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

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Two nontrivial absolute values induce the same topology exactly when one is a positive power of the other

Statement

Let $F$ be a field, and let $∣ \cdot ∣_{1}$ and $∣ \cdot ∣_{2}$ be nontrivial absolute values on $F$ . Then they induce the same topology on $F$ if and only if they are equivalent in the sense of Equivalent nontrivial absolute values.

Facts & Assumptions

Given: A field $F$ and nontrivial absolute values $∣ \cdot ∣_{1}$ and $∣ \cdot ∣_{2}$ on $F$ .

[L1]

Absolute values are multiplicative and have their open unit balls available, and equivalence means equality up to a positive power (Absolute values on a field, Equivalent nontrivial absolute values).

Proof

technique · direct

1.1

If $∣ x ∣_{2} = ∣ x ∣_{1}^{c}$ for some $c > 0$ , then $t \mapsto t^{c}$ is increasing on $R_{> 0}$ , so the conditions $∣ x - a ∣_{1} < r$ and $∣ x - a ∣_{2} < r^{c}$ define the same neighborhoods of every point $a$ . Hence the two absolute values induce the same topology.

L1givenalgebra

1.2

Assume conversely that the two topologies agree. Then $∣ x ∣_{1} < 1 ⟺ x^{n} \to 0 in ∣ \cdot ∣_{1} ⟺ x^{n} \to 0 in ∣ \cdot ∣_{2} ⟺ ∣ x ∣_{2} < 1,$ and the same argument with $x^{- 1}$ gives $∣ x ∣_{1} > 1 ⟺ ∣ x ∣_{2} > 1$ for every nonzero $x$ .

L1givenalgebra

2.1

Choose $a \in F^{\times}$ with $∣ a ∣_{1} > 1$ ; then step 1.2 gives $∣ a ∣_{2} > 1$ . Set $c : = \frac{\log ∣ a ∣_{2}}{\log ∣ a ∣_{1}} > 0.$ Fix $x \in F^{\times}$ . If $m / n < \log ∣ x ∣_{1} / \log ∣ a ∣_{1}$ , then $∣ x ∣_{1}^{n} < ∣ a ∣_{1}^{m}$ , so $∣ x^{n} a^{- m} ∣_{1} < 1$ and therefore $∣ x^{n} a^{- m} ∣_{2} < 1$ by step 1.2. This gives $n \log ∣ x ∣_{2} < m \log ∣ a ∣_{2}$ . Reversing the inequality yields the opposite bound. Rational approximation therefore forces $\frac{\log ∣ x ∣_{2}}{\log ∣ a ∣_{2}} = \frac{\log ∣ x ∣_{1}}{\log ∣ a ∣_{1}},$ so $∣ x ∣_{2} = ∣ x ∣_{1}^{c}$ .

step 1.2givenalgebra

3.1

Step 1.1 proves one direction and step 2.1 proves the other, so the two topologies agree exactly when the absolute values are equivalent.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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 7.1 (standard reference, not scraped)
Keith Conrad, Ostrowski's Theorem for Q, Section 1 (standard reference, not scraped)