False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: $μ_{n} (K)$ has $n$ elements in every field $K$

Statement

False claim. For every field $K$ and every $n \geq 1$ , the group $μ_{n} (K)$ of $n$ -th roots of unity in $K$ has exactly $n$ elements.

The witness below shows two different failures: over $Q$ there is no primitive cube root of unity in the field at all, while in characteristic $3$ the equation $x^{3} = 1$ is inseparable and has only one root.

Facts & Assumptions

Given: The groups $μ_{n} (K)$ of roots of unity.

[L1]

$μ_{n} (K)$ is cyclic of order dividing $n$ , and it has a primitive $n$ -th root of unity exactly when its order is $n$ ( $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ ).

[L2]

$[Q (ζ_{3}) : Q] = φ (3) = 2$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ).

[L3]

In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ).

Refutation

technique · direct

1.1

If $μ_{3} (Q)$ had three elements, then [L1] would give a primitive cube root of unity $ζ_{3}$ in $Q$ . But then $Q (ζ_{3}) = Q$ , contradicting [L2], which says this extension has degree $2$ . So $μ_{3} (Q)$ does not have three elements.

L1L2

1.2

If $K$ has characteristic $3$ , then [L3] gives $μ_{3} (K) = {1}$ , so again $μ_{3} (K)$ does not have three elements.

2.1

The false claim fails already at $n = 3$ , both over $Q$ and over every field of characteristic $3$ .

step 1.1step 1.2∎

Remarks

The two failures have different causes. Over $Q$ the polynomial $t^{3} - 1$ is separable but its nontrivial roots lie in a quadratic extension; in characteristic $3$ the polynomial itself collapses to $(t - 1)^{3}$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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

P. L. Clark, Field Theory (course notes/monograph), Proposition 9.4 (standard reference, not scraped)
K. Conrad, Cyclotomic Extensions (expository blurb), Section 1 (standard reference, not scraped)

FALSE: μn(K)μn​(K) has nn elements in every field KK