2Field extensions
II Galois Theory
2.1 Field extensions
Definition (Field extension). A field extension is an inclusion of a field
K ⊆ L
,
where
K
inherits the algebraic operations from
L
. We also write this as
L/K
.
Alternatively, we can define this by a injective homomorphism
K → L
. We say
L is an extension of K, and K is a subfield of L.
Example.
(i) R/Q is a field extension.
(ii) C/Q is a field extension.
(iii) Q(
√
2) = {a + b
√
2 : a, b ∈ Q} ⊆ R is a field extension over Q.
Given a field extension
L/K
, we want to quantify how much “bigger”
L
is compared to
K
. For example, to get from
Q
to
R
, we need to add a lot of
elements (since
Q
is countable and
R
is uncountable). On the other hand, to get
from R to C, we just need to add a single element
√
−1.
To do so, we can consider
L
as a vector space over
K
. We know that
L
already comes with an additive abelian group structure, and we can define scalar
multiplication by simply multiplying: if
a ∈ K, α ∈ L
, then
a · α
is defined as
multiplication in L.
Definition (Degree of field extension). The degree of
L
over
K
is [
L
:
K
] is the
dimension of
L
as a vector space over
K
. The extension is finite if the degree is
finite.
In this course, we are mostly concerned with finite extensions.
Example.
(i)
Consider
C/R
. This is a finite extension with degree [
C
:
R
] = 2 since we
have a basis of {1, i}.
(ii) The extension Q(
√
2)/Q has degree 2 since we have a basis of {1,
√
2}.
(iii) The extension R/Q is not finite.
We are going to use the following result a lot:
Theorem (Tower Law). Let F/L/K be field extensions. Then
[F : K] = [F : L][L : K]
Proof.
Assume [
F
:
L
] and [
L
:
K
] are finite. Let
{α
1
, ··· , α
m
}
be a basis for
L
over
K
, and
{β
1
, ··· , β
n
}
be a basis for
F
over
L
. Pick
γ ∈ F
. Then we can
write
γ =
X
i
b
i
β
i
, b
i
∈ L.
For each b
i
, we can write as
b
i
=
X
j
a
ij
α
j
, a
ij
∈ K.
So we can write
γ =
X
i
X
j
a
ij
α
j
β
i
=
X
i,j
a
ij
α
j
β
i
.
So
T
=
{α
j
β
i
}
i,j
spans
F
over
K
. To show that this is a basis, we have to show
that they are linearly independent. Consider the case where
γ
= 0. Then we
must have
b
i
= 0 since
{β
i
}
is a basis of
F
over
L
. Hence each
a
ij
= 0 since
{α
j
} is a basis of L over K.
This implies that T is a basis of F over K. So
[F : K] = |T | = nm = [F : L][L : K].
Finally, if [
F
:
L
] =
∞
or [
L
:
K
] =
∞
, then clearly [
F
:
K
] =
∞
as well. So
equality holds as well.
Recall that in IA Numbers and Sets, we defined a real number
x
to be
algebraic if it is a root of some polynomial in integer (or rational) coefficients.
We can do this for general field (extensions) as well.
Definition (Algebraic number). Let
L/K
be a field extension,
α ∈ L
. We
define
I
α
= {f ∈ K[t] : f (α) = 0} ⊆ K[t]
This is the set of polynomials for which
α
is a root. It is easy to show that
I
α
is
an ideal, since it is the kernel of the ring homomorphism
K
[
t
]
→ L
by
g 7→ g
(
α
).
We say
α
is algebraic over
K
if
I
α
= 0. Otherwise,
α
is transcendental over
K.
We say L is algebraic over K if every element of L is algebraic.
Example.
(i)
9
√
7
is algebraic over
Q
because
f
(
9
√
7
) = 0, where
f
=
t
9
−
7. In general,
any number written with radicals is algebraic over Q.
(ii) π is not algebraic over Q.
These are rather simple examples, and the following lemma will provide us a
way of generating much more examples.
Lemma. Let L/K be a finite extension. Then L is algebraic over K.
Proof.
Let
n
= [
L
:
K
], and let
α ∈ L
. Then 1
, α, α
2
, ··· , α
n
are linearly
dependent over
K
(since there are
n
+ 1 elements). So there exists some
a
i
∈ K
(not all zero) such that
a
n
α
n
+ a
n−1
α
n−1
+ ··· + a
1
α + a
0
= 0.
So we have a non-trivial polynomial that vanishes at
α
. So
α
is algebraic over
K.
Since α was arbitrary, L itself is algebraic.
If
L/K
is a field extension and
α ∈ L
is algebraic, then by definition, there is
some polynomial
f
such that
f
(
α
) = 0. It is a natural question to ask if there is
a “smallest” polynomial that does this job. Obviously we can find a polynomial
of smallest degree (by the well-ordering principle of the natural numbers), but
we can get something even stronger.
Since
K
is a field,
K
[
t
] is a PID (principal ideal domain). This, by definition,
implies we can find some (monic)
P
α
∈ K
[
t
] such that
I
α
=
⟨P
α
⟩
. In other words,
every element of I
α
is just a multiple of P
α
.
Definition (Minimal polynomial). Let
L/K
be a field extension,
α ∈ L
. The
minimal polynomial of
α
over
K
is a monic polynomial
P
α
such that
I
α
=
⟨P
α
⟩
.
Example.
(i) Consider R/Q, α =
3
√
2. Then the minimal polynomial is P
α
= t
3
− 2.
(ii) Consider C/R, α =
3
√
2. Then the minimal polynomial is P
α
= t −
3
√
2.
It should be intuitively obvious that by virtue of being “minimal”, the
minimal polynomial is irreducible.
Proposition. Let
L/K
be a field extension,
α ∈ L
algebraic over
K
, and
P
α
the minimal polynomial. Then P
α
is irreducible in K[t].
Proof.
Assume that
P
α
=
QR
in
K
[
t
]. So 0 =
P
α
(
α
) =
Q
(
α
)
R
(
α
). So
Q
(
α
) = 0
or
R
(
α
) = 0. Say
Q
(
α
) = 0. So
Q ∈ I
α
. So
Q
is a multiple of
P
α
. However, we
also know that
P
α
is a multiple of
Q
α
. This is possible only if
R
is a unit in
K[t], i.e. R ∈ K. So P
α
is irreducible.
It should also be clear that if
f
is irreducible and
f
(
α
) = 0, then
f
is the
minimal polynomial. Often, it is the irreducibility of P
α
that is important.
Apart from the minimal polynomial, we can also ask for the minimal field
containing α.
Definition (Field generated by
α
). Let
L/K
be a field extension,
α ∈ L
. We
define
K
(
α
) to be the smallest subfield of
L
containing
K
and
α
. We call
K
(
α
)
the field generated by α over K.
This definition by itself is rather abstract and not very helpful. Intuitively,
K
(
α
) is what we get when we add
α
to
K
, plus all the extra elements needed to
make
K
(
α
) a field (i.e. closed under addition, multiplication and inverse). We
can express this idea more formally by the following result:
Theorem. Let L/K a field extension, α ∈ L algebraic. Then
(i) K
(
α
) is the image of the (ring) homomorphism
ϕ
:
K
[
t
]
→ L
defined by
f 7→ f(α).
(ii) [K(α) : K] = deg P
α
, where P
α
is the minimal polynomial of α over K.
Note that the kernel of the homomorphism
ϕ
is (almost) by definition the
ideal ⟨P
α
⟩. So this theorem tells us
K[t]
⟨P
α
⟩
∼
=
K(α).
Proof.
(i)
Let
F
be the image of
ϕ
. The first step is to show that
F
is indeed a field.
Since
F
is the image of a ring homomorphism, we know
F
is a subring of
L. Given β ∈ F non-zero, we have to find an inverse.
By definition,
β
=
f
(
α
) for some
f ∈ K
[
t
]. The idea is to use B´ezout’s
identity. Since
β
= 0,
f
(
α
)
= 0. So
f ∈ I
α
=
⟨P
α
⟩
. So
P
α
∤ f
in
K
[
t
]. Since
P
α
is irreducible,
P
α
and
f
are coprime. Then there exists some
g, h ∈ K
[
t
]
such that
fg
+
hP
α
= 1. So
f
(
α
)
g
(
α
) =
f
(
α
)
g
(
α
) +
h
(
α
)
P
α
(
α
) = 1. So
βg(α) = 1. So β has an inverse. So F is a field.
From the definition of
F
, we have
K ⊆ F
and
α ∈ F
, using the constant
polynomials f = c ∈ K and the identity f = t.
Now, if
K ⊆ G ⊆ L
and
α ∈ G
, then
G
contains all the polynomial
expressions of α. Hence F ⊆ G. So K(α) = F .
(ii)
Let
n
=
deg P
α
. We show that
{
1
, α, α
2
, ··· , α
n−1
}
is a basis for
K
(
α
)
over K.
First note that since deg P
α
= n, we can write
α
n
=
n−1
X
i=0
a
i
α
i
.
So any other higher powers are also linear combinations of the
α
i
s (by
induction). This means that
K
(
α
) is spanned by 1
, ··· , α
n−1
as a
K
vector
space.
It remains to show that
{
1
, ··· , α
n−1
}
is linearly independent. Assume
not. Then for some b
i
, we have
n−1
X
i=0
b
i
α
i
= 0.
Let
f
=
P
b
i
t
i
. Then
f
(
α
) = 0. So
f ∈ I
α
=
⟨P
α
⟩
. However,
deg f <
deg P
α
. So we must have
f
= 0. So all
b
i
= 0. So
{
1
, ··· , α
n−1
}
is a basis
for K(α) over K. So [K(α) : K] = n.
Corollary. Let
L/K
be a field extension,
α ∈ L
. Then
α
is algebraic over
K
if
and only if K(α)/K is a finite extension.
Proof.
If
α
is algebraic, then [
K
(
α
) :
K
] =
deg P
α
< ∞
by above. So the
extension is finite.
If
K ⊆ K
(
α
) is a finite extension, then by previous lemma, the entire
K
(
α
)
is algebraic over K. So α is algebraic over K.
We can extend this definition to allow more elements in the generating set.
Definition (Field generated by elements). Let
L/K
be a field extension,
α
1
, ··· , α
n
⊆ L
. We define
K
(
α
1
, ··· , α
n
) to be the smallest subfield of
L
containing K and α
1
, ··· , α
n
.
We call K(α
1
, ··· , α
n
) the field generated by α
1
, ··· , α
n
over K.
And we can prove some similar results.
Theorem. Suppose that L/K is a field extension.
(i)
If
α
1
, ··· , α
n
∈ L
are algebraic over
K
, then
K
(
α
1
, ··· , α
n
)
/K
is a finite
extension.
(ii)
If we have field extensions
L/F/K
and
F/K
is a finite extension, then
F = K(α
1
, ··· , α
n
) for some α
1
, ··· , α
n
∈ L.
Proof.
(i)
We prove this by induction. Since
α
1
is algebraic over
K
,
K ⊆ K
(
α
1
) is a
finite extension.
For 1
≤ i < n
,
α
i+1
is algebraic over
K
. So
α
i+1
is also algebraic
over
K
(
α
1
, ··· , α
i
). So
K
(
α
1
, ··· , α
i
)
⊆ K
(
α
1
, ··· , α
i
)(
α
i+1
) is a finite
extension. But
K
(
α
1
, ··· , α
i
)(
α
i+1
) =
K
(
α
1
, ··· , α
i+1
). By the tower law,
K ⊆ K(α
i
, ··· , α
i+1
) is a finite extension.
(ii)
Since
F
is a finite dimensional vector space over
K
, we can take a basis
{α
1
, ··· , α
n
}
of
F
over
K
. Then it should be clear that
F
=
K
(
α
1
, ··· , α
n
).
When studying polynomials, the following result from IB Groups, Rings and
Modules is often helpful:
Proposition (Eisenstein’s criterion). Let
f
=
a
n
t
n
+
···
+
a
1
t
+
a
0
∈ Z
[
t
].
Assume that there is some prime number p such that
(i) p | a
i
for all i < n.
(ii) p ∤ a
n
(iii) p
2
∤ a
0
.
Then f is irreducible in Q[t].
Example. Consider the field extensions
Q ⊆ Q(
√
2) ⊆ Q(
√
2,
3
√
2) ⊆ R,
Q ⊆ Q(
3
√
2) ⊆ Q(
√
2,
3
√
2) ⊆ R.
We have [Q(
√
2) : Q] = 2 since {1,
√
2} is a basis of Q(
√
2) over Q.
How about [
Q
(
3
√
2
) :
Q
]? By the Eisenstein criterion, we know that
t
3
−
2 is
irreducible in
Q
[
t
]. So the minimal polynomial of
3
√
2
over
Q
is
t
3
−
2 which has
degree 3. So [Q(
3
√
2) : Q] = 3.
These results immediately tells that
3
√
2 ∈ Q
(
√
2
). Otherwise, this entails
that Q(
3
√
2) ⊆ Q(
√
2). Then the tower law says that
[Q(
√
2) : Q] = [Q(
√
2) : Q(
3
√
2)][Q(
3
√
2) : Q].
In particular, plugging the numbers in entails that that 3 is a factor of 2, which
is clearly nonsense. Similarly,
√
2 ∈ Q(
3
√
2).
How about the inclusion
Q
(
√
2
)
⊆ Q
(
√
2,
3
√
2
)? We now show that the
minimal polynomial P
3
√
2
of
3
√
2 over Q(
√
2) is t
3
− 2.
Suppose not. Then
t
3
−
2 is reducible, with the real
P
3
√
2
as one of its factors.
Let t
3
− 2 = P
3
√
2
· R for some non-unit polynomial R.
We know that
P
3
√
2
does not have degree 3 (or else it would be
t
3
−
2), and
not degree 1, since a degree 1 polynomial has a root. So it has degree 2. So
R
has degree 1. Then
R
has a root, i.e.
R
(
β
) = 0 for some
β ∈ Q
(
√
2
). So
β
3
− 2 = 0. Hence [Q(β) : Q] = 3. Again, by the tower law, we have
[Q(
√
2) : Q] = [Q(
√
2) : Q(β)][Q(β) : Q].
Again, this is nonsense since it entails that 3 is a factor of 2. So the minimal
polynomial is indeed t
3
− 2. So [Q(
√
2,
3
√
2) : Q] = 6 by the tower law.
Alternatively, we can obtain this result by noting that the tower law on
Q ⊆ Q
(
√
2
)
⊆ Q
(
√
2,
3
√
2
) and
Q ⊆ Q
(
3
√
2
)
⊆ Q
(
√
2,
3
√
2
) entails that 2 and 3 are
both factors of [
Q
(
√
2,
3
√
2
) :
Q
]. So it is at least 6. Then since
t
3
−
2
∈ Q
(
√
2
)[
t
]
has
3
√
2 as a root, the degree is at most 6. So it is indeed 6.