II Galois Theory - Solutions to polynomial equations

3Solutions to polynomial equations

II Galois Theory

3.1 Cyclotomic extensions

Definition (Cyclotomic extension). For a field

, we define the

th cyclotomic

extension to be the splitting field of t

− 1.

Note that if

is a field and

is the

th cyclotomic extension, then

Root

−1

(

) is a subgroup of multiplicative group

∗

L \ {

}

. Since this is a

finite subgroup of L

∗

, it is a cyclic group.

Moreover, if

char K

= 0 or 0

< char K ∤ n

, then (

−

′

n−1

and this

has no common roots with

−

1. So

−

1 has no repeated roots. In other

words, t

− 1 has n distinct roots. So as a group,

Root

−1

(L)

∼

Z/nZ.

In particular, this group has at least one element µ of order n.

Definition (Primitive root of unity). The

th primitive root of unity is an

element of order n in Root

−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

th cyclotomic monic polynomial

∈ Z[t] satisfying:

(i) For each n ∈ N, we have

− 1 =

d|n

(ii) Assume char K = 0 or 0 < char K ∤ n. Then

Root

(L) = {nth primitive roots of unity}.

Note that here we have an abuse of notation, since

is a polynomial in

[

], not

[

], but we can just use the canonical map

[

]

→ K

[

] mapping

1 to 1 and t to t.

Proof.

We do induction on

to construct

. When

= 1, let

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 =

d|n,d<n

By induction,

f ∈ Z

[

]. Moreover, if

d | n

and

d < n

, then

(

−

1) because

(

−

(

−

1). We would like to say that

also divides

−

1. However, we

have to be careful, since to make this conclusion, we need to show that

and

′

have no common roots for distinct d, d

′

| n (and d, , d

′

< n).

Indeed, by induction, ϕ

and ϕ

′

have no common roots because

Root

(L) = {dth primitive roots of unity},

Root

′

(L) = {d

′

th primitive roots of unity},

and these two sets are disjoint (or else the roots would not be primitive).

Therefore

and

′

have no common irreducible factors. Hence

f | t

−

1. So

we can write

− 1 = fϕ

where

∈ Q

[

]. Since

is monic,

has integer coefficients. So indeed

∈ Z[t]. So the first part is proven.

To prove the second part, note that by induction,

Root

(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

−

contains the remaining, primitive

th roots of unit.

Since

−

1 has no repeated roots, we know that

does not contain any extra

roots. So

Root

(L) = {nth primitive roots of unity}.

These

are what we use to “build up” the polynomials

−

1. These

will later serve as a technical tool to characterize the Galois group of the

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

be a field with

char K

= 0 or 0

< char K ∤ n

. Let

be the

th cyclotomic extension of

. 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

(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

−1

(L) = {1, µ, µ

, ··· , µ

n−1

}

is a cyclic group of order

generated by

. We first construct the homomorphism

Aut

(

)

→

(

Z/nZ

)

as follows: for each

ϕ ∈ Aut

(

is completely

determined by the value of

(

) since

(

). Since

is an automorphism, it

must take an

th primitive root of unity to another

th primitive root of unity.

(

) =

for some

such that (

i, n

) = 1. Now let

(

) =

i ∈

(

Z/nZ

)

. Note

that this is well-defined since if µ

= µ

, then i − j has to be a multiple of n.

Now it is easy to see that if

ϕ, ψ ∈ Aut

(

) are given by

(

) =

, and

(

) =

, then

ϕ ◦ ψ

(

) =

(

) =

. So

(

ϕψ

) =

(

)

(

). So

is a

group homomorphism.

Now we check that

is injective. If

(

) =

(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

(

), and let

a minimal polynomial of

over

. Since

is a root of

−

1, we know that

| t

−

1. Since

−

1 has no repeated roots,

has no repeated roots. So

is separable. Moreover,

splits over

−

1 splits over

. So the extension

is separable and normal, and hence Galois.

Applying the previous theorem, each irreducible factor

is a minimal

polynomial of some nth primitive root of unity, say λ. Then L = K(λ). So

deg g = deg P

= [K(λ) : K] = [L : K].

Example. We can calculate the following in Q[t].

(i) ϕ

= t − 1.

(ii) ϕ

= t + 1 since t

− 1 = ϕ

(iii) ϕ

= t

+ t + 1.

(iv) ϕ

= t

+ 1.

These are rather expected. Now take

. Then 1 =

−

1. So we might be

able to further decompose these polynomials. For example,

+ 1 =

t −

1 in

So we have

= t

+ 1 = t

− 1 = ϕ

So in

is not irreducible. Similarly, if we have too much time, we can show

that

= (t

+ t + 1)(t

+ t

+ 1).

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,

irreducible in K[t] if and only if θ is an isomorphism.

Proof.

(

⇒

) Suppose

is irreducible. Recall that

Root

(

) is exactly the

primitive roots of unity. So if

is an

th primitive root of unity, then

, the

minimal polynomial of

over

. In particular, if

is also an

th primitive

root of unity, then

. This implies that there is some

∈ Aut

(

) such

that ϕ

(µ) = λ.

Now if

i ∈

(

Z/nZ

)

, then taking

, this shows that we have

∈

Aut

(L) such that θ(ϕ

) =

i. So θ is surjective, and hence an isomorphism.

(

⇔

) Suppose that

is an isomorphism. We will reverse the above argument

and show that all roots have the same minimal polynomial. Let

be a

primitive root of unity, and pick

i ∈

(

Z/nZ

)

, and let

. Since

is an

isomorphism, there is some

∈ Aut

(

) such that

(

) =

, i.e.

(

) =

= λ. Then we must have P

= P

Since every

th primitive root of unity is of the form

(with (

i, n

) = 1), this

implies that all

th primitive roots have the same minimal polynomial. Since

the roots of

are all the

th primitive roots of unity, its irreducible factors are

exactly the minimal polynomials of the primitive roots. Moreover, ϕ

does not

have repeated roots. So ϕ

= P

. In particular, ϕ

is irreducible.

We want to apply this lemma to the case of rational numbers. We want to

show that

is an isomorphism. So we have to show that

is irreducible in

Q[t].

Theorem. ϕ

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

th primitive roots

have the same minimal polynomial. Moreover, let

be our favorite

th primitive

root. Then all other primitive roots

are of the form

, where (

i, n

) = 1. By

the fundamental theorem of arithmetic, we can write

as a product

···q

Hence it suffices to show that for all primes

q ∤ n

, we have

. Noting

that µ

is also an nth primitive root, this gives

= P

(µ

)

= P

= ··· = P

···q

= P

So we now let

be an

th primitive root,

be its minimal polynomial. Since

µ is a root of ϕ

, we can write P

| ϕ

inside Q[t]. So we can write

= P

Since

and

are monic,

is also monic. By Gauss’ lemma, we must have

, R ∈ Z[t].

Note that showing

is the same as showing

is a root of

, since

deg P

. So suppose it’s not. Since

is an

th primitive root of unity,

it is a root of

. So

must be a root of

. Now let

(

). Then

is a

root of S, and so P

| S.

We now reduce mod

. For any polynomial

f ∈ Z

[

], we write the result of

reducing the coefficients mod

. Then we have

R(t

)

R(t)

. Since

divides

(by Gauss’ lemma), we know

and

R(t)

have common roots. But

, and so this implies

has repeated roots. This is impossible since

divides

−

1, and since

q ∤ n

, we know the derivative of

−

1 does not

vanish at the roots. So we are done.

Corollary. Let

and

be the

th cyclotomic extension of

. Then the

injection θ : Gal(L/Q) → (Z/nZ)

is an isomorphism.

Example. Let

be a prime number, and

d ∈ N

. Consider

, a field

with

elements, and let

be the

th cyclotomic extension of

(where

p ∤ n

Then we have a homomorphism θ : Gal(L/F

) → (Z/nZ)

We have previously shown that

Gal

(

L/F

) must be a cyclic group. So if

(

Z/nZ

)

is non-cyclic, then

is not an isomorphism, and

is not irreducible

in F

[t].

For example, take p = q = 7 and n = 8. Then

(Z/8Z)

= {

is not cyclic, because manual checking shows that there is no element of order 4.

Hence

Gal

(

L/F

)

→

(

)

is not an isomorphism, and

is not irreducible

in F

[t].