2Valued fields
III Local Fields
2.3 Newton polygons
We are going to have a small digression to Newton polygons. We will not make
use of them in this course, but it is a cute visual devices that tell us about roots
of polynomials. It is very annoying to write down a formal definition, so we first
look at some examples. We will work with valuations rather than the absolute
value.
Example. Consider the valued field (Q
p
, v
p
), and the polynomial
t
4
+ p
2
t
4
− p
3
t
2
+ pt + p
3
.
We then plot the coefficients for each power of
t
, and then draw a “convex
polygon” so that all points lie on or above it:
power of t
valuation of coefficient
1 2 3 40
1
2
3
Example. Consider (Q
2
, v
2
) with the polynomial
4t
4
+ 5t
3
+
7
2
t +
9
2
.
Here there is no t
2
term, so we simply don’t draw anything.
power of t
valuation of coefficient
1 2 3 40
−1
1
2
We now go to come up with a formal definition.
Definition (Lower convex set). We say a set S ⊆ R
2
is lower convex if
(i) Whenever (x, y) ∈ S, then (x, z) ∈ S for all z ≥ y.
(ii) S is convex.
Definition
(Lower convex hull)
.
Given any set of points
T ⊆ R
2
, there is a
minimal lower convex set
S ⊇ T
(by the intersection of all lower convex sets
containing
T
– this is a non-empty definition because
R
2
satisfies the property).
This is known as the lower convex hull of the points.
Example.
The lower convex hull of the points (0
,
3)
,
(1
,
1)
,
(2
,
3)
,
(3
,
2)
,
(4
,
0) is
given by the region denoted below:
Definition
(Newton polygon)
.
Let
f
(
x
) =
a
0
+
a
1
x
+
···
+
a
n
x
n
∈ K
[
x
], where
(
K, v
) is a valued field. Then the Newton polygon of
f
is the lower convex hull
of {(i, v(a
i
)) : i = 0, ··· , n, a
i
6= 0}.
This is the formal definition, so in our first example, the Newton polygon
really should be the shaded area shown above, but most of the time, we only
care about the lower line.
Definition
(Break points)
.
Given a polynomial, the points (
i, v
(
a
i
)) lying on
the boundary of the Newton polygon are known as the break points.
Definition
(Line segment)
.
Given a polynomial, the line segment between two
adjacent break points is a line segment.
Definition
(Multiplicity/length)
.
The length or multiplicity of a line segment
is the horizontal length.
Definition (Slope). The slope of a line segment is its slope.
Example. Consider again (Q
2
, v
2
) with the polynomial
4t
4
+ 5t
3
+
7
2
t +
9
2
.
power of t
valuation of coefficient
The middle segment has length 2 and slope 1/2.
Example. In the following Newton polygon:
The second line segment has length 3 and slope −
1
3
.
It turns out the Newton polygon tells us something about the roots of the
polynomial.
Theorem. Let K be complete valued field, and v the valuation on K. We let
f(x) = a
0
+ a
1
x + ··· + a
n
x
n
∈ K[x].
Let
L
be the splitting field of
f
over
K
, equipped with the unique extension
w
of v.
If (
r, v
(
a
r
))
→
(
s, v
(
a
s
)) is a line segment of the Newton polygon of
f
with
slope −m ∈ R, then f has precisely s − r roots of valuation m.
Note that by lower convexity, there can be at most one line segment for each
slope. So this theorem makes sense.
Proof.
Dividing by
a
n
only shifts the polygon vertically, so we may wlog
a
n
= 1.
We number the roots of f such that
w(α
1
) = ··· = w(α
s
1
) = m
1
w(α
s
1
+1
) = ··· = w(α
s
2
) = m
2
.
.
.
w(α
s
t
) = ··· = w(α
n
) = m
t+1
,
where we have
m
1
< m
2
< ··· < m
t+1
.
Then we know
v(a
n
) = v(1) = 0
v(a
n−1
) = w
X
α
i
≥ min
i
w(α
i
) = m
1
v(a
n−2
) = w
X
α
i
α
j
≥ min
i6=j
w(α
i
α
j
) = 2m
1
.
.
.
v(a
n−s
1
) = w
X
i
1
6=...6=i
s
1
α
i
1
...α
i
s
1
= min w(α
i
1
···α
i
s
1
) = s
1
m
1
.
It is important that in the last one, we have equality, not an inequality, because
there is one term in the sum whose valuation is less than all the others.
We can then continue to get
v(α
n−s
1
−1
) ≥ min w(α
i
1
···α
i
s
1
+1
) = s
1
m
1
+ m
2
,
until we reach
v(α
n−s
1
−s
2
) = s
1
m
1
+ (s
2
− s
1
)m
2
.
We keep going on.
We draw the Newton polygon.
(n, 0)
(n − s
1
, s
1
m
1
)
(n − s
1
− s
2
, s
1
m
1
+ (s
2
− s
1
)m
1
)
···
We don’t know where exactly the other points are, but the inequalities imply
that the (i, v(a
i
)) are above the lines drawn. So this is the Newton polygon.
Counting from the right, the first line segment has length
n −
(
n − s
1
) =
s
1
and slope
0 − s
1
m
1
n − (n − s
1
)
= −m
1
.
In general, the
k
th segment has length (
n − s
k−1
)
−
(
n − s
k
) =
s
k
− s
k−1
, and
slope
s
1
m
1
+
P
k−2
i=1
(s
i+1
− s
i
)m
i+1
−
s
1
m
1
+
P
k−1
i=1
(s
i+1
− s
i
)m
i+1
s
k
− s
k−1
=
−(s
k
− s
k−1
)m
k
s
k
− s
k−1
= −m
k
.
and the others follow similarly.
Corollary.
If
f
is irreducible, then the Newton polygon has a single line segment.
Proof.
We need to show that all roots have the same valuation. Let
α, β
be in
the splitting field
L
. Then there is some
σ ∈ Aut
(
L/K
) such that
σ
(
α
) =
β
.
Then w(α) = w(σ(α)) = β. So done.
Note that Eisenstein’s criterion is a (partial) converse to this!