A proof of a version of Hensel’s lemma 93
Proof of Proposition 2.1. By Lemma 2.1 there is a translation-invariant ultrametric d on R
such that
I
n
=
λ ∈ R; d(λ, 0) ≤
1
2
n
for each integer n ≥ 1, and the operation of addition in R is continuous. Moreover, if (λ
0
, µ
0
) ∈
R × R and n = 1, 2, . . . are arbitrary, the relations λ ∈ λ
0
+ I
n
, µ ∈ µ
0
+ I
n
imply
λµ − λ
0
µ
0
= λµ − λ
0
µ + λ
0
µ − λ
0
µ
0
= µ(λ − λ
0
) + λ
0
(µ − µ
0
) ∈ I
n
+ I
n
⊂ I
n
,
proving the continuity of the mapping
(λ, µ) ∈ R × R 7−→ λµ ∈ R
at (λ
0
, µ
0
).
Now, assume that (R, d) is complete and let f(X),
¯
f(X), λ, θ be as in the statement of the
proposition. In order to conclude the proof we shall apply Newton’s approximation method, as
in p. 44 of [6]. Let us ﬁrst observe that, if h(X) ∈ R[X] and γ ∈ R, then h(γ) =
¯
h(¯γ).
To prove the uniqueness, assume the existence of a µ ∈ R so that ¯µ = θ and f(µ) = 0. Since
¯
λ = θ is a simple root of
¯
f(X), there is a g(X) ∈ R[X] such that f(X) =
(X − λ) g(X) and ¯g(θ) 6= 0; thus
0 = f(µ) = (µ − λ) g(µ).
Therefore, since g(µ) = ¯g(θ) 6= 0, we conclude that g(µ) is an invertible element of R; hence
λ = µ.
To prove the existence, we claim that there is a sequence (λ
n
) n ≥ 1 in R so that
¯
λ
n
= θ,
f(λ
n
) ∈ I
n
and λ
n+1
− λ
n
∈ I
n
for each integer n ≥ 1. Indeed, let λ
1
∈ R be such that
¯
λ
1
= θ. Then f(λ
1
) =
¯
f(θ) = 0, that is, f(λ
1
) ∈ I
1
. Now, let n ≥ 1 be arbitrary, and
suppose the existence of a λ
n
∈ R such that
¯
λ
n
= θ and f(λ
n
) ∈ I
n
. Then, for every h ∈ I
n
,
(λ
n
+ h) − λ
n
∈ I
n
and (λ
n
+ h) =
¯
λ
n
+
¯
h = θ. We shall show the existence of an h ∈ I
n
with
f(λ
n
+ h) ∈ I
n+1
. In fact, by Taylor’s formula [4, p. 387], there is a ξ ∈ R so that
f(λ
n
+ h) = f(λ
n
) + hf
0
(λ
n
) + h
2
ξ.
And, by hypothesis, h
2
ξ = h(hξ) ∈ I
n
I
n
⊂ I
n
I
1
⊂ I
n+1
. But, since θ is a simple root of
¯
f(X), f
0
(λ
n
) = (
¯
f)
0
(θ) 6= 0, that is, f
0
(λ
n
) is an invertible element of R. Thus, by taking
h = −f(λ
n
)(f
0
(λ
n
))
−1
∈ I
n
and λ
n+1
= λ
n
+ h, we arrive at λ
n+1
= θ, f(λ
n+1
) ∈ I
n+1
and
λ
n+1
− λ
n
∈ I
n
, as desired.
Finally,
f(λ
n
)
n≥1
converges to 0 in R, because d
f(λ
n
), 0
≤
1
2
n
for n = 1, 2, . . . . On the
other hand, for n, ` = 1, 2, . . . ,
d(λ
n+`
, λ
n
) ≤ max{d(λ
n+`
, λ
n+`−1
), . . . , d(λ
n+1
, λ
n
)} ≤ max
1
2
n+`−1
, . . . ,
1
2
n
=
1
2
n
,
and hence (λ
n
) n ≥ 1 is a Cauchy sequence in (R, d). By the completeness of (R, d), there is a
λ ∈ R for which (λ
n
) n ≥ 1 converges. Consequently, in view of the continuity of the mappings
(α, β) ∈ R × R 7−→ α + β ∈ R and (α, β) ∈ R × R 7−→ αβ ∈ R,
Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 90–95