Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 90–95
A proof of a version of Hensel’s lemma
Una prueba de una versi´on del lema de Hensel
Dinam´erico P. Pombo Jr. (dpombojr@gmail.com)
Instituto de Matem´atica e Estat´ıstica
Universidade Federal Fluminense
Rua Professor Marcos Waldemar de Freitas Reis, s/n
o
Bloco G, Campus do Gragoat´a
24210-201 Niteri, RJ Brasil
Abstract
By using a few basic facts, a proof of a known version of Hensel’s lemma in the context
of local rings is presented.
Key words and phrases: local rings, discrete valuation rings, Hensel’s lemma.
Resumen
Usando algunos pocos hechos b´asicos, se presenta una demostraci´on de una versi´on del
lema de Hensel en el contexto de los anillos locales.
Palabras y frases clave: anillos locales, anillos de valoraci´on discretos, lema de Hensel.
1 Introduction
A classical and fundamental result, known as Hensel’s lemma, is discussed in [1], [3], [5], [6] and
[7], for instance. A quite general form of Hensel’s lemma may be found in Chapter III of [2],
although special cases of it may also be very important, as the one valid in the framework of local
rings and presented in Chapter II of [6]. The main purpose of this note is to offer an elementary
proof of the last-mentioned form of Hensel’s lemma, as well as to derive a few consequences of it.
2 A proof of a version of Hensel’s lemma
Definition 2.1 (cf. [2, p. 80]). A commutative ring R with and identity element 1 6= 0 is said to
be a local ring if it contains a unique maximal ideal I
1
, namely, the set of non-invertible elements
of R. If K is the quotient ring R/I
1
, which is a field,
λ ∈ R 7−→
¯
λ ∈ K
will denote the canonical surjection. For f(X) = a
0
+ a
1
X + · · · + a
n
X
n
∈ R[X], we will write
¯
f(X) = ¯a
0
+ ¯a
1
X + · · · + ¯a
n
X
n
∈ K[X].
Received 29/06/2021. Revised 30/06/2021. Accepted 26/07/2021.
MSC (2010): Primary 12J25, 13F30; Secondary 13H99, 13J10, 13B25.
Corresponding author: Dinam´erico P. Pombo Jr.
A proof of a version of Hensel’s lemma 91
Example 2.1 (cf. [6]). Let R be a discrete valuation ring and I
1
the maximal ideal of R, which
may be written as I
1
= π R. We have that
I
1
= π R ⊃ I
2
= π
2
R ⊃ · · · ⊃ I
n
= π
n
R ⊃ I
n+1
= π
n+1
R ⊃ ....
is a decreasing sequence of ideals of R such that I
n
I
1
⊂ I
n+1
for each integer n ≥ 1 and
T
n≥1
I
n
= {0}.
Example 2.2 (cf. [3]). Let K be a field endowed with a non-trivial discrete valuation | · |,
R = {λ ∈ R; |λ| ≤ 1} the ring of integers of (K, | · |) and I
1
= {λ ∈ R; |λ| < 1} the maximal ideal
of R. Let µ ∈ I
1
be such that |µ| = sup{|λ|; λ ∈ I
1
}. Then
I
1
= µ R ⊃ I
2
= µ
2
R ⊃ · · · ⊃ I
n
= µ
n
R ⊃ I
n+1
= µ
n+1
R ⊃ ....
is a decreasing sequence of ideals of R such that I
n
I
1
⊂ I
n+1
for each integer n ≥ 1 and
T
n≥1
I
n
= {0}.
It may be seen that every discrete valuation ring may be regarded as the ring of integers of a
field endowed with a non-trivial discrete valuation.
Let us recall that, if X is a non-empty set, a mapping
d: X × X −→ R
+
is an ultrametric on X if the following conditions hold for all x, y, z ∈ X:
(a) d(x, y) = 0 if and only if x = y;
(b) d(x, y) = d(y, x);
(c) d(x, y) ≤ max{d(x, z), d(z, y)}.
By induction,
d(x
1
, x
n
) ≤ max{d(x
1
, x
2
), . . . , d(x
n−1
, x
n
)}
for n = 2, 3, . . . and x
1
, . . . , x
n
∈ X. And, since max{d(x, z), d(z, y)} ≤ d(x, z) + d(z, y), d is a
metric on X.
We shall present an elementary proof of the following form of Hensel’s lemma [6, p. 43]:
Proposition 2.1. Let R be a local ring and I
1
its maximal ideal, and assume the existence of a
decreasing sequence I
1
⊃ I
2
⊃ · · · ⊃ I
n
⊃ I
n+1
⊃ .... of ideals of R such that I
n
I
1
⊂ I
n+1
for
each integer n ≥ 1 and
T
n≥1
I
n
= {0}. Then there exists a translation-invariant ultrametric d on
R such that I
n
=
λ ∈ R; d(λ, 0) ≤
1
2
n
for each integer n ≥ 1 (thus (I
n
) n ≥ 1 is a fundamental
system of neighborhoods of 0 in R with respect to the topology defined by d ) and the mappings
(λ, µ) ∈ R × R 7−→ λ + µ ∈ R and (λ, µ) ∈ R × R 7−→ λµ ∈ R
are continuous. Moreover, if the metric space (R, d) is complete and if f(X) ∈ R[X] is such
that
¯
f(X) admits a simple root θ in K, then there exists a unique root λ of f (X) in R such that
¯
λ = θ.
Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 90–95
92 Dinam´erico P. Pombo Jr.
In order to prove Proposition 2.1 we shall need an auxiliary result:
Lemma 2.1. Let (G, +) be a commutative group and H
1
⊃ H
2
⊃ · · · ⊃ H
n
⊃ H
n+1
⊃ · · · a
decreasing sequence of subgroups of G such that
T
n≥1
H
n
= {0}. Then there exists a translation-
invariant ultrametric d on G such that H
n
=
x ∈ G; d(x, 0) ≤
1
2
n
for each integer n ≥ 1
(thus (H
n
) n ≥ 1 is a fundamental system of neighborhoods of 0 in G with respect to the topology
defined by d ) and the mapping
(x, y) ∈ G × G 7−→ x + y ∈ G
is continuous.
Proof of Lemma 2.1. We shall use a classical argument. Put H
0
= G and let g : G → R
+
be
the mapping given by g(0) = 0 and g(x) =
1
2
n
if x ∈ H
n
\H
n+1
(n = 0, 1, 2, . . . ). Obviously,
g(x) > 0 if g ∈ G\{0}, g(−x) = g(x) if x ∈ G and
H
n
=
x ∈ G; g(x) ≤
1
2
n
for n = 0, 1, 2, . . . . Moreover, g(x+y) ≤ max{g(x), g(y)} for all x, y ∈ G, which is clear if x = 0 or
y = 0. Indeed, if x, y ∈ G\{0}, x ∈ H
k
\H
k+1
, y ∈ H
`
\H
`+1
, with ` ≥ k ≥ 0, then g(x) =
1
2
k
and
g(y) =
1
2
`
≤
1
2
k
· But, since H
`
⊂ H
k
, x + y ∈ H
k
, and hence g(x + y) ≤
1
2
k
= max{g(x), g(y)}.
Therefore the mapping
d: G × G −→ R
+
,
defined by d(x, y) = g(x − y), is a translation-invariant ultrametric on G such that
H
n
=
t ∈ G; d(t, 0) ≤
1
2
n
for each integer n ≥ 0. Consequently,
x + H
n
=
t ∈ G; d(t, x) ≤
1
2
n
if x ∈ G and n = 0, 1, 2, . . . are arbitrary.
Finally, if x
0
, y
0
∈ G and n = 0, 1, 2, . . . are arbitrary,
(x
0
+ H
n
) + (y
0
+ H
n
) ⊂ (x
0
+ y
0
) + H
n
,
proving the continuity of the mapping
(x, y) ∈ G × G 7−→ x + y ∈ G
at (x
0
, y
0
).
Now, let us turn to the
Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 90–95
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
94 Dinam´erico P. Pombo Jr.
f(λ
n
)
n ≥ 1 converges to f(λ); thus f(λ) = 0.
Now, let us consider K = R/I
1
endowed with the discrete ultrametric d
0
, given by d
0
(s, s) = 0
and d
0
(s, t) = 1 if s 6= t (s, t ∈ K). Since the canonical surjection
λ ∈ (R, d) 7−→
¯
λ ∈ (K, d
0
)
is continuous (
¯
I
1
= {0}) and (λ
n
) n ≥ 1 converges to λ, (
¯
λ
n
) n ≥ 1 converges to
¯
λ. Therefore
¯
λ = θ.
Corollary 2.1. Let R be a discrete valuation ring which is complete under the ultrametric d
given in Proposition 2.1. Let f(X) ∈ R[X] be such that
¯
f(X) ∈ K[X] admits a simple root θ.
Then there exists a unique root λ of f(X) such that
¯
λ = θ.
Proof. Follows immediately from Proposition 2.1, by recalling Example 2.1.
Remark 2.1. Let (K, | · |) and I
n
(n = 1, 2, . . . ) be as in Example 2.2. Then
e
d(λ, µ) = |λ − µ|
is an ultrametric on K, and hence its restriction to R × R is an ultrametric on R (which we shall
also denote by
e
d). Since, for n = 1, 2, . . . ,
λ ∈ R;
e
d(λ, 0) = |λ| ≤
1
2
n
= I
n
=
λ ∈ R; d(λ, 0) ≤
1
2
n
,
d being as in Proposition 2.1, it follows that
e
d and d are equivalent.
Corollary 2.2. Let (K, | · |) and µ be as in Example 2.2, and assume that (K,
e
d) is complete. If
f(X) ∈ R[X] and
¯
f(X) ∈ K[X] admits a simple root θ, then there is a unique root λ of f (X) so
that |λ − ξ| ≤ |µ| (where ξ ∈ R and
¯
ξ = θ).
Proof. Follows immediately from Remark 2.1 and Proposition 2.1.
Corollary 2.3 (cf. [5, p. 16]). Let p be a prime number, Z
p
= {λ ∈ Q
p
; |λ|
p
≤ 1} the ring of
p-adic integers and f(X) ∈ Z
p
[X]. If there is an a
0
∈ Z
p
such that |f(a
0
)|
p
< 1 and |f
0
(a
0
)|
p
= 1,
then there is a unique a ∈ Z
p
such that f(a) = 0 and |a − a
0
|
p
≤
1
p
·
Proof. Since the condition “|f(a
0
)| < 1 ” is equivalent to the condition “
¯
f(¯a
0
) = f(a
0
) = 0 ”, and
the condition “|f
0
(a
0
)|
p
= 1 ” is equivalent to the condition “(
¯
f)
0
(¯a
0
) = f
0
(a
0
) 6= 0”, Theorem
6, p. 391 of [4] guarantees that ¯a
0
is a simple root of
¯
f(X). Therefore the result follows from
Corollary 2.2.
Example 2.3 (cf. [3, p. 52]). Let p be a prime number, p 6= 2, and let b ∈ Z
p
with |b|
p
= 1.
If there is an a
0
∈ Z
p
such that |a
2
0
− b| p < 1, then b = a
2
for a unique a ∈ Z
p
such that
|a − a
0
| p ≤
1
p
·
Indeed, put f(X) = X
2
− b ∈ Z
p
[X]. Then |f (a
0
)| p = |a
2
0
− b| p < 1 and |f
0
(a
0
)| p =
|2a
0
| p = |2|
p
|a
0
| p = |a
0
| p = 1 (the relation |a
2
0
− b| p < 1 = |b|
p
= 1 implies
|a
0
| p
2
=
|(a
2
0
− b) + b|
p
= |b|
p
= 1). Thus the result follows from Corollary 2.9.
In the same vein one shows that if p is a prime number, p 6= 3, c ∈ Z
p
, |c|
p
= 1, and there is
an f
0
∈ Z
p
such that |f
3
0
− c| p < 1, then c = f
3
for a unique f ∈ Z
p
such that |f − f
0
| p ≤
1
p
·
Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 90–95
A proof of a version of Hensel’s lemma 95
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Divulgaciones Matem´aticas Vol. 22, No. 1 (2021), pp. 90–95
