278 VI Hensel’sp-adic Numbers
be a polynomial with coefficients c 0 ,...,cn∈R and let
f 1 (x)=ncnxn−^1 +(n− 1 )cn− 1 xn−^2 +···+c 1be its formal derivative. If|f(a 0 )|<|f 1 (a 0 )|^2 for some a 0 ∈R, then the equation
f(a)= 0 has a unique solution a∈R such that|a−a 0 |<|f 1 (a 0 )|.
Proof We consider first the existence ofaand postpone discussion of its uniqueness.
Put
σ:=|f 1 (a 0 )|> 0 ,θ 0 :=σ−^2 |f(a 0 )|< 1 ,and letDθdenote the set
{a∈R:|f 1 (a)|=σ,|f(a)|≤θσ^2 }.Thusa 0 ∈Dθ 0 andDθ′⊆Dθifθ′≤θ. We are going to show that, ifθ∈( 0 , 1 ),then
the ‘Newton’ map
Ta=a∗:=a−f(a)/f 1 (a)mapsDθintoDθ 2.
We can write
f(x+y)=f(x)+f 1 (x)y+···+fn(x)yn,wheref 1 (x)has already been defined andf 2 (x),...,fn(x)are also polynomials with
coefficients fromR. We substitute
x=a,y=b:=−f(a)/f 1 (a),wherea∈Dθ.Then|fj(a)|≤1, sincea∈Randfj(x)∈ R[x](j= 1 ,...,n).
Furthermore
|b|=σ−^1 |f(a)|≤θσ <σ.Thusb∈R.Sincef(a)+f 1 (a)b=0, it follows thata∗=a+bsatisfies
|f(a∗)|≤max
2 ≤j≤n|fj(a)bj|≤|b|^2 =σ−^2 |f(a)|^2 ≤θ^2 σ^2.Similarly, sincef 1 (a+b)−f 1 (a)can be written as a polynomial inbwith coefficients
fromRand with no constant term,
|f 1 (a+b)−f 1 (a)|≤|b|<σ=|f 1 (a)|and hence|f 1 (a∗)|=σ. This completes the proof thatTDθ⊆Dθ 2.
Now putak=Tka 0 ,sothat
ak+ 1 −ak=−f(ak)/f 1 (ak).It follows by induction from what we have proved that