 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
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 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 , , ,  and
, for instance. A quite general form of Hensel’s lemma may be found in Chapter III of ,
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 . 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. ). 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
n1
I
n
= {0}.
Example 2.2 (cf. ). 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
n1
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
n1
, 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
n1
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
n1
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 . 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() 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
)
n1
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
[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
¯
fa
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  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 