1Solving equations
II Galois Theory
1 Solving equations
Galois theory grew of the desire to solve equations. In particular, to solve
polynomial equations. To begin with, we will come up with general solutions to
polynomial equations of up to degree 4. However, this is the best we can do, as
we will later show in the course — there is no general solution to polynomial
equations of degree 5 or above.
Before we start, we will define some notations that we will frequently use.
If
R
is a ring, then
R
[
t
] is the polynomial ring over
R
in the variable
t
.
Usually, we take
R
=
Q
and consider polynomials
f
(
t
)
∈ Q
[
t
]. The objective
is then to find roots to the equation
f
(
t
) = 0. Often, we want to restrict our
search domain. For example, we might ask if there is a root in
Q
. We will thus
use Root
f
(X) to denote the set of all roots of f in X.
Linear equations
Suppose that
f
=
t
+
a ∈ Q
[
t
] (with
a ∈ Q
). This is easy to solve — we have
Root
f
(Q) = {−a}.
Quadratic equations
Consider a simple quadratic
f
=
t
2
+ 1
∈ Q
[
t
]. Then
Root
f
(
Q
) =
∅
since the
square of all rationals are positive. However, in the complex plane, we have
Root
f
(C) = {
√
−1, −
√
−1}.
In general, let
f
=
t
2
+
at
+
b ∈ Q
[
t
]. Then as we all know, the roots are
given by
Root
f
(C) =
(
−a ±
√
a
2
− 4b
2
)
Cubic equations
Let f = t
3
+ c ∈ Q[t]. The roots are then
Root
f
(C) = {
3
√
−c, µ
3
√
−c, µ
2
3
√
−c},
where
µ
=
−1+
√
−3
2
is the 3rd root of unity. Note that
µ
is defined by the
equation µ
3
− 1 = 0, and satisfies µ
2
+ µ + 1 = 0.
In general, let
f
=
t
3
+
at
2
+
bt
+
c ∈ Q
[
t
], and let
Root
f
(
C
) =
{α
1
, α
2
, α
3
}
,
not necessarily distinct.
Our objective is to solve
f
= 0. Before doing so, we have to make it explicit
what we mean by “solving” the equation. As in solving the quadratic, we want
to express the roots α
1
, α
2
and α
3
in terms of “radicals” involving a, b and c.
Unlike the quadratic case, there is no straightforward means of coming up
with a general formula. The result we currently have is the result of many
many years of hard work, and the substitutions we make seemingly come out of
nowhere. However, after a lot of magic, we will indeed come up with a general
formula for it.
We first simplify our polynomial by assuming
a
= 0. Given any polynomial
f
=
t
3
+
at
2
+
bt
+
c
, we know
a
is the negative of the sum of the roots. So we
can increase each root by
a
3
so that the coefficient of
t
2
vanishes. So we perform
the change of variables
t 7→ t −
a
3
, and get rid of the coefficient of
t
2
. So we can
assume a = 0.
Let µ be as above. Define
β = α
1
+ µα
2
+ µ
2
α
3
γ = α
1
+ µ
2
α
2
+ µα
3
These are the Lagrange resolvers. We obtain
βγ = α
2
1
+ α
2
2
+ α
2
3
+ (µ + µ
2
)(α
1
α
2
+ α
2
α
3
+ α
1
α
3
)
Since µ
2
+ µ + 1 = 0, we have µ
2
+ µ = −1. So we can simplify to obtain
= (α
1
+ α
2
+ α
3
)
2
− 3(α
1
α
2
+ α
2
α
3
+ α
1
α
3
)
We have α
1
+ α
2
+ α
3
= −a = 0, while b = α
1
α
2
+ α
2
α
3
+ α
1
α
3
. So
= −3b
Cubing, we obtain
β
3
γ
3
= −27b
3
.
On the other hand, recalling again that α
1
+ α
2
+ α
3
= 0, we have
β
3
+ γ
3
= (α
1
+ µα
2
+ µ
2
α
3
)
3
+ (α
1
+ µ
2
α
2
+ µα
3
)
3
+ (α
1
+ α
2
+ α
3
)
3
= 3(α
3
1
+ α
3
2
+ α
3
3
) + 18α
1
α
2
α
3
We have
α
1
α
2
α
3
=
−c
, and since
α
3
i
+
bα
i
+
c
= 0 for all
i
, summing gives
α
3
1
+ α
3
2
+ α
3
3
+ 3c = 0. So
= −27c
Hence, we obtain
(t − β
3
)(t − γ
3
) = t
2
+ 27ct − 27b
3
.
We already know how to solve this equation using the quadratic formula. We
obtain
{β
3
, γ
3
} =
(
−27c ±
p
(27c)
2
+ 4 × 27b
3
2
)
We now have
β
3
and
γ
3
in terms of radicals. So we can find
β
and
γ
in terms of
radicals. Finally, we can solve for α
i
using
0 = α
1
+ α
2
+ α
3
β = α
1
+ µα
2
+ µ
2
α
3
γ = α
1
+ µ
2
α
2
+ µα
3
In particular, we obtain
α
1
=
1
3
(β + γ)
α
2
=
1
3
(µ
2
β + µγ)
α
3
=
1
3
(µβ + µ
2
γ)
So we can solve a cubic in terms of radicals.
This was a lot of magic involved, and indeed this was discovered through a
lot of hard work throughout many many years. This is also not a very helpful
result since we have no idea where these substitutions came from and why they
intuitively work.
Quartic equations
Assume
f
=
t
4
+
at
3
+
bt
2
+
ct
+
d ∈ Q
[
t
]. Let
Root
f
(
C
) =
{α
1
, α
2
, α
3
, α
4
}
.
Can we express all these in terms of radicals? Again the answer is yes, but the
procedure is much more complicated.
We can perform a similar change of variable to assume a = 0. So α
1
+ α
2
+
α
3
+ α
4
= 0.
This time, define
β = α
1
+ α
2
γ = α
1
+ α
3
λ = α
1
+ α
4
Doing some calculations, we see that
β
2
= −(α
1
+ α
2
)(α
3
+ α
4
)
γ
2
= −(α
1
+ α
3
)(α
2
+ α
4
)
λ
2
= −(α
1
+ α
4
)(α
2
+ α
3
)
Now consider
g = (t −β
2
)(t − γ
2
)(t − λ
2
)
= t
3
+ 2bt
2
+ (b
2
− 4d)t − c
2
This we know how to solve, and so we are done.
Quintics and above
So far so good. But how about polynomials of higher degrees? In general, let
f ∈ Q
[
t
]. Can we write down all the roots of
f
in terms of radicals? We know
that the answer is yes if deg f ≤ 4.
Unfortunately, for
deg f ≥
5, the answer is no. Of course, this “no” means
no in general. For example,
f
= (
t −
1)(
t −
2)
···
(
t −
5)
∈ Q
[
t
] has the obvious
roots in terms of radicals.
There isn’t an easy proof of this result. The general idea is to first associate
a field extension
F ⊇ Q
for our polynomial
f
. This field
F
will be obtained
by adding all roots of
f
. Then we associate a Galois group
G
to this field
extension. We will then prove a theorem that says
f
has a solution in terms
of radicals if and only if the Galois group is “soluble”, where “soluble” has a
specific algebraic definition in group theory we will explore later. Finally, we
find specific polynomials whose Galois group is not soluble.