5Ramification theory for local fields
III Local Fields
5.3 Totally ramified extensions
We now quickly look at totally ramified extensions. Recall the following irre-
ducibility criterion:
Theorem
(Eisenstein criterion)
.
Let
K
be a local field, and
f
(
x
) =
x
n
+
a
n−1
x
n−1
+
···
+
a
0
∈ O
K
[
x
]. Let
π
K
be the uniformizer of
K
. If
π
K
|
a
n−1
, ··· , a
0
and π
2
K
- a
0
, then f is irreducible.
Proof.
Left as an exercise. You’ve probably seen this already in a much more
general context, but in this case there is a neat proof using Newton polygons.
We will need to use the following characterization of the ramification index:
Proposition.
Let
L/K
be an extension of local fields, and
v
K
be the normalized
valuation. Let
w
be the unique extension of
v
K
to
L
. Then the ramification
index e
L/K
is given by
e
−1
L/K
= w(π
L
) = min{w(x) : x ∈ m
L
},
Proof.
We know
w
and
v
L
differ by a constant. To figure out what this is, we
have
1 = w(π
K
) = e
−1
L/K
v
L
(π
K
).
So for any x ∈ L, we have
w(x) = e
−1
L/K
v
L
(x).
In particular, putting x = π
L
, we have
w(π
L
) = e
−1
L/K
v
L
(π
L
) = e
−1
L/K
.
The equality
w(π
L
) = min{w(x) : x ∈ m
L
},
is trivially true because the minimum is attained by π
L
.
Definition
(Eisenstein polynomial)
.
A polynomial
f
(
x
)
∈ O
K
[
x
] satisfying the
assumptions of Eisenstein’s criterion is called an Eisenstein polynomial.
We can now state the proposition:
Proposition.
Let
L/K
be a totally ramified extension of local fields. Then
L = K(π
L
) and the minimal polynomial of π
L
over K is Eisenstein.
Conversely, if
L
=
K
(
α
) and the minimal polynomial of
α
over
K
is Eisenstein,
then L/K is totally ramified and α is a uniformizer of L.
Proof.
Let
n
= [
L
:
K
],
v
K
be the valuation of
K
, and
w
the unique extension
to L. Then
[K(π
L
) : K]
−1
≤ e
−1
K(π
L
)/K
= min
x∈m
K(π
L
)
w(c) ≤
1
n
,
where the last inequality follows from the fact that π
L
∈ m
L(π
L
)
.
But we also know that
[K(π
L
) : K] ≤ [L : K].
So we know that L = K(π
L
).
Now let
f
(
x
) =
x
n
+
a
n−1
x
n−1
+
···
+
a
0
∈ O
K
[
x
] be the minimal polynomial
of π
L
/K. Then we have
π
n
L
= −(a
0
+ a
1
π
L
+ ···+ a
n−1
π
n−1
L
).
So we have
1 = w(π
n
L
) = w(a
0
+ a
1
π
L
+ ··· + a
n−1
π
n−1
L
) = min
i=0,...,n−1
v
k
(a
i
) +
i
n
.
This implies that v
K
(a
i
) ≥ 1 for all i, and v
K
(x
0
) = 1. So it is Eisenstein.
For the converse, if K = K(α) and n = [L : K], take
g(x) = x
n
+ b
n−1
x
n−1
+ .. + b
0
∈ O
K
[x]
be the minimal polynomial of
α
. So all roots have the same valuation. So we
have
1 = w(b
0
) = n · w(α).
So we have w(α) =
1
n
. So we have
e
−1
L/K
= min
x∈m
L
w(x) ≤
1
n
= [L : K]
−1
.
So [L : K] = e
L/K
= n. So L/K is totally ramified and α is a uniformizer.
In fact, more is true. We have
O
L
=
O
K
[
π
L
], since every element in
O
L
can
be written as
X
i≥0
a
i
π
i
L
,
where a
i
is a lift of an element in k
L
= k
K
, which can be chosen to be in O
K
.