The $N$-adic Numbers

Lemma 14.2.5   Let $N$ be a positive integer. Then for any nonzero rational number $\alpha$ there exists a unique $e\in\mathbf{Z}$ and integers $a$, $b$, with $b$ positive, such that $\alpha = N^e \cdot \frac{a}{b}$ with $N\nmid a$, $\gcd(a,b)=1$, and $\gcd(N,b)=1$.

Proof. Write $\alpha = c/d$ with $c,d\in\mathbf{Z}$ and $d>0$. First suppose $d$ is exactly divisible by a power of $N$, so for some $r$ we have $N^r\mid d$ but $\gcd(N,d/N^r)=1$. Then

$$\frac{c}{d} = N^{-r} \frac{c}{d/N^r}.$$

If $N^s$ is the largest power of $N$ that divides $c$, then $e=s-r$, $a=c/N^s$, $b=d/N^r$ satisfy the conclusion of the lemma.

By unique factorization of integers, there is a smallest multiple $f$ of $d$ such that $fd$ is exactly divisible by $N$. Now apply the above argument with $c$ and $d$ replaced by $cf$ and $df$. $\qedsymbol$

Definition 14.2.6 ($N$-adic valuation)   Let $N$ be a positive integer. For any positive $\alpha\in\mathbf{Q}$, the $N$-adic valuation of $\alpha$ is $e$, where $e$ is as in Lemma 14.2.5. The $N$-adic valuation of 0 is $\infty$.

We denote the $N$-adic valuation of $\alpha$ by $\ord _N(\alpha)$. (Note: Here we are using ``valuation'' in a different way than in the rest of the text. This valuation is not an absolute value, but the logarithm of one.)

Definition 14.2.7 ($N$-adic metric)   For $x,y\in\mathbf{Q}$ the $N$-adic distance between $x$ and $y$ is

$$d_N(x,y) = N^{-\ord _N(x-y)}.$$

We let $d_N(x,x) = 0$, since $\ord _N(x-x)=\ord _N(0)=\infty$.

For example, $x,y\in\mathbf{Z}$ are close in the $N$-adic metric if their difference is divisible by a large power of $N$. E.g., if $N=10$ then $93427$ and $13427$ are close because their difference is $80000$, which is divisible by a large power of $10$.

Proposition 14.2.8   The distance $d_N$ on  $\mathbf {Q}$ defined above is a metric. Moreover, for all $x,y,z\in\mathbf{Q}$ we have

$$d(x,z) \leq \max(d(x,y),d(y,z)).$$

(This is the ``nonarchimedean'' triangle inequality.)

Proof. The first two properties of Definition 14.2.1 are immediate. For the third, we first prove that if $\alpha,\beta\in\mathbf{Q}$ then

$$\ord _N(\alpha+\beta)\geq \min(\ord _N(\alpha),\ord _N(\beta)).$$

Assume, without loss, that $\ord _N(\alpha) \leq \ord _N(\beta)$ and that both $\alpha$ and $\beta$ are nonzero. Using Lemma 14.2.5 write $\alpha=N^e(a/b)$ and $\beta=N^f(c/d)$ with $a$ or $c$ possibly negative. Then

$$\alpha + \beta = N^e \left(\frac{a}{b} + N^{f-e}\frac{c}{d}\right) = N^e \left(\frac{ad+bcN^{f-e}}{bd}\right).$$

Since $\gcd(N,bd)=1$ it follows that $\ord _N(\alpha+\beta)\geq e$. Now suppose $x,y,z\in\mathbf{Q}$. Then

$$x-z = (x-y) + (y-z),$$

so

$$\ord _N(x-z) \geq \min (\ord _N(x-y), \ord _N(y-z)),$$

hence $d_N(x,z) \leq \max(d_N(x,y), d_N(y,z))$. $\qedsymbol$

We can finally define the $N$-adic numbers.

Definition 14.2.9 (The $N$-adic Numbers)   The set of $N$-adic numbers, denoted $\mathbf {Q}_N$, is the completion of  $\mathbf {Q}$ with respect to the metric $d_N$.

The set $\mathbf {Q}_N$ is a ring, but it need not be a field as you will show in Exercises 11 and 12. It is a field if and only if $N$ is prime. Also, $\mathbf {Q}_N$ has a ``bizarre'' topology, as we will see in Section 14.2.3.