2Valued fields
III Local Fields
2.1 Hensel’s lemma
We return to the discussion of general valued fields. We are now going to introduce
an alternative to the absolute value that contains the same information, but is
presented differently.
Definition
(Valuation)
.
Let
K
be a field. A valuation on
K
is a function
v : K → R ∪{∞} such that
(i) v(x) = 0 iff x = 0
(ii) v(xy) = v(x) + v(y)
(iii) v(x + y) ≥ min{v(x), v(y)}.
Here we use the conventions that r + ∞ = ∞ and r ≤ ∞ for all r ∈ ∞.
In some sense, this definition is sortof pointless, since if
v
is a valuation,
then the function
x = c
−v(x)
for any
c >
1 is a (nonarchimedean) absolute value. Conversely, if
 · 
is a
valuation, then
v(x) = −log
c
x
is a valuation.
Despite this, sometimes people prefer to talk about the valuation rather than
the absolute value, and often this is more natural. As we will later see, in certain
cases, there is a canonical normalization of
v
, but there is no canonical choice
for the absolute value.
Example. For x ∈ Q
p
, we define
v
p
(x) = −log
p
x
p
.
This is a valuation, and if x ∈ Z
p
, then v
p
(x) = n iff p
n
 x.
Example. Let K be a field, and define
k((T )) =
(
∞
X
i=n
a
i
T
i
: a
i
∈ k, n ∈ Z
)
.
This is the field of formal Laurent series over k. We define
v
X
a
i
T
i
= min{i : a
i
6= 0}.
Then v is a valuation of k((T )).
Recall that for a valued field K, the valuation ring is given by
O
K
= {x ∈ K : x ≤ 1} = {x ∈ K : v(x) ≥ 0}.
Since this is a subring of a field, and the absolute value is multiplicative, we
notice that the units in
O
are exactly the elements of absolute value 1. The
remaining elements form an ideal (since the field is nonarchimedean), and thus
we have a maximal ideal
m = m
K
= {x ∈ K : x < 1}
The quotient
k = k
K
= O
K
/m
K
is known as the residue field.
Example. Let K = Q
p
. Then O = Z
p
, and m = pZ
p
. So
k = O/m = Z
p
/pZ
p
∼
=
Z/pZ.
Definition
(Primitive polynomial)
.
If
K
is a valued field and
f
(
x
) =
a
0
+
a
1
x
+
··· + a
n
x
n
∈ K[x] is a polynomial, we say that f is primitive if
max
i
a
i
 = 1.
In particular, we have f ∈ O[x].
The point of a primitive polynomial is that such a polynomial is naturally,
and nontrivially, an element of
k
[
x
]. Moreover, focusing on such polynomials is
not that much of a restriction, since any polynomial is a constant multiple of a
primitive polynomial.
Theorem
(Hensel’s lemma)
.
Let
K
be a complete valued field, and let
f ∈ K
[
x
]
be primitive. Put
¯
f = f mod m ∈ k[x]. If there is a factorization
¯
f(x) = ¯g(x)
¯
h(x)
with (¯g,
¯
h) = 1, then there is a factorization
f(x) = g(x)h(x)
in O[x] with
¯g = g,
¯
h = h mod m,
with deg g = deg ¯g.
Note that requiring
deg g
=
deg ¯g
is the best we can hope for — we cannot
guarantee deg h = deg
¯
h, since we need not have deg f = deg
¯
f.
This is one of the most important results in the course.
Proof.
Let
g
0
, h
0
be arbitrary lifts of
¯g
and
¯
h
to
O
[
x
] with
deg ¯g
=
g
0
and
deg
¯
h = h
0
. Then we have
f = g
0
h
0
mod m.
The idea is to construct a “Taylor expansion” of the desired
g
and
h
term by
term, starting from
g
0
and
h
0
, and using completeness to guarantee convergence.
To proceed, we use our assumption that
¯g,
¯
h
are coprime to find some
a, b ∈ O
[
x
]
such that
ag
0
+ bh
0
≡ 1 mod m. (†)
It is easier to work modulo some element
π
instead of modulo the ideal
m
, since
we are used to doing Taylor expansion that way. Fortunately, since the equations
above involve only finitely many coefficients, we can pick an
π ∈ m
with absolute
value large enough (i.e. close enough to 1) such that the above equations hold
with m replaced with π. Thus, we can write
f = g
0
h
0
+ πr
0
, r
0
∈ O[x].
Plugging in (†), we get
f = g
0
h
0
+ πr
0
(ag
0
+ bh
0
) + π
2
(something).
If we are lucky enough that
deg r
0
b < deg g
0
, then we group as we learnt in
secondary school to get
f = (g
0
+ πr
0
b)(h
0
+ πr
0
a) + π
2
(something).
We can then set
g
1
= g
0
+ πr
0
b
h
1
= h
0
+ πr
0
a,
and then we can write
f = g
1
h
1
+ π
2
r
1
, r
1
∈ O[x], deg g
1
= deg ¯g. (∗)
If it is not true that deg r
0
b ≤ deg g
0
, we use the division algorithm to write
r
0
b = qg
0
+ p.
Then we have
f = g
0
h
0
+ π((r
0
a + q)g
0
+ ph
0
),
and then proceed as above.
Given the factorization (
∗
), we replace
r
1
by
r
1
(
ag
0
+
bh
0
), and then repeat
the procedure to get a factorization
f ≡ g
2
h
2
mod π
3
, deg g
2
= deg ¯g.
Inductively, we constrict g
k
, h
k
such that
f ≡ g
k
h
k
mod π
k+1
g
k
≡ g
k−1
mod π
k
h
k
≡ h
k−1
mod π
k
deg g
k
= deg ¯g
Note that we may drop the terms of
h
k
whose coefficient are in
π
k+1
O
, and the
above equations still hold. Moreover, we can then bound
deg h
k
≤ deg f −deg g
k
.
It now remains to set
g = lim
k→∞
g
k
, h = lim
k→∞
h
k
.
Corollary.
Let
f
(
x
) =
a
0
+
a
1
x
+
···
+
a
n
x
n
∈ K
[
x
] where
K
is complete and
a
0
, a
n
6= 0. If f is irreducible, then
a
`
 ≤ max(a
0
, a
n
)
for all `.
Proof.
By scaling, we can wlog
f
is primitive. We then have to prove that
max
(
a
0
, a
n

) = 1. If not, let
r
be minimal such that
a
r

= 1. Then 0
< r < n
.
Moreover, we can write
f(x) ≡ x
r
(a
r
+ a
r+1
x + ···+ a
n
x
n−r
) mod m.
But then Hensel’s lemma says this lifts to a factorization of
f
, a contradiction.
Corollary (of Hensel’s lemma). Let f ∈ O[x] be monic, and K complete. If f
mod m
has a simple root
¯α ∈ k
, then
f
has a (unique) simple root
α ∈ O
lifting
¯α.
Example.
Consider
x
p−1
−
1
∈ Z
p
[
x
]. We know
x
p−1
splits into distinct linear
factors over
F
p
[
x
]. So all roots lift to
Z
p
. So
x
p−1
−
1 splits completely in
Z
p
.
So Z
p
contains all p roots of unity.
Example. Since 2 is a quadratic residue mod 7, we know
√
2 ∈ Q
7
.