didn't explain induced measures on quotients well in the last chapter, hopefully the following discussion will help clarify what is going on.

The real numbers $\mathbf{R}^+$ under addition is a locally compact topological group. Normalized Haar measure $\mu$ has the property that $\mu([a,b]) = b-a$, where $a\leq b$ are real numbers and $[a,b]$ is the closed interval from $a$ to $b$. The subset $\mathbf{Z}^+$ of $\mathbf{R}^+$ is discrete, and the quotient $S^1 = \mathbf{R}^+/\mathbf{Z}^+$ is a compact topological group, which thus has a Haar measure. Let $\overline{\mu}$ be the Haar measure on $S^1$ normalized so that the natural quotient $\pi:\mathbf{R}^+\to S^1$ preserves the measure, in the sense that if $X\subset \mathbf{R}^+$ is a measurable set that maps injectively into $S^1$, then $\mu(X) = \overline{\mu}(\pi(X))$. This determine $\overline{\mu}$ and we have $\overline{\mu}(S^1)=1$ since $X=[0,1)$ is a measurable set that maps bijectively onto $S^1$ and has measure $1$. The situation for the map $\AA _K \to \AA _K/K^+$ is pretty much the same.

Lemma 18.4.1   There is a constant $C>0$ that depends only on the global field $K$ with the following property:

Whenever $\mathbf{x}=\{x_v\}_v \in \AA _K$ is such that

$$\prod_v \left\vert x_v\right\vert _v > C,$$ (18.8)

then there is a nonzero principal adele $a \in K\subset \AA _K$ such that

$$\left\vert a\right\vert _v \leq \left\vert x_v\right\vert _v$$   for all $v$$$.$$

Proof. This proof is modelled on Blichfeldt's proof of Minkowski's Theorem in the Geometry of Numbers, and works in quite general circumstances.

First we show that (18.4.1) implies that $\left\vert x_v\right\vert _v=1$ for almost all $v$. Because $\mathbf{x}$ is an adele, we have $\left\vert x_v\right\vert _v\leq 1$ for almost all $v$. If $\left\vert x_v\right\vert _v<1$ for infinitely many $v$, then the product in (18.4.1) would have to be 0. (We prove this only when $K$ is a finite extension of $\mathbf {Q}$.) Excluding archimedean valuations, this is because the normalized valuation $\left\vert x_v\right\vert _v = \left\vert\Norm (x_v)\right\vert _p$, which if less than $1$ is necessarily $\leq 1/p$. Any infinite product of numbers $1/p_i$ must be 0, whenever $p_i$ is a sequence of primes.

Let $c_0$ be the Haar measure of $\AA _K^+/K^+$ induced from normalized Haar measure on $\AA _K^+$, and let $c_1$ be the Haar measure of the set of $\mathbf{y}=\{y_v\}_v \in \AA _K^+$ that satisfy

$$\left\vert y_v\right\vert _v \leq \frac{1}{2} v ,$$

$$\left\vert y_v\right\vert _v \leq \frac{1}{2} v ,$$

$$\left\vert y_v\right\vert _v \leq 1 v .$$

(As we will see, any positive real number $\leq 1/2$ would suffice in the definition of $c_1$ above. For example, in Cassels's article he uses the mysterious $1/10$. He also doesn't discuss the subtleties of the complex archimedean case separately.)

Then $0<c_0<\infty$ since $\AA _K/K^+$ is compact, and $0<c_1<\infty$ because the number of archimedean valuations $v$ is finite. We show that

$$C=\frac{c_0}{c_1}$$

will do. Thus suppose $\mathbf{x}$ is as in (18.4.1).

The set $T$ of $\mathbf{t}=\{t_v\}_v\in \AA _K^+$ such that

$$\left\vert t_v\right\vert _v \leq \frac{1}{2}\left\vert x_v\right\vert _v v ,$$

$$\left\vert t_v\right\vert _v \leq \frac{1}{2}\sqrt{\left\vert x_v\right\vert _v} v ,$$

$$\left\vert t_v\right\vert _v \leq \left\vert x_v\right\vert _v \quad v$$

has measure

$$c_1 \cdot \prod_{v} \left\vert x_v\right\vert _v > c_1 \cdot C = c_0.$$ (18.9)

(Note: If there are complex valuations, then the some of the $\left\vert x_v\right\vert _v$'s in the product must be squared.)

Because of (18.4.2), in the quotient map $\AA _K^+ \to \AA _K^+/K^+$ there must be a pair of distinct points of $T$ that have the same image in $\AA _K^+/K^+$, say

$$\mathbf{t}' = \{t'_v\}_v \in T$$   and$$\quad \mathbf{t}'' = \{t''_v\}_v\in T$$

and

$$a = \mathbf{t}' - \mathbf{t}'' \in K^+$$

is nonzero. Then

$$\left\vert a\right\vert _v = \left\vert t'_v - t''_v\right\ve... ...\right\vert _v & \text{if $v$ is non-archimedean,} \end{cases}$$

for all $v$. In the case of complex archimedean $v$, we must be careful because the normalized valuation $\left\vert \cdot \right\vert _v$ is the square of the usual archimedean complex valuation $\left\vert \cdot \right\vert _\infty$ on $\mathbf{C}$, so e.g., it does not satisfy the triangle inequality. In particular, the quantity $\left\vert t_v'-t''_v\right\vert _v$ is at most the square of the maximum distance between two points in the disc in $\mathbf{C}$ of radius $\frac{1}{2}\sqrt{\left\vert x_v\right\vert _v}$, where by distance we mean the usual distance. This maximum distance in such a disc is at most $\sqrt{\left\vert x_v\right\vert _v}$, so $\left\vert t_v'-t''_v\right\vert _v$ is at most $\left\vert x_v\right\vert _v$, as required. Thus $a$ satisfies the requirements of the lemma. $\qedsymbol$

Corollary 18.4.2   Let $v_0$ be a normalized valuation and let $\delta_v>0$ be given for all $v\neq v_0$ with $\delta_v=1$ for almost all $v$. Then there is a nonzero $a\in K$ with

$$\left\vert a\right\vert _v \leq \delta_v$$   (all $v&ne#neq;v_0$)$$.$$

Proof. This is just a degenerate case of Lemma 18.4.1. Choose $x_v\in K_v$ with $0< \left\vert x_v\right\vert _v \leq \delta_v$ and $\left\vert x_v\right\vert _v=1$ if $\delta_v=1$. We can then choose $x_{v_0}\in K_{v_0}$ so that

$$\prod_{\text{all $v$ including $v_0$}} \left\vert x_v\right\vert _v > C.$$

Then Lemma 18.4.1 does what is required. $\qedsymbol$

Remark 18.4.3   The character group of the locally compact group $\AA _K^+$ is isomorphic to $\AA _K^+$ and $K^+$ plays a special role. See Chapter XV of [Cp86], Lang's [Lan64], Weil's [Wei82], and Godement's Bourbaki seminars 171 and 176. This duality lies behind the functional equation of $\zeta$ and $L$-functions. Iwasawa has shown [Iwa53] that the rings of adeles are characterized by certain general topologico-algebraic properties.

We proved before that $K$ is discrete in $\AA _K$. If one valuation is removed, the situation is much different.

Theorem 18.4.4 (Strong Approximation)   Let $v_0$ be any normalized nontrivial valuation of the global field $K$. Let $\AA _{K,v_0}$ be the restricted topological product of the $K_v$ with respect to the $\O_v$, where $v$ runs through all normalized valuations $v\neq v_0$. Then $K$ is dense in $\AA _{K,v_0}$.

Proof. This proof was suggested by Prof. Kneser at the Cassels-Frohlich conference.

Recall that if $\mathbf{x}=\{x_v\}_v\in \AA _{K,v_0}$ then a basis of open sets about $\mathbf{x}$ is the collection of products

$$\prod_{v\in S} B(x_v,\varepsilon _v) \times \prod_{v\not\in S, v\neq v_0} \O_v,$$

where $B(x_v,\varepsilon _v)$ is an open ball in $K_v$ about $x_v$, and $S$ runs through finite sets of normalized valuations (not including $v_0$). Thus denseness of $K$ in $\AA _{K,v_0}$ is equivalent to the following statement about elements. Suppose we are given (i) a finite set $S$ of valuations $v\neq v_0$, (ii) elements $x_v\in K_v$ for all $v\in S$, and (iii) an $\varepsilon >0$. Then there is an element $b\in K$ such that $\left\vert b-x_v\right\vert _v<\varepsilon$ for all $v\in S$ and $\left\vert b\right\vert _v\leq 1$ for all $v\not\in S$ with $v\neq v_0$.

By the corollary to our proof that $\AA _K^+/K^+$ is compact (Corollary 18.3.6), there is a $W\subset \AA _K$ that is defined by inequalities of the form $\left\vert y_v\right\vert _v\leq \delta_v$ (where $\delta_v=1$ for almost all $v$) such that ever $\mathbf{z}\in \AA _K$ is of the form

$$\mathbf{z} = \mathbf{y}+ c, \qquad \mathbf{y}\in W, \quad c\in K.$$ (18.10)

By Corollary 18.4.2, there is a nonzero $a\in K$ such that

$$\left\vert a\right\vert _v < \frac{1}{\delta_v}\cdot \varepsilon v\in S,$$

$$\left\vert a\right\vert _v \leq \frac{1}{\delta_v} \qquad v\not\in S, v\neq v_0.$$

Hence on putting $\mathbf{z} = \frac{1}{a}\cdot \mathbf{x}$ in (18.4.3) and multiplying by $a$, we see that every $\mathbf{x}\in \AA _K$ is of the shape

$$\mathbf{x}= \mathbf{w} + b,\qquad \mathbf{w}\in a\cdot W, \quad b\in K,$$

where $a\cdot W$ is the set of $a\mathbf{y}$ for $\mathbf{y}\in W$. If now we let $\mathbf{x}$ have components the given $x_v$ at $v\in S$, and (say) 0 elsewhere, then $b=\mathbf{x}-\mathbf{w}$ has the properties required. $\qedsymbol$

Remark 18.4.5   The proof gives a quantitative form of the theorem (i.e., with a bound for $\left\vert b\right\vert _{v_0}$). For an alternative approach, see [Mah64].

In the next chapter we'll introduce the ideles $\AA _K^*$. Finally, we'll relate ideles to ideals, and use everything so far to give a new interpretation of class groups and their finiteness.