3Solutions to polynomial equations

II Galois Theory



3.1 Cyclotomic extensions
Definition (Cyclotomic extension). For a field
K
, we define the
n
th cyclotomic
extension to be the splitting field of t
n
1.
Note that if
K
is a field and
L
is the
n
th cyclotomic extension, then
Root
t
n
1
(
L
) is a subgroup of multiplicative group
L
=
L \ {
0
}
. Since this is a
finite subgroup of L
, it is a cyclic group.
Moreover, if
char K
= 0 or 0
< char K n
, then (
t
n
1)
=
nt
n1
and this
has no common roots with
t
n
1. So
t
n
1 has no repeated roots. In other
words, t
n
1 has n distinct roots. So as a group,
Root
t
n
1
(L)
=
Z/nZ.
In particular, this group has at least one element µ of order n.
Definition (Primitive root of unity). The
n
th primitive root of unity is an
element of order n in Root
t
n
1
(L).
These elements correspond to the elements of the multiplicative group of
units in Z/nZ, written (Z/nZ)
×
.
The next theorem tells us some interesting information about these roots
and some related polynomials.
Theorem. For each
d N
, there exists a
d
th cyclotomic monic polynomial
ϕ
d
Z[t] satisfying:
(i) For each n N, we have
t
n
1 =
Y
d|n
ϕ
d
.
(ii) Assume char K = 0 or 0 < char K n. Then
Root
ϕ
n
(L) = {nth primitive roots of unity}.
Note that here we have an abuse of notation, since
ϕ
n
is a polynomial in
Z
[
t
], not
K
[
t
], but we can just use the canonical map
Z
[
t
]
K
[
t
] mapping
1 to 1 and t to t.
Proof.
We do induction on
n
to construct
ϕ
n
. When
n
= 1, let
ϕ
1
=
t
1. Then
(i) and (ii) hold in this case, trivially.
Assume now that (i) and (ii) hold for smaller values of n. Let
f =
Y
d|n,d<n
ϕ
d
.
By induction,
f Z
[
t
]. Moreover, if
d | n
and
d < n
, then
ϕ
d
|
(
t
n
1) because
(
t
d
1)
|
(
t
n
1). We would like to say that
f
also divides
t
n
1. However, we
have to be careful, since to make this conclusion, we need to show that
ϕ
d
and
ϕ
d
have no common roots for distinct d, d
| n (and d, , d
< n).
Indeed, by induction, ϕ
d
and ϕ
d
have no common roots because
Root
ϕ
d
(L) = {dth primitive roots of unity},
Root
ϕ
d
(L) = {d
th primitive roots of unity},
and these two sets are disjoint (or else the roots would not be primitive).
Therefore
ϕ
d
and
ϕ
d
have no common irreducible factors. Hence
f | t
n
1. So
we can write
t
n
1 = fϕ
n
,
where
ϕ
n
Q
[
t
]. Since
f
is monic,
ϕ
n
has integer coefficients. So indeed
ϕ
n
Z[t]. So the first part is proven.
To prove the second part, note that by induction,
Root
f
(L) = {non-primitive nth roots of unit},
since all nth roots of unity are dth primitive roots of unity for some smaller d.
Since
fϕ
n
=
t
n
1,
ϕ
n
contains the remaining, primitive
n
th roots of unit.
Since
t
n
1 has no repeated roots, we know that
ϕ
n
does not contain any extra
roots. So
Root
ϕ
n
(L) = {nth primitive roots of unity}.
These
ϕ
n
are what we use to “build up” the polynomials
t
n
1. These
will later serve as a technical tool to characterize the Galois group of the
n
th
cyclotomic extension of Q.
Before we an reach that, we first take a tiny step, and prove something that
works for arbitrary fields first.
Theorem. Let
K
be a field with
char K
= 0 or 0
< char K n
. Let
L
be the
n
th cyclotomic extension of
K
. Then
L/K
is a Galois extension, and there is an
injective homomorphism θ : Gal(L/K) (Z/nZ)
×
.
In addition, every irreducible factor of ϕ
n
(in K[t]) has degree [L : K].
The important thing about our theorem is the homomorphism
θ : Gal(L/K) (Z/nZ)
×
.
In general, we don’t necessarily know much about
Gal
(
L/K
), but the group
(
Z/nZ
)
×
is well-understood. In particular, we now know that
Gal
(
L/K
) is
abelian.
Proof. Let µ be an nth primitive root of unity. Then
Root
t
n
1
(L) = {1, µ, µ
2
, ··· , µ
n1
}
is a cyclic group of order
n
generated by
µ
. We first construct the homomorphism
θ
:
Aut
K
(
L
)
(
Z/nZ
)
×
as follows: for each
ϕ Aut
K
(
L
),
ϕ
is completely
determined by the value of
ϕ
(
µ
) since
L
=
K
(
µ
). Since
ϕ
is an automorphism, it
must take an
n
th primitive root of unity to another
n
th primitive root of unity.
So
ϕ
(
µ
) =
µ
i
for some
i
such that (
i, n
) = 1. Now let
θ
(
ϕ
) =
¯
i
(
Z/nZ
)
×
. Note
that this is well-defined since if µ
i
= µ
j
, then i j has to be a multiple of n.
Now it is easy to see that if
ϕ, ψ Aut
K
(
L
) are given by
ϕ
(
µ
) =
µ
i
, and
ψ
(
µ
) =
µ
j
, then
ϕ ψ
(
µ
) =
ϕ
(
µ
j
) =
µ
ij
. So
θ
(
ϕψ
) =
¯
ij
=
θ
(
ϕ
)
θ
(
ψ
). So
θ
is a
group homomorphism.
Now we check that
θ
is injective. If
θ
(
ϕ
) =
¯
1
(note that (
Z/nZ
)
×
is a
multiplicative group with unit 1), then ϕ(µ) = µ. So ϕ = id.
Now we show that
L/K
is Galois. Recall that
L
=
K
(
µ
), and let
P
µ
be
a minimal polynomial of
µ
over
K
. Since
µ
is a root of
t
n
1, we know that
P
µ
| t
n
1. Since
t
n
1 has no repeated roots,
P
µ
has no repeated roots. So
P
µ
is separable. Moreover,
P
µ
splits over
L
as
t
n
1 splits over
L
. So the extension
is separable and normal, and hence Galois.
Applying the previous theorem, each irreducible factor
g
of
ϕ
n
is a minimal
polynomial of some nth primitive root of unity, say λ. Then L = K(λ). So
deg g = deg P
λ
= [K(λ) : K] = [L : K].
Example. We can calculate the following in Q[t].
(i) ϕ
1
= t 1.
(ii) ϕ
2
= t + 1 since t
2
1 = ϕ
1
ϕ
2
.
(iii) ϕ
3
= t
2
+ t + 1.
(iv) ϕ
4
= t
2
+ 1.
These are rather expected. Now take
K
=
F
2
. Then 1 =
1. So we might be
able to further decompose these polynomials. For example,
t
+ 1 =
t
1 in
F
2
.
So we have
ϕ
4
= t
2
+ 1 = t
2
1 = ϕ
1
ϕ
2
.
So in
F
2
,
ϕ
4
is not irreducible. Similarly, if we have too much time, we can show
that
ϕ
15
= (t
4
+ t + 1)(t
4
+ t
3
+ 1).
So
ϕ
15
is not irreducible. However, they are irreducible over the rationals, as we
will soon see.
So far, we know
Gal
(
L/K
) is an abelian group, isomorphic to a subgroup of
(
Z/nZ
)
×
. However, we are greedy and we want to know more. The following
lemma tells us when this θ is an isomorphism.
Lemma. Under the notation and assumptions of the previous theorem,
ϕ
n
is
irreducible in K[t] if and only if θ is an isomorphism.
Proof.
(
) Suppose
ϕ
n
is irreducible. Recall that
Root
ϕ
n
(
L
) is exactly the
n
th
primitive roots of unity. So if
µ
is an
n
th primitive root of unity, then
P
µ
, the
minimal polynomial of
µ
over
K
is
ϕ
n
. In particular, if
λ
is also an
n
th primitive
root of unity, then
P
µ
=
P
λ
. This implies that there is some
ϕ
λ
Aut
K
(
L
) such
that ϕ
λ
(µ) = λ.
Now if
¯
i
(
Z/nZ
)
×
, then taking
λ
=
µ
i
, this shows that we have
ϕ
λ
Aut
K
(L) such that θ(ϕ
λ
) =
¯
i. So θ is surjective, and hence an isomorphism.
(
) Suppose that
θ
is an isomorphism. We will reverse the above argument
and show that all roots have the same minimal polynomial. Let
µ
be a
n
th
primitive root of unity, and pick
¯
i
(
Z/nZ
)
×
, and let
λ
=
µ
i
. Since
θ
is an
isomorphism, there is some
ϕ
λ
Aut
K
(
L
) such that
θ
(
ϕ
λ
) =
¯
i
, i.e.
ϕ
λ
(
µ
) =
µ
i
= λ. Then we must have P
µ
= P
λ
.
Since every
n
th primitive root of unity is of the form
µ
i
(with (
i, n
) = 1), this
implies that all
n
th primitive roots have the same minimal polynomial. Since
the roots of
ϕ
n
are all the
n
th primitive roots of unity, its irreducible factors are
exactly the minimal polynomials of the primitive roots. Moreover, ϕ
n
does not
have repeated roots. So ϕ
n
= P
µ
. In particular, ϕ
n
is irreducible.
We want to apply this lemma to the case of rational numbers. We want to
show that
θ
is an isomorphism. So we have to show that
ϕ
n
is irreducible in
Q[t].
Theorem. ϕ
n
is irreducible in Q[t]. In particular, it is also irreducible in Z[t].
Proof.
As before, this can be achieved by showing that all
n
th primitive roots
have the same minimal polynomial. Moreover, let
µ
be our favorite
n
th primitive
root. Then all other primitive roots
λ
are of the form
λ
=
µ
i
, where (
i, n
) = 1. By
the fundamental theorem of arithmetic, we can write
i
as a product
i
=
q
1
···q
n
.
Hence it suffices to show that for all primes
q n
, we have
P
µ
=
P
µ
q
. Noting
that µ
q
is also an nth primitive root, this gives
P
µ
= P
µ
q
1
= P
(µ
q
1
)
q
2
= P
µ
q
1
q
2
= ··· = P
µ
q
1
···q
r
= P
µ
i
.
So we now let
µ
be an
n
th primitive root,
P
µ
be its minimal polynomial. Since
µ is a root of ϕ
n
, we can write P
µ
| ϕ
n
inside Q[t]. So we can write
ϕ
n
= P
µ
R,
Since
ϕ
n
and
P
µ
are monic,
R
is also monic. By Gauss’ lemma, we must have
P
µ
, R Z[t].
Note that showing
P
µ
=
P
µ
q
is the same as showing
µ
q
is a root of
P
µ
, since
deg P
µ
=
deg P
µ
q
. So suppose it’s not. Since
µ
q
is an
n
th primitive root of unity,
it is a root of
ϕ
n
. So
µ
q
must be a root of
R
. Now let
S
=
R
(
t
q
). Then
µ
is a
root of S, and so P
µ
| S.
We now reduce mod
q
. For any polynomial
f Z
[
t
], we write the result of
reducing the coefficients mod
q
as
¯
f
. Then we have
¯
S
=
R(t
q
)
=
R(t)
q
. Since
¯
P
µ
divides
¯
S
(by Gauss’ lemma), we know
¯
P
µ
and
R(t)
have common roots. But
¯
ϕ
n
=
¯
P
µ
¯
R
, and so this implies
¯
ϕ
n
has repeated roots. This is impossible since
¯
ϕ
n
divides
t
n
1, and since
q n
, we know the derivative of
t
n
1 does not
vanish at the roots. So we are done.
Corollary. Let
K
=
Q
and
L
be the
n
th cyclotomic extension of
Q
. Then the
injection θ : Gal(L/Q) (Z/nZ)
×
is an isomorphism.
Example. Let
p
be a prime number, and
q
=
p
d
,
d N
. Consider
F
q
, a field
with
q
elements, and let
L
be the
n
th cyclotomic extension of
F
q
(where
p n
).
Then we have a homomorphism θ : Gal(L/F
q
) (Z/nZ)
×
.
We have previously shown that
Gal
(
L/F
q
) must be a cyclic group. So if
(
Z/nZ
)
×
is non-cyclic, then
θ
is not an isomorphism, and
ϕ
n
is not irreducible
in F
q
[t].
For example, take p = q = 7 and n = 8. Then
(Z/8Z)
×
= {
¯
1,
¯
3,
¯
5,
¯
7}
is not cyclic, because manual checking shows that there is no element of order 4.
Hence
θ
:
Gal
(
L/F
7
)
(
Z/
8
Z
)
×
is not an isomorphism, and
ϕ
8
is not irreducible
in F
7
[t].