3Solutions to polynomial equations
II Galois Theory
3.5 Insolubility of general equations of degree 5 or more
We now want to do something more interesting. Instead of looking at a particular
example, we want to say there is no general formula for solving polynomial
equations of degree 5 or above. First we want to define certain helpful notions.
Definition (Field of symmetric rational functions). Let
K
be a field,
L
=
K
(
x
1
, ··· , x
n
), the field of rational functions over
K
. Then there is an injective
homomorphism S
n
→ Aut
K
(L) given by permutations of x
i
.
We define the field of symmetric rational functions
F
=
L
S
n
to be the fixed
field of S
n
.
There are a few important symmetric rational functions that we care about
more.
Definition (Elementary symmetric polynomials). The elementary symmetric
polynomials are e
1
, e
2
, ··· , e
n
defined by
e
i
=
X
1≤l
1
<l
2
<···<l
i
≤n
x
ℓ
1
···x
ℓ
i
.
It is easy to see that
e
1
= x
1
+ x
2
+ ··· + x
n
e
2
= x
1
x
2
+ x
1
x
3
+ ··· + x
n−1
x
n
e
n
= x
1
···x
n
.
Obviously, e
1
, ··· , e
n
∈ F .
Theorem (Symmetric rational function theorem). Let
K
be a field,
L
=
K
(
x
1
, ··· , x
n
). Let
F
the field fixed by the automorphisms that permute the
x
i
.
Then
(i) L is the splitting field of
f = t
n
− e
1
t
n−1
+ ··· + (−1)
n
e
n
over F .
(ii) F = L
S
n
⊆ L is a Galois group with Gal(L/F ) isomorphic to S
n
.
(iii) F = K(e
1
, ··· , e
n
).
Proof.
(i) In L[t], we have
f = (t − x
1
) ···(t − x
n
).
So L is the splitting field of f over F .
(ii) By Artin’s lemma, L/K is Galois and Gal(L/F )
∼
=
S
n
.
(iii)
Let
E
=
K
(
e
1
, ··· , e
n
). Clearly,
E ⊆ F
. Now
E ⊆ L
is a Galois extension,
since L is the splitting field of f over E and f has no repeated roots.
By the fundamental theorem of Galois theory, since we have the Galois ex-
tensions
E ⊆ F ⊆ L
, we have
Gal
(
L/F
)
≤ Gal
(
L/E
). So
S
n
≤ Gal
(
L/E
).
However, we also know that
Gal
(
L/E
) is a subgroup of
S
n
, we must have
Gal(L/E) = Gal(L/F) = S
n
. So we must have E = F .
Definition (General polynomial). Let
K
be a field,
u
1
, ··· , u
n
variables. The
general polynomial over K of degree n is
f = t
n
+ u
1
t
n−1
+ ··· + u
n
.
Technically, this is a polynomial in the polynomial ring
K
(
u
1
, ··· , u
n
)[
t
]. How-
ever, we say this is the general polynomial over
K
be cause we tend to think of
these u
i
as representing actual elements of K.
We say the general polynomial over
K
of degree
n
can be solved by radicals
if f can be solved by radicals over K(u
1
, ··· , u
n
).
Example. The general polynomial of degree 2 over Q is
t
2
+ u
1
t + u
2
.
This can be solved by radicals because its roots are
−u
1
±
p
u
2
1
− 4u
2
2
.
Theorem. Let
K
be a field with
char K
= 0. Then the general polynomial
polynomial over K of degree n cannot be solved by radicals if n ≥ 5.
Proof. Let
f = t
n
+ u
1
t
n−1
+ ··· + u
n
.
be our general polynomial of degree
n ≥
5. Let
N
be a splitting field of
f
over
K(u
1
, ··· , u
n
). Let
Root
f
(N) = {α
1
, ··· , α
n
}.
We know the roots are distinct because
f
is irreducible and the field has charac-
teristic 0. So we can write
f = (t − α
1
) ···(t − α
n
) ∈ N[t].
We can expand this to get
u
1
= −(α
1
+ ··· + α
n
)
u
2
= α
1
α
2
+ α
1
α
3
+ ··· + α
n−1
α
n
.
.
.
u
i
= (−1)
i
(ith elementary symmetric polynomial in α
1
, ··· , α
n
).
Now let
x
1
, ··· , x
n
be new variables, and
e
i
the
i
th elementary symmetric
polynomial in
x
1
, ··· , x
n
. Let
L
=
K
(
x
1
, ··· , x
n
), and
F
=
K
(
e
1
, ··· , e
n
). We
know that F ⊆ L is a Galois extension with Galois group isomorphic to S
n
.
We define a ring homomorphism
θ : K[u
1
, ··· , u
n
] → K[e
1
, ··· , e
n
] ⊆ K[x
1
, ··· , x
n
]
u
i
7→ (−1)
i
e
i
.
This is our equations of u
i
in terms α
i
, but with x
i
instead of α
i
.
We want to show that
θ
is an isomorphism. Note that since the homomorphism
just renames
u
i
into
e
i
, the fact that
θ
is an isomorphism means there are no
“hidden relations” between the
e
i
. It is clear that
θ
is a surjection. So it suffices
to show θ is injective. Suppose θ(h) = 0. Then
h(−e
1
, ··· , (−1)
n
e
n
) = 0.
Since the x
i
are just arbitrary variables, we now replace x
i
with α
i
. So we get
h(−e
1
(α
1
, ··· , α
n
), ··· , (−1)
n
(e
n
(α
1
, ··· , α
n
))) = 0.
Using our expressions for u
i
in terms of e
i
, we have
h(u
1
, ··· , u
n
) = 0,
But
h
(
u
1
, ··· , u
n
) is just
h
itself. So
h
= 0. Hence
θ
is injective. So it is an
isomorphism. This in turns gives an isomorphism between
K(u
1
, ··· , u
n
) → K(e
1
, ··· , e
n
) = F.
We can extend this to their polynomial rings to get isomorphisms between
K(u
1
, ··· , u
n
)[t] → F [t].
In particular, this map sends our original f to
f 7→ t
n
− e
1
t
n−1
+ ··· + (−1)
n
e
n
= g.
Thus, we get an isomorphism between the splitting field of
f
over
K
(
u
1
, ··· , u
n
)
and the splitting field g over F .
The splitting field of
f
over
K
(
u
1
, ··· , u
n
) is just
N
by definition. From the
symmetric rational function theorem, we know that the splitting field of
g
over
F is just L, and So N
∼
=
L. So we have an isomorphism
Gal(N/K(u
1
, ··· , u
n
)) → Gal(L/F )
∼
=
S
n
.
Since S
n
is not soluble, f is not soluble.
This is our second main goal of the course, to prove that general polynomials
of degree 5 or more are not soluble by radicals.
Recall that we proved that all radical extensions are soluble. We now prove
the converse.
Theorem. Let
K
be a field with
char K
= 0. If
L/K
is a soluble extension,
then it is a radical extension.
Proof.
Let
L ⊆ E
be such that
K ⊆ E
is Galois and
Gal
(
E/K
) is soluble. We
can replace
L
with
E
, and assume that in fact
L/K
is a soluble Galois extension.
So there is a sequence of groups
{0} = G
r
◁ ··· ◁ G
1
◁ G
0
= Gal(L/K)
such that G
i
/G
i+1
is cyclic.
By the fundamental theorem of Galois theory, we get a sequence of field
extension given by L
i
= L
G
i
:
K = L
0
⊆ ··· ⊆ L
r
= L.
Moreover, we know that
L
i
⊆ L
i+1
is a Galois extension with Galois group
Gal(L
i+1
/L
i
)
∼
=
G
i
/G
i+1
. So Gal(L
i+1
/L
i
) is cyclic.
Let
n
= [
L
:
K
]. Recall that we proved a previous theorem that if
Gal
(
L
i+1
/L
i
) is cyclic, and
L
i
contains a primitive
k
th root of unity (with
k
= [
L
i+1
:
L
i
]), then
L
i
⊆ L
i+1
is a Kummer extension. However, we do not
know of
L
i
contains the right root of unity. Hence, the trick here is to add an
nth primitive root of unity to each field in the sequence.
Let
µ
an
n
th primitive root of unity. Then if we add the
n
th primitive root
to each item of the sequence, we have
L
0
(µ) ··· L
i
(µ) L
i+1
(µ) ··· L
r
(µ)
K = L
0
··· L
i
L
i+1
··· L
r
= L
⊆
⊆
⊆ ⊆
⊆
⊆
⊆
⊆
⊆
⊆ ⊆ ⊆ ⊆ ⊆
We know that
L
0
⊆ L
0
(
µ
) is a cyclotomic extension by definition. We will now
show that
L
i
(
µ
)
⊆ L
i+1
(
µ
) is a Kummer extension for all
i
. Then
L/K
is radical
since L ⊆ L
r
(µ).
Before we do anything, we have to show
L
i
(
µ
)
⊆ L
i+1
(
µ
) is a Galois extension.
To show this, it suffices to show L
i
⊆ L
i+1
(µ) is a Galois extension.
Since
L
i
⊆ L
i+1
is Galois,
L
i
⊆ L
i+1
is normal. So
L
i+1
is the splitting of
some
h
over
L
i
. Then
L
i+1
(
µ
) is just the splitting field of (
t
n
−
1)
h
. So
L
i
⊆
L
i+1
(
µ
) is normal. Also,
L
i
⊆ L
i+1
(
µ
) is separable since
char K
=
char L
i
= 0.
Hence L
i
⊆ L
i+1
(µ) is Galois, which implies that L
i
(µ) ⊆ L
i+1
(µ) is Galois.
We define a homomorphism of groups
Gal(L
i+1
(µ)/L
i
(µ)) → Gal(L
i+1
/L
i
)
by restriction. This is well-defined because L
i+1
is the splitting field of some h
over
L
i
, and hence any automorphism of
L
i+1
(
µ
) must send roots of
h
to roots
of h, i.e. L
i+1
to L
i+1
.
Moreover, we can see that this homomorphism is injective. If
ϕ 7→ ϕ|
L
i+1
=
id
,
then it fixes everything in
L
i+1
. Also, since it is in
Gal
(
L
i+1
(
µ
)
/L
i
(
µ
)), it fixes
L
i
(µ). In particular, it fixes µ. So ϕ must fix the whole of L
i+1
(µ). So ϕ = id.
By injectivity, we know that
Gal
(
L
i+1
(
µ
)
/L
i
(
µ
)) is isomorphic to a subgroup
of
Gal
(
L
i+1
/L
i
). Hence it is cyclic. By our previous theorem, it follows that
L
i
(µ) ⊆ L
i+1
(µ) is a Kummer extension. So L/K is radical.
Corollary. Let
K
be a field with
char K
= 0 and
h ∈ K
[
t
]. Let
L
be the
splitting of
h
over
K
. Then
h
can be solved by radicals if and only if
Gal
(
L/K
)
is soluble.
Proof. (⇒) Proved before.
(
⇐
) Since
L/K
is a Galois extension,
L/K
is a soluble extension. So it is a
radial extension. So h can be solved by radicals.
Corollary. Let
K
be a field with
char K
= 0. Let
f ∈ K
[
t
] have
deg f ≤
4.
Then f can be solved by radicals.
Proof. Exercise.
Note that in the case where
K
=
Q
, we have proven this already by given
explicit solutions in terms of radicals in the first lecture.