Weak Approximation

The following theorem asserts that inequivalent valuations are in fact almost totally indepedent. For our purposes it will be superseded by the strong approximation theorem (Theorem 20.4.4).

Theorem 16.3.1 (Weak Approximation)   Let $\left\vert \cdot \right\vert _n$, for $1\leq n \leq N$, be inequivalent nontrivial valuations of a field $K$. For each $n$, let $K_n$ be the topological space consisting of the set of elements of $K$ with the topology induced by $\left\vert \cdot \right\vert _n$. Let $\Delta$ be the image of $K$ in the topological product

$$A=\prod_{1\leq n\leq N} K_n$$

equipped with the product topology. Then $\Delta$ is dense in $A$.

The conclusion of the theorem may be expressed in a less topological manner as follows: given any $a_n\in K$, for $1\leq n \leq N$, and real $\varepsilon >0$, there is an $b\in K$ such that simultaneously

$$\left\vert a_n - b\right\vert _n < \varepsilon \qquad (1\leq n\leq N).$$

If $K=\mathbf{Q}$ and the $\left\vert \cdot \right\vert{}$ are $p$-adic valuations, Theorem 16.3.1 is related to the Chinese Remainder Theorem (Theorem 9.1.3), but the strong approximation theorem (Theorem 20.4.4) is the real generalization.

Proof. We note first that it will be enough to find, for each $n$, an element $c_n\in K$ such that

$$\left\vert c_n\right\vert _n > 1$$    and $$\quad \left\vert c_n\right\vert _m < 1$$    for $$n\neq m,$$

where $1\leq n,m\leq N$. For then as $r\to+\infty$, we have

$$\frac{c_n^r}{1+c_n^r} = \frac{1}{1+\left(\frac{1}{c_n}\right)^{r}... ...spect to } \left\vert \cdot \right\vert _m, \text{ for } m\neq n. \end{cases}$$

It is then enough to take

$$b = \sum_{n=1}^N \frac{c_n^r}{1+c_n^r} \cdot a_n$$

By symmetry it is enough to show the existence of $c=c_1$ with

$$\left\vert c\right\vert _1 > 1$$   and$$\qquad \left\vert c\right\vert _n<1$$   for$$\quad 2\leq n\leq N.$$

We will do this by induction on $N$.

First suppose $N=2$. Since $\left\vert \cdot \right\vert _1$ and $\left\vert \cdot \right\vert _2$ are inequivalent (and all absolute values are assumed nontrivial) there is an $a\in K$ such that

$$\left\vert a\right\vert _1 < 1 \qquad \left\vert a\right\vert _2 \geq 1$$ (16.2)

and similarly a $b$ such that

$$\left\vert b\right\vert _1 \geq 1$$   and$$\qquad \left\vert b\right\vert _2 < 1.$$

Then $$c=\frac{b}{a}$$ will do.

Remark 16.3.2   It is not completely clear that one can choose an $a$ such that (16.3.1) is satisfied. Suppose it were impossible. Then because the valuations are nontrivial, we would have that for any $a\in K$ if $\left\vert a\right\vert _1<1$ then $\left\vert a\right\vert _2<1$. This implies the converse statement: if $a\in K$ and $\left\vert a\right\vert _2<1$ then $\left\vert a\right\vert _1<1$. To see this, suppose there is an $a\in K$ such that $\left\vert a\right\vert _2<1$ and $\left\vert a\right\vert _1\geq 1$. Choose $y\in K$ such that $\left\vert y\right\vert _1<1$. Then for any integer $n>0$ we have $\left\vert y/a^n\right\vert _1<1$, so by hypothesis $\left\vert y/a^n\right\vert _2<1$. Thus $\left\vert y\right\vert _2 < \left\vert a\right\vert _2^n < 1$ for all $n$. Since $\left\vert a\right\vert _2<1$ we have $\left\vert a\right\vert _2^n\to 0$ as $n\to \infty$, so $\left\vert y\right\vert _2=0$, a contradiction since $y\neq 0$. Thus $\left\vert a\right\vert _1<1$ if and only if $\left\vert a\right\vert _2<1$, and we have proved before that this implies that $\left\vert \cdot \right\vert _1$ is equivalent to $\left\vert \cdot \right\vert _2$.

Next suppose $N\geq 3$. By the case $N-1$, there is an $a\in K$ such that

$$\left\vert a\right\vert _1 > 1$$   and$$\qquad \left\vert a\right\vert _n<1$$   for$$\quad 2\leq n\leq N-1.$$

By the case for $N=2$ there is a $b\in K$ such that

$$\left\vert b\right\vert _1>1$$   and$$\qquad\left\vert b\right\vert _N<1.$$

Then put

$$c = \begin{cases} a & \text{if } \left\vert a\right\vert _N<1\\ ... ... \frac{a^r}{1+a^r}}\cdot b & \text{if }\left\vert a\right\vert _N>1 \end{cases}$$

where $r\in\mathbf{Z}$ is sufficiently large so that $\left\vert c\right\vert _1>1$ and $\left\vert c\right\vert _n<1$ for $2\leq n\leq N$. $\qedsymbol$

Example 16.3.3   Suppose $K=\mathbf{Q}$, let $\left\vert \cdot \right\vert _1$ be the archimedean absolute value and let $\left\vert \cdot \right\vert _2$ be the $2$-adic absolute value. Let $a_1=-1$, $a_2=8$, and $\varepsilon =1/10$, as in the remark right after Theorem 16.3.1. Then the theorem implies that there is an element $b\in \mathbf{Q}$ such that

$$\left\vert-1-b\right\vert _1 < \frac{1}{10}$$   and$$\qquad \left\vert 8-b\right\vert _2 < \frac{1}{10}.$$

As in the proof of the theorem, we can find such a $b$ by finding a $c_1, c_2\in\mathbf{Q}$ such that $\left\vert c_1\right\vert _1>1$ and $\left\vert c_1\right\vert _2<1$, and a $\left\vert c_2\right\vert _1<1$ and $\left\vert c_2\right\vert _2>1$. For example, $c_1=2$ and $c_2=1/2$ works, since $\left\vert 2\right\vert _1 = 2$ and $\left\vert 2\right\vert _2 = 1/2$ and $\left\vert 1/2\right\vert _1=1/2$ and $\left\vert 1/2\right\vert _2=2$. Again following the proof, we see that for sufficiently large $r$ we can take

$$b_r = \frac{c_1^r}{1+c_1^r} \cdot a_1 + \frac{c_2^r}{1+c_2^r} \cdot a_2$$

$$=\frac{2^r}{1+2^r} \cdot (-1) + \frac{(1/2)^r}{1+(1/2)^r} \cdot 8.$$

We have $b_1 = 2$, $b_2 = 4/5$, $b_3 = 0$, $b_4 = -8/17$, $b_5 = -8/11$, $b_6 = -56/55$. None of the $b_i$ work for $i<6$, but $b_6$ works.