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

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The p-adic absolute value is nonarchimedean

Statement

Let $p$ be a prime and let $∣ \cdot ∣_{p}$ be the absolute value of The p-adic absolute value on the rationals. Then for all $x, y \in Q$ one has

$∣ x y ∣_{p} = ∣ x ∣_{p} ∣ y ∣_{p}, ∣ x + y ∣_{p} \leq \max {∣ x ∣_{p}, ∣ y ∣_{p}} .$

So $∣ \cdot ∣_{p}$ is a nonarchimedean absolute value on $Q$ .

Facts & Assumptions

Given: A prime $p$ and rational numbers $x, y$ .

[L1]

On $Q^{\times}$ , the valuation $v_{p}$ is well defined, additive under multiplication, and satisfies $v_{p} (x + y) \geq \min {v_{p} (x), v_{p} (y)}$ whenever $x, y, x + y$ are nonzero (The $p$ -adic valuation extends to the nonzero rationals by $v_{p} (a / b) : = v_{p} (a) - v_{p} (b) \in Z$ , independently of the representation; it satisfies $v_{p} (x y) = v_{p} (x) + v_{p} (y)$ , and $v_{p} (x + y) \geq \min {v_{p} (x), v_{p} (y)}$ whenever $x$ , $y$ and $x + y$ are nonzero).

[L2]

The $p$ -adic absolute value is defined by $∣ x ∣_{p} = p^{- v_{p} (x)}$ for nonzero $x$ , with $∣ 0 ∣_{p} = 0$ (The p-adic absolute value on the rationals).

Proof

technique · direct

1.1

If one of $x$ or $y$ is zero then $∣ x y ∣_{p} = 0 = ∣ x ∣_{p} ∣ y ∣_{p}$ . If both are nonzero, [L1] and [L2] give $∣ x y ∣_{p} = p^{- v_{p} (x y)} = p^{- v_{p} (x) - v_{p} (y)} = ∣ x ∣_{p} ∣ y ∣_{p} .$

L1L2givenalgebra

1.2

If $x + y = 0$ , then $∣ x + y ∣_{p} = 0 \leq \max {∣ x ∣_{p}, ∣ y ∣_{p}}$ . If $x, y, x + y$ are nonzero, [L1] gives $v_{p} (x + y) \geq \min {v_{p} (x), v_{p} (y)},$ so exponentiating by $p^{- t}$ , which reverses order, yields $∣ x + y ∣_{p} \leq \max {∣ x ∣_{p}, ∣ y ∣_{p}} .$

L1L2givenalgebra

2.1

The multiplicative law is step 1.1 and the strong triangle inequality is step 1.2, so $∣ \cdot ∣_{p}$ is nonarchimedean.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

23 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

Andrew V. Sutherland, 18.782 Lecture 5 (standard reference, not scraped)
J. S. Milne, Algebraic Number Theory, Chapter 7 (standard reference, not scraped)