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Kronecker root-of-unity criterion

Statement

Let K be a number field and let 0≠α∈OK be an algebraic integer all of whose complex conjugates satisfy ∣σ(α)∣≤1. Then α is a root of unity.

Facts & Assumptions

Given: A number field K of degree n=[K:Q], the set Σ=Hom⁡Q(K,C) of its n embeddings into C, and an element 0≠α∈OK with ∣σ(α)∣≤1 for every σ∈Σ.

[F1]

K/Q is separable, because Q has characteristic zero, hence is perfect, and algebraic extensions of perfect fields are separable (Fields of characteristic zero, finite fields, and algebraically closed fields are perfect, Every algebraic extension of a perfect field is separable); so the norm is the product over the distinct embeddings, NK/Q(α)=∏σ∈Σσ(α), with ∣Σ∣=n=[K:Q] (Norm and trace from embeddings, with the inseparable exponent in the norm formula, Archimedean embeddings and signature).

[F2]

For α∈OK the norm NK/Q(α) is an integer (Trace and norm of an algebraic integer); if α≠0 then multiplication by α is an invertible linear map, so NK/Q(α)≠0 (The norm NK/F and trace Tr⁡K/F of a finite field extension, Ring of integers).

[F3]

An element β is conjugate to α over Q exactly when β is a complex root of the minimal polynomial mα∈Q[X] (Conjugate algebraic elements over a field). Sending an embedding τ:Q(α)→C to τ(α) is a bijection onto the set of distinct complex roots of mα (F-embeddings of F(α) into an algebraically closed field correspond to the distinct roots of mα); restriction Σ→Hom⁡Q(Q(α),C) is surjective (Restriction partitions embeddings in a finite tower into extension fibres); and for σ∈Σ one has mα(σ(α))=σ(mα(α))=0. Hence the set of complex roots of mα is exactly {σ(α):σ∈Σ}.

[F4]

Complex modulus is multiplicative, ∣zw∣=∣z∣ ∣w∣, and ∣z∣=0 only for z=0 (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive).

[F5]

Sums and products of elements integral over Z are integral (Integral elements over a nonzero base ring form a subring), so for m≥1 the power αm is a nonzero element of OK, the integral closure of Z in K (Ring of integers).

[F6]

For every m≥1 the minimal polynomial mαm of the algebraic element αm has coefficients in Z, and deg⁡mαm=[Q(αm):Q] divides [K:Q]=n (Minimal-polynomial criterion for algebraic integers, An element is algebraic over F if and only if its simple extension F(a)/F is finite, The degree of an intermediate field divides the degree of a finite extension).

[F7]

Every complex root w of mαm is the image of αm under a Q-embedding of Q(αm) into C (F-embeddings of F(α) into an algebraically closed field correspond to the distinct roots of mα). Such an embedding extends to a Q-embedding σ:K→C (Restriction partitions embeddings in a finite tower into extension fibres), so w=σ(α)m.

[F8]

For fixed n≥1 and R=1 there are only finitely many monic integer polynomials of degree at most n whose complex roots, counted with multiplicity, all have modulus at most 1 (Bounded roots give finitely many monic integer polynomials); this is the supplier consumed here, and the exact obligation used is this instance R=1.

[F9]

A nonzero polynomial of degree at most n over the integral domain C has at most n distinct roots (A nonzero polynomial of degree n over an integral domain has at most n distinct roots).

[F10]

An element ζ of K is a root of unity exactly when ζN=1 for some N≥1 (The group μn(K) of n-th roots of unity in a field, and primitive n-th roots of unity).

Proof

technique · the norm bounds every conjugate of $\alpha$ above by $1$ and below by $1$ simultaneously; the powers $\alpha^{m}$ then have norm-controlled minimal polynomials of bounded degree, and only finitely many such polynomials exist
1.1F2

Since 0≠α∈OK, the norm NK/Q(α) is a nonzero integer, so ∣NK/Q(α)∣≥1.

1.2F1

The set Σ of Q-embeddings K→C has n elements and NK/Q(α)=∏σ∈Σσ(α).

1.3F3given

The complex roots of mα are exactly the numbers σ(α) with σ∈Σ; each is a conjugate of α, so by the hypothesis ∣σ(α)∣≤1 for every σ∈Σ.

2.1F4step 1.2step 1.3

By multiplicativity of the modulus, ∣NK/Q(α)∣=∏σ∈Σ∣σ(α)∣, and every factor is at most 1 by step 1.3, so ∣NK/Q(α)∣≤1.

3.1step 1.1step 1.3step 2.1

Steps 1.1 and 2.1 give 1≤∣NK/Q(α)∣≤1, so ∣NK/Q(α)∣=1; a product of finitely many real numbers in [0,1] equals 1 only if every factor equals 1, so ∣σ(α)∣=1 for every σ∈Σ, and every complex root of mα has modulus exactly 1.

4.1F4F5F6F7step 3.1

Let m≥1. Then mαm∈Z[X] is monic of degree at most n. By [F7], every complex root w of mαm equals σ(α)m for some σ∈Σ; hence ∣w∣=∣σ(α)∣m=1 by step 3.1.

5.1F8F9step 4.1

By [F8] there are only finitely many monic integer polynomials of degree at most n whose complex roots all have modulus at most 1, and each of them has at most n distinct complex roots by [F9]; hence the union of the complex root sets of these finitely many polynomials is finite.

6.1step 4.1step 5.1

For every m≥1, αm is a complex root of mαm, so the set {αm:m≥1} is contained in the finite union of step 5.1 and is finite.

7.1F10step 6.1∎

Two distinct powers therefore coincide: αk=αm for integers k>m≥1, and since α≠0 this gives αk−m=1, so α is a root of unity.

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Sources