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!