Number Theory: An Introduction to Mathematics

5 Hensel’s Lemma 283

Proposition 19 shows that if a quadratic polynomialat^2 +bt+cis irreducible,
then|b|≤max{|a|,|c|}, with strict inequality if|a| =|c|. Proposition 19 will now be
used to extend an absolute value on a given field to a finite extension of that field.

Proposition 20Let F be a field with a complete non-archimedean absolute value||.
If the field E is a finite extension of F , then the absolute value on F can be extended
to an absolute value on E.

Proof We will not only show that an extension of the absolute value exists, but we
will provide an explicit expression for it.
RegardEas a vector space overFof finite dimensionn, and with anya∈ E
associate the linear transformationLa:E → Edefined byLa(x) = ax.Then
detLa∈Fand we claim thatan extended absolute value is given by the formula

|a|=|detLa|^1 /n.

Evidently|a|≥0, and equality holds only ifa =0, sinceax =0forsome
x =0 impliesa =0. Furthermore|ab|=|a||b|,sinceLab =LaLband hence
detLab=(detLa)(detLb).Ifa∈F,thenLa=aInand hence the proposed absolute
value coincides with the original absolute value onF. It only remains to show that

|a−b|≤max(|a|,|b|) for alla,b∈F.

In fact we may suppose|a|≤|b|and then, by dividing byb, we see that it is sufficient
to show that 0<|a|≤1 implies| 1 −a|≤1.
To simplify notation, writeA=Laand let

f(t)=det(tI−A)=tn+cn− 1 tn−^1 +···+c 0

be thecharacteristic polynomialofA.Thenci∈Ffor alliandc 0 =(− 1 )ndetA.Let
g(t)be the monic polynomial inF[t] of least positive degree such thatg(a)=0.
Theng(t)is irreducible, since the fieldEhas no zero divisors. Evidentlyg(t)is
also theminimal polynomialofA. But, for an arbitrary linear transformation of an
n-dimensional vector space, the characteristic polynomial divides then-th power of
the minimal polynomial (see M. Deuring,Algebren, p.4). It follows in the present case
thatf(t)=g(t)rfor some positive integerr.
Suppose

g(t)=tm+bm− 1 tm−^1 +···+b 0

and leta ∈ Esatisfy|a|≤1 with respect to the proposed absolute value. Then
|c 0 |=|detA|≤1 and hence, sincebr 0 =c 0 ,|b 0 |≤1. Sincegis irreducible, it
follows from Proposition 19 that|bj|≤1forallj.Since

g( 1 )= 1 +bm− 1 +···+b 0 ,

this implies|g( 1 )|≤1 and hence|f( 1 )|≤1. Sincef( 1 )=det(I−A), this proves
that| 1 −a|≤1.

Finally we show that there is no other extension toEof the given absolute value
onFbesides the one constructed in the proof ofProposition 20.