Number Theory: An Introduction to Mathematics

282 VI Hensel’sp-adic Numbers

Put

gj(x)=xm+

m∑− 1

i= 0

α(ij)xi, hj(x)=

n∑−m

i= 0

βi(j)xi.

By (iii), the sequences(α
(j)
i )and(β

(j)
i )are fundamental sequences for eachiand
hence, sinceFis complete, there existαi,βi∈Rsuch that

α(ij)→αi,βi(j)→βiasj→∞.

If

g(x)=xm+

m∑− 1

i= 0

αixi, hj(x)=

n∑−m

i= 0

βixi,

then, for eachj≥1,

g−gj∈πjR[x], h−hj∈πjR[x].

Since

f−gh=f−gjhj−(g−gj)h−gj(h−hj),

it follows thatf−gh∈πjR[x] for eachj≥1. Hencef =gh. It is obvious thatg
andhhave the other required properties.

As an application of this form of Hensel’s lemma we prove

Proposition 19Let F be a field with a complete non-archimedean absolute value||
and let

f(t)=cntn+cn− 1 tn−^1 +···+c 0 ∈F[t].

If c 0 cn= 0 and, for some m such that 0 <m<n,

|c 0 |≤|cm|, |cn|≤|cm|,

with at least one of the two inequalities strict, then f is reducible over F.

Proof Suppose first that|c 0 |<|cm|and|cn|≤|cm|. Evidently we may choosemso
that|cm|=max 0 <i<n|ci|and|ci|<|cm|for 0≤i<m. By multiplyingfbyc−m^1 we
may further assume that, ifRis the valuation ring ofF,thenf(t)∈R[t],cm=1and
|ci|<1for0≤i<m. Hence

f ̄(t)=tm(c ̄ntn−m+ ̄cn− 1 tn−m−^1 +···+ 1 ).

Since the two factors are relatively prime, it follows from Proposition 18 thatf is
reducible.
If|cn| < |cm|and|c 0 |≤|cm|, then the same argument also applies to the
polynomialtnf(t−^1 ).