Statement Hensel's lemma
1 statement
1.1 general statement
1.2 alternative statement
1.3 derivation
statement
many equivalent statements of hensel s lemma exist. arguably common statement following.
general statement
assume
k
{\displaystyle k}
field complete respect normalised discrete valuation
v
{\displaystyle v}
. suppose, furthermore,
o
k
{\displaystyle {\mathcal {o}}_{k}}
ring of integers of
k
{\displaystyle k}
(i.e. elements of
k
{\displaystyle k}
non-negative valuation), let
π
∈
k
{\displaystyle \pi \in k}
such
v
(
π
)
=
1
{\displaystyle v(\pi )=1}
, let
k
=
o
k
/
π
{\displaystyle k={\mathcal {o}}_{k}/\pi }
denote residue field. let
f
(
x
)
∈
o
k
[
x
]
{\displaystyle f(x)\in {\mathcal {o}}_{k}[x]}
polynomial coefficients in
o
k
{\displaystyle {\mathcal {o}}_{k}}
. if reduction
f
¯
(
x
)
∈
k
[
x
]
{\displaystyle {\overline {f}}(x)\in k[x]}
has simple root (i.e. there exists
k
0
∈
k
{\displaystyle k_{0}\in k}
such
f
¯
(
k
0
)
=
0
{\displaystyle {\overline {f}}(k_{0})=0}
,
f
′
¯
(
k
0
)
≠
0
{\displaystyle {\overline {f }}(k_{0})\neq 0}
), there exists unique
a
∈
o
k
{\displaystyle a\in {\mathcal {o}}_{k}}
such
f
(
a
)
=
0
{\displaystyle f(a)=0}
, reduction
a
¯
=
k
0
{\displaystyle {\overline {a}}=k_{0}}
in
k
{\displaystyle k}
.
alternative statement
another way of stating (in less generality) is: let
f
(
x
)
{\displaystyle f(x)}
polynomial integer (or p-adic integer) coefficients, , let m,k positive integers such m ≤ k. if r integer such that
f
(
r
)
≡
0
(
mod
p
k
)
{\displaystyle f(r)\equiv 0{\pmod {p^{k}}}}
,
f
′
(
r
)
≢
0
(
mod
p
)
{\displaystyle f (r)\not \equiv 0{\pmod {p}}}
then there exists integer s such that
f
(
s
)
≡
0
(
mod
p
k
+
m
)
{\displaystyle f(s)\equiv 0{\pmod {p^{k+m}}}}
,
r
≡
s
(
mod
p
k
)
.
{\displaystyle r\equiv s{\pmod {p^{k}}}.}
furthermore, s unique modulo p, , can computed explicitly integer such that
s
=
r
−
f
(
r
)
⋅
a
{\displaystyle s=r-f(r)\cdot a}
a
{\displaystyle a}
integer satisfying
a
≡
[
f
′
(
r
)
]
−
1
(
mod
p
m
)
.
{\displaystyle a\equiv [f (r)]^{-1}{\pmod {p^{m}}}.}
note
f
(
r
)
≡
0
(
mod
p
k
)
{\displaystyle f(r)\equiv 0{\pmod {p^{k}}}}
condition
s
≡
r
(
mod
p
k
)
{\displaystyle s\equiv r{\pmod {p^{k}}}}
met. aside, if
f
′
(
r
)
≡
0
(
mod
p
)
{\displaystyle f (r)\equiv 0{\pmod {p}}}
, 0, 1, or several s may exist (see hensel lifting below).
derivation
the lemma derives considering taylor expansion of f around r.
r
≡
s
(
mod
p
k
)
{\displaystyle r\equiv s{\pmod {p^{k}}}}
, see s has of form s = r + tp integer t.
let
f
(
r
+
t
p
k
)
=
∑
n
=
0
n
c
n
(
t
p
k
)
n
{\displaystyle f(r+tp^{k})=\sum _{n=0}^{n}c_{n}\ \left(tp^{k}\right)^{n}}
,
c
0
=
f
(
r
)
,
c
1
=
f
′
(
r
)
{\displaystyle c_{0}=f(r),\,c_{1}=f (r)}
, hence
f
(
r
+
t
p
k
)
=
f
(
r
)
+
t
p
k
f
′
(
r
)
+
p
2
k
t
2
g
(
t
)
{\displaystyle f(r+tp^{k})=f(r)+t\,p^{k}\,f (r)+p^{2k}\,t^{2}\,g(t)}
polynomial
g
(
t
)
{\displaystyle g(t)}
integer coefficients.
reducing both sides modulo
p
k
+
m
,
m
≤
k
{\displaystyle p^{k+m},m\leq k}
, see
f
(
s
)
≡
0
(
mod
p
k
+
m
)
{\displaystyle f(s)\equiv 0{\pmod {p^{k+m}}}}
hold, need
0
≡
f
(
r
+
t
p
k
)
≡
f
(
r
)
+
t
p
k
f
′
(
r
)
(
mod
p
k
+
m
)
{\displaystyle 0\equiv f(r+tp^{k})\equiv f(r)+t\,p^{k}\,f (r){\pmod {p^{k+m}}}}
then note
f
(
r
)
=
z
p
k
{\displaystyle f(r)=zp^{k}}
integer
z
{\displaystyle z}
since
r
{\displaystyle r}
root of
f
(
x
)
≡
0
(
mod
p
k
)
{\displaystyle f(x)\equiv 0{\pmod {p^{k}}}}
. thus,
0
≡
(
z
+
t
f
′
(
r
)
)
p
k
(
mod
p
k
+
m
)
{\displaystyle 0\equiv (z+tf (r))p^{k}{\pmod {p^{k+m}}}}
,
which say
0
≡
z
+
t
f
′
(
r
)
(
mod
p
m
)
.
{\displaystyle 0\equiv z+tf (r){\pmod {p^{m}}}.}
solving
t
{\displaystyle t}
in
z
/
p
m
z
{\displaystyle \mathbb {z} /p^{m}\mathbb {z} }
gives explicit formula s mentioned above. assumption
f
′
(
r
)
{\displaystyle f (r)}
not divisible p ensures
f
′
(
r
)
{\displaystyle f (r)}
has inverse mod
p
m
{\displaystyle p^{m}}
unique. hence solution t exists uniquely modulo
p
m
{\displaystyle p^{m}}
, , s exists uniquely modulo
p
k
+
m
{\displaystyle p^{k+m}}
.
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