proof of norm and trace of algebraic number

Theorem 1.

Let K be a number fieldMathworldPlanetmath and αK.  The norm N(α) and the trace T(α) of α in the field extension K/Q both are rational numbers and especially rational integers in the case α is an algebraic integerMathworldPlanetmath.  If β is another element of K, then N(αβ)=N(α)N(β) and T(α+β)=T(α)+T(β).   If   [K:Q]=n  and  aQ, then N(a)=an and T(a)=na.

Before proving this theorem, a lemma will be stated and proven.


Let K be a number field with [K:Q]=n, αK such that [Q(α):Q]=d, and N*(α) and T*(α) denote the absolute norm ( and absolute trace ( of α, respectively.   Then d divides n, N(α)=(N*(α))nd, and T(α)=ndT*(α).


Note that d divides n because n=[K:]=[K:(α)][(α):]=[K:(α)]d.

Note also that each of the d embeddingsPlanetmathPlanetmathPlanetmath of (α) into extends to exactly nd embeddings of K into . Thus,

N(α)=σ emb. of Kσ(α)=σ emb. of Kσ|(α)(α)=(τ emb. of (α)τ(α))nd=(N*(α))nd


T(α)=σ emb. of Kσ(α)=σ emb. of Kσ|(α)(α)=ndτ emb. of (α)τ(α)=ndT*(α).

Now, the above theorem will be proven.

Proof of theorem 1. Let f(x)[x] be the minimal polynomial for α over . Then degf=d, where d is as in the previous lemma. Note that |N*(α)| is equal to the absolute valueMathworldPlanetmathPlanetmathPlanetmath of the constant term of f and that T*(α) is equal to the opposite of the coefficient of xd-1 of f. Thus, N*(α),T*(α). Therefore, N(α)=(N*(α))nd and T(α)=ndT*(α). Moreover, if α is an algebraic integer, then f(x)[x], N*(α),T*(α), N(α)=(N*(α))nd, and T(α)=ndT*(α).

If a, then d=1, N(a)=(N*(a))n=an, and T(a)=nT*(a)=na.

Finally, if α,βK, then

N(αβ)=σ emb. of Kσ(αβ)=σ emb. of Kσ(α)σ(β)=(σ emb. of Kσ(α))(σ emb. of Kσ(β))=N(α)N(β)


T(α+β)=σ emb. of Kσ(α+β)=σ emb. of Kσ(α)+σ(β)=σ emb. of Kσ(α)+σ emb. of Kσ(β)=T(α)+T(β).

Theorem 2.

An algebraic integer ε is a unit if and only if its N*(ε)=±1,.   Thus, in the minimal polynomial of an algebraic unit is always  ±1.


Let K=(ε). Since ε is an algebraic integer, d=[K:] is finite. Let 𝒪K denote the ring of integersMathworldPlanetmath of K.

If N*(ε)=±1, then let f(x)[x] be the minimal polynomial of ε over . Let a1,,ad-1 such that f(x)=xd+j=1d-1ajxj±1. Then 0=f(ε)=εd+j=1d-1ajεj±1. Thus, ε(εd-1+j=1d-1ajεj-1)=±1. Since εd-1+j=1d-1ajεj-1𝒪K, it follows that ε is a unit in 𝒪K.

Conversely, let ε be a unit in 𝒪K. Let υ𝒪K with ευ=1. Since N*(ε)N*(υ)=N*(ευ)=N*(1)=1 and N*(ε),N*(υ), it follows that N*(ε)=±1. ∎


  • 1 Marcus, Daniel A. Number Fields. New York: Springer-Verlag, 1977.
Title proof of norm and trace of algebraic number
Canonical name ProofOfNormAndTraceOfAlgebraicNumber
Date of creation 2013-03-22 15:58:53
Last modified on 2013-03-22 15:58:53
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 27
Author Wkbj79 (1863)
Entry type Proof
Classification msc 11R04