2Field extensions
II Galois Theory
2.2 Ruler and compass constructions
Before we develop our theory further, we first look into a rather unexpected
application of field extensions. We are going to look at some classic problems in
geometry, and solve them using what we’ve learnt so far. In particular, we want
to show that certain things cannot be constructed using a compass and a ruler
(as usual, we assume the ruler does not have markings on it).
It is often easy to prove that certain things are constructible — just exhibit
an explicit construction of it. However, it is much more difficult to show that
things are not constructible. Two classical examples are
(i)
Doubling the cube: Given a cube, can we construct the side of another
cube whose volume is double the volume of the original cube?
(ii)
Trisecting an angle: Given an angle, can we divide the angle into three
equal angles?
The idea here is to associate with each possible construction a field extension,
and then prove certain results about how these field extensions should behave.
We then show that if we could, say, double the cube, then this construction
would inevitable break some of the properties it should have.
Firstly, we want to formulate our problem in a more convenient way. In
particular, we will view the plane as
R
2
, and describe lines and circles by
equations. We also want to describe “compass and ruler” constructions in a
more solid way.
Definition (Constructible points). Let
S ⊆ R
2
be a set of (usually finite) points
in the plane.
A “ruler” allows us to do the following: if
P, Q ∈ S
, then we can draw the
line passing through P and Q.
A “compass” allows us to do the following: if
P, Q, Q
′
∈ S
, then we can draw
the circle with center at P and radius of length QQ
′
.
Any point
R ∈ R
2
is 1-step constructible from
S
if
R
belongs to the in-
tersection of two distinct lines or circles constructed from
S
using rulers and
compasses.
A point
R ∈ R
2
is constructible from
S
if there is some
R
1
, ··· , R
n
=
R ∈ R
2
such that R
i+1
is 1-step constructible from S ∪ {R
1
, ··· , R
i
} for each i.
Example. Let
S
=
{
(0
,
0)
,
(1
,
0)
}
. What can we construct? It should be easy
to see that (
n,
0) for all
n ∈ Z
are all constructible from
S
. In fact, we can show
that all points of the form (m, n) ∈ Z are constructible from S.
(0, 0) (1, 0)
Definition (Field of S). Let S ⊆ R
2
be finite. Define the field of S by
Q(S) = Q({coordinates of points in S}) ⊆ R,
where we put in the
x
coordinate and
y
coordinate separately into the generating
set.
For example, if S = {(
√
2,
√
3)}, then Q(S) = Q(
√
2,
√
3).
The key theorem we will use to prove our results is
Theorem. Let S ⊆ R
2
be finite. Then
(i) If R is 1-step constructible from S, then [Q(S ∪ {R}) : Q(S)] = 1 or 2.
(ii)
If
T ⊆ R
2
is finite,
S ⊆ T
, and the points in
T
are constructible from
S
,
Then [Q(S ∪ T ) : Q(S)] = 2
k
for some k (where k can be 0).
Proof.
By assumption, there are distinct lines or circles
C, C
′
constructed from
S
using ruler and compass, such that
R ∈ C ∩ C
′
. By elementary geometry,
C
and C
′
can be given by the equations
C : a(x
2
+ y
2
) + bx + cy + d = 0,
C
′
: a
′
(x
2
+ y
2
) + b
′
x + c
′
y + d
′
= 0.
where
a, b, c, d, a
′
, b
′
, c
′
, d
′
∈ Q
(
S
). In particular, if we have a line, then we can
take a = 0.
Let
R
= (
r
1
, r
2
). If
a
=
a
′
= 0 (i.e.
C
and
C
′
are lines), then solving the two
linear equations gives r
1
, r
2
∈ Q(S). So [Q(S ∪ {R}) : Q(S)] = 1.
So we can now assume wlog that a = 0. We let
p = a
′
b − ab
′
, q = a
′
c − ac
′
, ℓ = a
′
d − ad
′
,
which are the coefficients when we perform
a
′
×C −a ×C
′
. Then by assumption,
p
= 0 or
q
= 0. Otherwise,
c
and
c
′
would be the same curve. wlog
p
= 0. Then
since (r
1
, r
2
) satisfy both equations of C and C
′
, they satisfy
px + qy + ℓ = 0.
In other words, pr
1
+ qr
2
+ ℓ = 0. This tells us that
r
1
= −
qr
2
+ ℓ
p
. (∗)
If we put
r
1
, r
2
into the equations of
C
and
C
′
and use (
∗
), we get an equation
of the form
αr
2
2
+ βr
2
+ γ = 0,
where
α, β, γ ∈ Q
(
S
). So we can find
r
2
(and hence
r
1
using linear relations)
using only a single radical of degree 2. So
[Q(S ∪ {R}) : Q(S)] = [Q(S)(r
2
) : Q(S)] = 1 or 2,
since the minimal polynomial of r
2
over Q(S) has degree 1 or 2.
Then (ii) follows directly from induction, using the tower law.
Corollary. It is impossible to “double the cube”.
Proof.
Consider the cube with unit side length, i.e. we are given the set
S
=
{
(0
,
0)
,
(1
,
0)
}
. Then doubling the cube would correspond to constructing a side
of length
ℓ
such that
ℓ
3
= 2, i.e.
ℓ
=
3
√
2
. Thus we need to construct a point
R = (
3
√
2, 0) from S.
If we can indeed construct this R, then we need
[Q(S ∪ {R}) : Q(S)] = 2
k
for some k. But we know that Q(S) = Q and Q(S ∪ {R}) = Q(
3
√
2), and that
[Q(
3
√
2) : Q] = 3.
This is a contradiction since 3 is not a power of 2.