5 Hensel’s Lemma 279|f(ak)|≤θ^2k
0 σ(^2).
Sinceθ 0 <1and|ak+ 1 −ak|=σ−^1 |f(ak)|, this shows that{ak}is a fundamental
sequence. Hence, sinceFis complete,ak→afor somea∈R. Evidentlyf(a)= 0
and|f 1 (a)|=σ. Since, for everyk≥1,
|ak−a 0 |≤max
1 ≤j≤k
|aj−aj− 1 |≤θ 0 σ,
we also have|a−a 0 |≤θ 0 σ<σ.
To prove uniqueness, assumef(a ̃)=0forsomea ̃=asuch that| ̃a−a 0 |<σ.If
we putb= ̃a−a,then
0 =f(a ̃)−f(a)=f 1 (a)b+···+fn(a)bn.
Fromb= ̃a−a 0 −(a−a 0 )we obtain|b|<σ.Sinceb=0and|fj(a)|≤1, it
follows that, forj≥2,
|fj(a)bj|≤|b|^2 <σ|b|=|f 1 (a)b|.
But this implies
|f(a ̃)−f(a)|=|f 1 (a)b|> 0 ,
which is a contradiction.
As an application of Proposition 15 we willdetermine which elements of the field
Qpofp-adic numbers are squares. Sinceb=a^2 impliesb=p^2 vb′,wherev∈Zand
|b′|p=1, we may restrict attention to the case|b|p=1.
Proposition 16Suppose b∈Qpand|b|p= 1.
If p= 2 ,thenb=a^2 for some a∈Qpif and only if|b−a 02 |p< 1 for some a 0 ∈Z.
If p= 2 ,thenb=a^2 for some a∈Q 2 if and only if|b− 1 | 2 ≤ 2 −^3.
Proof Suppose first thatp=2. Ifb=a^2 for somea∈Qp,then|a|p=1and
|a−a 0 |p<1forsomea 0 ∈Z,sinceZis dense inZp. Hence|a 0 |p=1and
|b−a 02 |p=|a−a 0 |p|a+a 0 |p≤|a−a 0 |p< 1.
Conversely, suppose|b−a 02 |p<1forsomea 0 ∈Z.Then|a^20 |p =1andso
|a 0 |p=1. In Proposition 15 takeF=Qpandf(x)=x^2 −b. The hypotheses of the
proposition are satisfied, since|f(a 0 )|p<1and|f 1 (a 0 )|p=| 2 a 0 |p=1, and hence
b=a^2 for somea∈Qp.
Suppose next that p =2. Ifb =a^2 for somea ∈ Q 2 ,then|a| 2 = 1and
|a−a 0 | 2 ≤ 2 −^3 for somea 0 ∈Z,sinceZis dense inZ 2. Hence|a 0 | 2 =1and
|b−a 02 | 2 =|a−a 0 | 2 |a+a 0 | 2 ≤|a−a 0 | 2 ≤ 2 −^3.
Sincea 0 is odd, we havea 0 ≡±1 mod 4 anda^20 ≡1 mod 8. Hence
|b− 1 | 2 ≤max{|b−a 02 | 2 ,|a^20 − 1 | 2 }≤ 2 −^3.