Product Formula for Norms on Non-zero Rationals/Lemma

Theorem

Let $\Q$ be the set of rational numbers.

Let $\Bbb P$ denote the set of prime numbers.

Let $z \in \Z_{\ne 0}$.

Then the following infinite product converges:

$\size z \times \ds \prod_{p \mathop \in \Bbb P}^{} \norm z_p = 1$

where:

$\size {\,\cdot\,}$ is the absolute value on $\Q$

$\norm {\,\cdot\,}_p$ is the $p$-adic norm on $\Q$ for prime number $p$

Proof

Case 1 : $z \in \Z_{>0}$

Let $z \in \Z_{>0}$.

From Fundamental Theorem of Arithmetic, we can factor $z$ as a product of one or more primes:

$z = p_1^{b_1} p_2^{b_2} \dots p_k^{b_k}$

Then for every prime number $q$:

$\norm z_q = \begin{cases}

p_i^{-b_i} & : \exists i \in \closedint 1 k :q = p_i \\ 1 & : \forall i \in \closedint 1 k : q \ne p_i \\ \end {cases}$

By definition of absolute value on $\Q$:

$\size z = p_1^{b_1} p_2^{b_2} \dots p_k^{b_k} $

For $n \ge \max \set{p_1, p_2 \dots p_k}$:

\(\ds \size z \times \prod_{p \mathop \in \Bbb P \mathop : p \mathop \le n } \norm z_p\)

\(=\)

\(\ds \paren {p_1^{b_1} p_2^{b_2} \dots p_k^{b_k} } \times \paren {p_1^{-b_1} p_2^{-b_2} \dots p_k^{-b_k} }\)

\(\ds \)

\(=\)

\(\ds 1\)

Hence:

\(\ds \size z \times \prod_{p \mathop\in \Bbb P} \norm z_p\)

\(=\)

\(\ds \lim_{n \mathop \to \infty} \paren{\size z \times \prod_{p \mathop \in \Bbb P \mathop : p \mathop \le n} \norm z_p}\)

Definition of Infinite Product

\(\ds \)

\(=\)

\(\ds 1\)

Eventually Constant Sequence Converges to Constant

$\Box$

Case 2 : $z \in \Z_{<0}$

Let $z \in \Z_{<0}$.

Hence

$-z \in \Z_{>0}$.

We have:

\(\ds \size z \times \prod_{p \mathop\in \Bbb P} \norm z_p\)

\(=\)

\(\ds \size {-z} \times \prod_{p \mathop \in \Bbb P} \norm {-z}_p\)

Norm of Negative

\(\ds \)

\(=\)

\(\ds 1\)

Case 1

$\Box$

In either case:

$\size z \times \ds \prod_{p \mathop \in \Bbb P} \norm z_p = 1$

$\blacksquare$