TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04

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The product formula for the rationals

Statement

For every nonzero rational number $x$ ,

$\prod_{v \leq \infty} ∣ x ∣_{v} = 1,$

where $v$ runs over the archimedean place and all prime places, and all but finitely many factors are equal to $1$ .

Facts & Assumptions

Given: A nonzero rational number $x$ .

[L1]

The finite places of $Q$ are represented by the normalized $p$ -adic absolute values (Places of the rationals, The p-adic absolute value on the rationals).

[L2]

A nonzero rational has a finite prime factorization in lowest terms (For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list).

Proof

technique · direct

1.1

Write $x = \pm \prod_{i = 1}^{r} p_{i}^{e_{i}}$ with distinct primes $p_{i}$ and integers $e_{i}$ , as supplied by [L2]. Then $∣ x ∣_{p_{i}} = p_{i}^{- e_{i}}$ for the listed primes and $∣ x ∣_{q} = 1$ for every other prime $q$ .

L1L2givenalgebra

2.1

The archimedean factor is $∣ x ∣_{\infty} = \prod_{i = 1}^{r} p_{i}^{e_{i}}$ , so $∣ x ∣_{\infty} \prod_{i = 1}^{r} ∣ x ∣_{p_{i}} = (\prod_{i = 1}^{r} p_{i}^{e_{i}}) (\prod_{i = 1}^{r} p_{i}^{- e_{i}}) = 1.$ Since the omitted prime factors are all $1$ , this is exactly the full product formula.

step 1.1algebra∎

Depends on

Used by

Weak approximation for rational places Theorem

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Number Theory, Proposition 7.2 (standard reference, not scraped)