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 20.4.1 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 20.4.1   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

$\displaystyle \prod_v \left\vert x_v\right\vert _v > C,$ (20.8)

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

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

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 (20.4.1) 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 (20.4.1) 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

$\displaystyle \left\vert y_v\right\vert _v$ $\displaystyle \leq \frac{1}{2}$   if $ v$ is real archimedean$\displaystyle ,
$\displaystyle \left\vert y_v\right\vert _v$ $\displaystyle \leq \frac{1}{2}$   if $ v$ is complex archimedean$\displaystyle ,
$\displaystyle \left\vert y_v\right\vert _v$ $\displaystyle \leq 1  $   if $ v$ is non-archimedean$\displaystyle .$    

(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

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

will do. Thus suppose $ \mathbf{x}$ is as in (20.4.1).

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

$\displaystyle \left\vert t_v\right\vert _v$ $\displaystyle \leq \frac{1}{2}\left\vert x_v\right\vert _v  $   if $ v$ is real archimedean$\displaystyle ,
$\displaystyle \left\vert t_v\right\vert _v$ $\displaystyle \leq \frac{1}{2}\sqrt{\left\vert x_v\right\vert _v}  $   if $ v$ is complex archimedean$\displaystyle ,
$\displaystyle \left\vert t_v\right\vert _v$ $\displaystyle \leq \left\vert x_v\right\vert _v \quad $   if $ v$ is non-archimedean    

has measure

$\displaystyle c_1 \cdot \prod_{v} \left\vert x_v\right\vert _v > c_1 \cdot C = c_0.$ (20.9)

(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 (20.4.2), 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

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


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

is nonzero. Then

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

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 20.4.2   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

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

Proof. This is just a degenerate case of Lemma 20.4.1. 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

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

Then Lemma 20.4.1 does what is required. $ \qedsymbol$

Remark 20.4.3   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 20.4.4 (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

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

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 20.3.6), 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

$\displaystyle \mathbf{z} = \mathbf{y}+ c, \qquad \mathbf{y}\in W, \quad c\in K.$ (20.10)

By Corollary 20.4.2, there is a nonzero $ a\in K$ such that

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

Hence on putting $ \mathbf{z} = \frac{1}{a}\cdot \mathbf{x}$ in (20.4.3) and multiplying by $ a$, we see that every $ \mathbf{x}\in \AA _K$ is of the shape

$\displaystyle \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 20.4.5   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.

William Stein 2004-05-06