2Field extensions
II Galois Theory
2.9 Finite fields
We’ll have a slight digression and look at finite fields. We adopt the notation
where
p
is always a prime number, and
Z
p
=
Z/⟨p⟩
. It turns out finite fields are
rather simple, as described in the lemma below:
Lemma. Let K be a finite field with q = |K| element. Then
(i) q = p
d
for some d ∈ N, where p = char K > 0.
(ii)
Let
f
=
t
q
− t
. Then
f
(
α
) = 0 for all
α ∈ K
. Moreover,
K
is the splitting
field of f over F
p
.
This means that a finite field is completely determined by the number of
elements.
Proof.
(i)
Consider the set
{m·
1
K
}
m∈Z
, where 1
K
is the unit in
K
and
m·
represents
repeated addition. We can identify this with
F
p
. So we have the extension
F
p
⊆ K. Let d = [K : F
p
]. Then q = |K| = p
d
.
(ii)
Note that
K
∗
=
K \ {
0
}
is a finite multiplicative group with order
q −
1.
Then by Lagrange’s theorem, α
q−1
= 1 for all α ∈ K
∗
. So α
q
− α = 0 for
all α = 0. The α = 0 case is trivial.
Now every element in
K
is a root of
f
. So we need to check that all roots
of
f
are in
K
. Note that the derivative
f
′
=
qt
q−1
−
1 =
−
1 (since
q
is a
power of the characteristic). So
f
′
(
α
) =
−
1
= 0 for all
α ∈ K
. So
f
and
f
′
have no common roots. So
f
has no repeated roots. So
K
contains
q
distinct roots of f. So K is a splitting field.
Lemma. Let q = p
d
, q
′
= p
d
′
, where d, d
′
∈ N. Then
(i)
There is a finite field
K
with exactly
q
elements, which is unique up to
isomorphism. We write this as F
q
.
(ii) We can embed F
q
⊆ F
q
′
iff d | d
′
.
Proof.
(i)
Let
f
=
t
q
− t
, and let
K
be a splitting field of
f
over
F
p
. Let
L
=
Root
f
(
K
). The objective is to show that
L
=
K
. Then we will have
|K|
=
|L|
=
|Root
f
(
K
)
|
=
deg f
=
q
, because the proof of the previous
lemma shows that f has no repeated roots.
To show that
L
=
K
, by definition, we have
L ⊆ K
. So we need to show
every element in
K
is in
L
. We do so by showing that
L
itself is a field.
Then since
L
contains all the roots of
f
and is a subfield of the splitting
field K, we must have K = L.
It is straightforward to show that L is a field: if α, β ∈ L, then
(α + β)
q
= α
q
+ β
q
= α + β.
So α + β ∈ L. Similarly, we have
(αβ)
q
= α
q
β
q
= αβ.
So αβ ∈ L. Also, we have
(α
−1
)
q
= (α
q
)
−1
= α
−1
.
So α
−1
∈ L. So L is in fact a field.
Since any field of size
q
is a splitting field of
f
, and splitting fields are
unique to isomorphism, we know that K is unique.
(ii)
Suppose
F
q
⊆ F
q
′
. Then let
n
= [
F
q
′
:
F
q
]. So
q
′
=
q
n
. So
d
′
=
nd
. So
d | d
′
.
On the other hand, suppose
d | d
′
. Let
d
′
=
dn
. We let
f
=
t
q
′
− t
. Then
for any α ∈ F
q
, we have
f(α) = α
q
′
− α = α
q
n
− α = (···((α
q
)
q
)
q
···)
q
− α = α − α = 0.
Since
F
q
′
is the splitting field of
f
, all roots of
f
are in
F
q
′
. So we know
that F
q
⊆ F
′
q
.
Note that if
¯
F
p
is the algebraic closure of
F
p
, then
F
q
⊆
¯
F
p
for every
q
=
p
d
.
We then have
[
k∈N
F
p
k
=
¯
F
p
,
because any α ∈
¯
F
p
is algebraic over F
p
, and so belongs to some F
q
.
Definition. Consider the extension
F
q
n
/F
q
, where
q
is a power of
p
. The
Frobenius Fr
q
: F
q
n
→ F
q
n
is defined by α 7→ α
q
.
This is a homomorphism precisely because the field is of characteristic zero.
In fact, Fr
q
∈ Aut
F
q
(F
q
n
), since α
q
= α for all α ∈ F
q
.
The following two theorems tells us why we care about the Frobenius.
Theorem. Consider
F
q
n
/F
q
. Then
Fr
q
is an element of order
n
as an element
of Aut
F
q
(F
q
n
).
Proof.
For all
α ∈ F
q
n
, we have
Fr
n
q
(
α
) =
α
q
n
=
α
. So the order of
Fr
q
divides
n.
If m | n, then the set
{α ∈ F
q
n
: Fr
m
q
(α) = α} = {α ∈ F
q
n
: α
q
m
= α} = F
q
m
.
So if m is the order of Fr
q
, then F
q
m
= F
q
n
. So m = n.
Theorem. The extension
F
q
n
/F
q
is Galois with Galois group
Gal
(
F
q
n
/F
q
) =
Aut
F
q
(F
q
n
)
∼
=
Z/nZ, generated by Fr
q
.
Proof.
The multiplicative group
F
∗
q
n
=
F
q
n
\ {
0
}
is finite. We have previously
seen that multiplicative groups of finite fields are cyclic. So let
α
be a generator
of this group. Then
F
q
n
=
F
q
(
α
). Let
P
α
be the minimal polynomial of
α
over
F
q
. Then since Aut
F
q
(F
q
n
) has an element of order n, we get
n ≤ |Aut
F
q
(F
q
n
)| = |Hom
F
q
(F
q
(α), F
q
n
)|.
Since F
q
(α) is generated by one element, we know
|Hom
F
q
(F
q
(α), F
q
n
)| = |Root
P
α
(F
q
n
)|
So we have
n ≤ |Root
P
α
(F
q
n
)| ≤ deg P
α
= [F
q
n
: F
q
] = n.
So we know that
|Aut
F
q
(F
q
n
)| = [F
q
n
: F
q
] = n.
So F
q
n
/F
q
is a Galois extension.
Since
|Aut
F
q
(
F
q
n
)
|
, it has to be generated by
Fr
q
, since this has order
n
. In
particular, this group is cyclic.
We see that finite fields are rather nice — there is exactly one field of order
p
d
for each
d
and prime
p
, and these are all of the finite fields. All extensions
are Galois and the Galois group is a simple cyclic group.
Example. Consider F
4
/F
2
. We can write
F
2
= {0, 1} ⊆ F
4
= {0, 1, α, α
2
},
where α is a generator of F
∗
4
. Define ϕ ∈ Aut
F
2
(F
4
) by ϕ(α) = α
2
. Then
Aut
F
2
(F
4
) = {id, ϕ}
since it has order 2.
Note that we can also define the Frobenius
Fr
p
:
¯
F
p
→
¯
F
p
, where
α 7→ α
p
.
Then
F
p
d
is the elements of
¯
F
p
fixed by
Fr
d
p
. So we can recover this subfield by
just looking at the Frobenius.