Number Theory: An Introduction to Mathematics

5 Hensel’s Lemma 281

Proof Putn=∂(f)andm=∂(φ).Then∂(ψ)=∂(f ̄)−∂(φ)≤n−m.Thereexist
polynomialsg 1 ,h 1 ∈R[x], withg 1 monic,∂(g 1 )=mand∂(h 1 )≤n−m, such that
g ̄ 1 =φ,h ̄ 1 =ψ.Sinceφ,ψare relatively prime, there exist polynomialsχ,ω∈k[x]
such that

χφ+ωψ= 1 ,

and there exist polynomialsu,v∈R[x] such thatu ̄=χ,v ̄=ω. Thus

f−g 1 h 1 ∈M[x], ug 1 +vh 1 − 1 ∈M[x].

Iff=g 1 h 1 , there is nothing more to do. Otherwise, letπbe the coefficient off−g 1 h 1
or ofug 1 +vh 1 −1 which has maximum absolute value. Then

f−g 1 h 1 ∈πR[x], ug 1 +vh 1 − 1 ∈πR[x].

We are going to construct inductively polynomialsgj,hj∈R[x] such that
(i)g ̄j=φ,h ̄j=ψ;
(ii)gjis monic and∂(gj)=m,∂(hj)≤n−m;
(iii)gj−gj− 1 ∈πj−^1 R[x],hj−hj− 1 ∈πj−^1 R[x];
(iv) f−gjhj∈πjR[x].

This holds already forj=1 withg 0 =h 0 =0. Assume that, for somek≥2, it
holds for allj<kand putf−gjhj=πjj,wherej∈R[x]. Sinceg 1 is monic,
the Euclidean algorithm provides polynomialsqk,rk∈R[x] such that

k− 1 v=qkg 1 +rk,∂(rk)<∂(g 1 )=m.

Letwk ∈ R[x] be a polynomial of minimal degree such that all coefficients of
k− 1 u+qkh 1 −wkhave absolute value at most|π|.Then

wkg 1 +rkh 1 −k− 1 =(ug 1 +vh 1 − 1 )k− 1 −(k− 1 u+qkh 1 −wk)g 1 ∈πR[x].

We will show that∂(wk)≤n−m. Indeed otherwise

∂(wkg 1 )>n≥∂(rkh 1 −k− 1 )

and hence, sinceg 1 is monic,wkg 1 +rkh 1 −k− 1 has the same leading coefficient as
wk. Consequently the leading coefficient ofwkis inπR. Thus the polynomial obtained
fromwkby omitting the term of highest degree satisfies the same requirements aswk,
which is a contradiction.
If we put

gk=gk− 1 +πk−^1 rk, hk=hk− 1 +πk−^1 wk,

then (i)–(iii) are evidently satisfied forj=k. Moreover

f−gkhk=−πk−^1 (wkgk− 1 +rkhk− 1 −k− 1 )−π^2 k−^2 rkwk

and

wkgk− 1 +rkhk− 1 −k− 1

=wkg 1 +rkh 1 −k− 1 +wk(gk− 1 −g 1 )+rk(hk− 1 −h 1 )∈πR[x].

Hence also (iv) is satisfied forj=k.