2Field extensions
II Galois Theory
2.6 Separable extensions
Here we will define what it means for an extension to be separable. This is
done via defining separable polynomials, and then an extension is separable if
all minimal polynomials are separable.
At first, the definition of separability might seem absurd — surely every
polynomial should be separable. Indeed, polynomials that are not separable
tend to be weird, and our theories often break without separability. Hence it is
important to figure out when polynomials are separable, and when they are not.
Fortunately, we will end up with a result that tells us exactly when a polynomial
is not separable, and this is just a very small, specific class. In particular, in
fields of characteristic zero, all polynomials are separable.
Definition (Separable polynomial). Let
K
be a field,
f ∈ K
[
t
] non-zero, and
L
a splitting field of
f
. For an irreducible
f
, we say it is separable if
f
has no
repeated roots, i.e.
|Root
f
(
L
)
|
=
deg f
. For a general polynomial
f
, we say it is
separable if all its irreducible factors in K[t] are separable.
It should be obvious from definition that if
P
is separable and
Q | P
, then
Q
is also separable.
Note that some people instead define a separable polynomial to be one with
no repeated roots, so (
x−
2)
2
over
Q
would not be separable under this definition.
Example. Any linear polynomial t − a (with a ∈ K) is separable.
This is, however, not a very interesting example. To get to more interesting
examples, we need even more preparation.
Definition (Formal derivative). Let
K
be a field,
f ∈ K
[
t
]. (Formal) differenti-
ation the K-linear map K[t] → K[t] defined by t
n
7→ nt
n−1
.
The image of a polynomial f is the derivative of f , written f
′
.
This is similar to how we differentiate real or complex polynomials (in case
that isn’t obvious).
The following lemma summarizes the properties of the derivative we need.
Lemma. Let K be a field, f, g ∈ K[t]. Then
(i) (f + g)
′
= f
′
+ g
′
, (fg)
′
= fg
′
+ f
′
g.
(ii)
Assume
f
= 0 and
L
is a splitting field of
f
. Then
f
has a repeated root in
L
if and only if
f
and
f
′
have a common (non-constant) irreducible factor
in K[t] (if and only if f and f
′
have a common root in L).
This will allow us to show when irreducible polynomials are separable.
Proof.
(i) (f + g)
′
= f
′
+ g
′
is true by linearity.
To show that (
fg
)
′
=
fg
′
+
f
′
g
, we use linearity to reduce to the case
where
f
=
t
n
, g
=
t
m
. Then both sides are (
n
+
m
)
t
n+m−1
. So this holds.
(ii)
First assume that
f
has a repeated root. So let
f
= (
t −α
)
2
h ∈ L
[
t
] where
α ∈ L
. Then
f
′
= 2(
t − α
)
h
+ (
t − α
)
2
h
′
= (
t − α
)(2
h
+ (
t − α
)
h
′
). So
f
(
α
) =
f
′
(
α
) = 0. So
f
and
f
′
have common roots. However, we want a
common irreducible factor in
K
[
t
], not
L
[
t
]. So we let
P
α
be the minimal
polynomial of α over K. Then P
α
| f and P
α
| f
′
. So done.
Conversely, suppose
e
is a common irreducible factor of
f
and
f
′
in
K
[
t
],
with deg e > 0. Pick α ∈ Root
e
(L). Then α ∈ Root
f
(L) ∩Root
f
′
(L).
Since α is a root of f, we can write f = (t −α)q ∈ L[t] for some q. Then
f
′
= (t −α)q
′
+ q.
Since (t −α) | f
′
, we must have (t − α) | q. So (t −α)
2
| f.
Recall that the characteristic of a field
char K
is the minimum
p
such that
p ·
1
K
= 0. If no such
p
exists, we say
char K
= 0. For example,
Q
has
characteristic 0 while Z
p
has characteristic p.
Corollary. Let K be a field, f ∈ K[t] non-zero irreducible. Then
(i) If char K = 0, then f is separable.
(ii)
If
char K
=
p >
0, then
f
is not separable iff
deg f >
0 and
f ∈ K
[
t
p
]. For
example, t
2p
+ 3t
p
+ 1 is not separable.
Proof.
By definition, for irreducible
f
,
f
is not separable iff
f
has a repeated
root. So by our previous lemma,
f
is not separable if and only if
f
and
f
′
have a common irreducible factor of positive degree in
K
[
t
]. However, since
f
is
irreducible, its only factors are 1 and itself. So this can happen if and only if
f
′
= 0.
To make it more explicit, we can write
f = a
n
t
n
+ ··· + a
1
t + a
0
.
Then we can write
f
′
= na
n
t
n−1
+ ··· + a
1
.
Now f
′
= 0 if and only if all coefficients ia
i
= 0 for all i.
(i)
Suppose
char K
= 0, then if
deg f
= 0, then
f
is trivially separable. If
deg f >
0, then
f
is not separable iff
f
′
= 0 iff
ia
i
= 0 for all
i
iff
a
i
= 0
for
i ≥
1. But we cannot have a polynomial of positive degree with all its
coefficients zero (apart from the constant term). So f must be separable.
(ii) If deg f = 0, then f is trivially separable. So assume deg f > 0.
Then
f
is not separable
⇔ f
′
= 0
⇔ ia
i
= 0 for
i ≥
0
⇔ a
i
= 0 for all
i ≥ 1 not multiples of p ⇔ f ∈ K[t
p
].
Using this, it should be easy to find lots of examples of separable polynomials.
Definition (Separable elements and extensions). Let
K ⊆ L
be an algebraic
field extension. We say
α ∈ L
is separable over
K
if
P
α
is separable, where
P
α
is the minimal polynomial of α over K.
We say
L
is separable over
K
(or
K ⊆ L
is separable) if all
α ∈ L
are
separable.
Example.
–
The extensions
Q ⊆ Q
(
√
2
) and
R ⊆ C
are separable because
char Q
=
char R = 0. So we can apply our previous corollary.
–
Let
L
=
F
p
(
s
) be the field of rational functions in
s
over
F
p
(which is the
fraction field of
F
p
[
s
]), and
K
=
F
p
(
s
p
). We have
K ⊆ L
, and
L
=
K
(
s
).
Since
s
p
∈ K
,
s
is a root of
t
p
− s
p
∈ K
[
t
]. So
s
is algebraic over
K
and
hence
L
is algebraic over
K
. In fact
P
s
=
t
p
−s
p
is the minimal polynomial
of s over K.
Now
t
p
−s
p
= (
t−s
)
p
since the field has characteristic
p
. So
Root
t
p
−s
p
(
L
) =
{s}. So P
s
is not separable.
As mentioned in the beginning, separable extensions are nice, or at least
non-weird. One particular nice result about separable extensions is that all finite
separable extensions are simple, i.e. if
K ⊆ L
is finite separable, then
L
=
K
(
α
)
for some
α ∈ L
. This is what we will be working towards for the remaining of
the section.
Example. Consider
Q ⊆ Q
(
√
2,
√
3
). This is a separable finite extension. So
we should be able to generate
Q
(
√
2,
√
3
) by just one element, not just two. In
fact, we can use α =
√
2 +
√
3, since we have
α
3
= 11
√
2 + 9
√
3 = 2
√
2 + 9α.
So since α
3
∈ Q(α), we know that
√
2 ∈ Q(α). So we also have
√
3 ∈ Q(α).
In general, it is not easy to find an
α
that works, but we our later result will
show that such an α exists.
Before that, we will prove some results about the K-homomorphisms.
Lemma. Let
L/F/K
be finite extensions, and
E/K
be a field extension. Then
for all α ∈ L, we have
|Hom
K
(F (α), E)| ≤ [F (α) : F ]|Hom
K
(F, E)|.
Note that if
P
α
is the minimal polynomial of
α
over
F
, then [
F
(
α
) :
F
] =
deg P
α
. So we can interpret this intuitively as follows: for each
ψ ∈ Hom
K
(
F, E
),
we can obtain a
K
-homomorphism in
Hom
K
(
F
(
α
)
, E
) by sending things in
F
according to
ψ
, and then send
α
to any root of
P
α
. Then there are at
most [
F
(
α
) :
F
]
K
-homomorphisms generated this way. Moreover, each
K
-
homomorphism in
Hom
K
(
F
(
α
)
, E
) can be created this way. So we get this
result.
Proof.
We show that for each
ψ ∈ Hom
K
(
F, E
), there are at most [
F
(
α
) :
F
]
K
-isomorphisms in
Hom
K
(
F
(
α
)
, E
) that restrict to
ψ
in
F
. Since each
K
-
isomorphism in
Hom
K
(
F
(
α
)
, E
) has to restrict to something, it follows that
there are at most [
F
(
α
) :
F
]
|Hom
K
(
F, E
)
| K
-homomorphisms from
F
(
α
) to
E
.
Now let
P
α
be the minimal polynomial for
α
in
F
, and let
ψ ∈ Hom
K
(
F, E
).
To extend ψ to a morphism F (α) → E, we need to decide where to send α. So
there should be some sort of correspondence
Root
P
α
(E) ←→ {ϕ ∈ Hom
K
(F (α), E) : ϕ|
F
= ψ}.
Except that the previous sentence makes no sense, since
P
α
∈ F
[
t
] but we are
not told that F is a subfield of E. So we use our ψ to “move” our things to E.
We let
M
=
ψ
(
F
)
⊆ E
, and
q ∈ M
[
t
] be the image of
P
α
under the
homomorphism
F
[
t
]
→ M
[
t
] induced by
ψ
. As we have previously shown, there
is a one-to-one correspondence
Root
q
(E) ←→ Hom
M
(M[t]/⟨q⟩, E).
What we really want to show is the correspondence between
Root
q
(
E
) and the
K
-homomorphisms
F
[
t
]
/⟨P
α
⟩ → E
that restrict to
ψ
on
F
. Let’s ignore the
quotient for the moment and think: what does it mean for
ϕ ∈ Hom
K
(
F
[
t
]
, E
) to
restrict to
ψ
on
F
? We know that any
ϕ ∈ Hom
L
(
F
[
t
]
, E
) is uniquely determined
by the values it takes on
F
and
t
. Hence if
ϕ|
F
=
ψ
, then our
ϕ
must send
F
to
ψ
(
F
) =
M
, and can send
t
to anything in
E
. This corresponds exactly to
the
M
-homomorphisms
M
[
t
]
→ E
that does nothing to
M
and sends
t
to that
“anything” in E.
The situation does not change when we put back the quotient. Changing
from
M
[
t
]
→ E
to
M
[
t
]
/⟨q⟩ → E
just requires that the image of
t
must be
a root of
q
. On the other hand, using
F
[
t
]
/⟨P
α
⟩
instead of
F
[
t
] requires that
ϕ
(
P
α
(
t
)) = 0. But we know that
ϕ
(
P
α
) =
ψ
(
P
α
) =
q
. So this just requires
q(t) = 0 as well. So we get the one-to-one correspondence
Hom
M
(M[t]/⟨q⟩, E) ←→ {ϕ ∈ Hom
K
(F [t]/⟨P
α
⟩, E) : ϕ|
F
= ψ}.
Since F [t]/⟨P
α
⟩ = F (α), there is a one-to-one correspondence
Root
q
(E) ←→ {ϕ ∈ Hom
K
(F (α), E) : ϕ|
F
= ψ}.
So done.
Theorem. Let L/K and E/K be field extensions. Then
(i) |Hom
K
(L, E)| ≤ [L : K]. In particular, |Aut
K
(L)| ≤ [L : K].
(ii) If equality holds in (i), then for any intermediate field K ⊆ F ⊆ L:
(a) We also have |Hom
K
(F, E)| = [F : K].
(b) The map Hom
K
(L, E) → Hom
K
(F, E) by restriction is surjective.
Proof.
(i) We have previously shown we can find a sequence of field extensions
K = F
0
⊆ F
1
⊆ ··· ⊆ F
n
= L
such that for each
i
, there is some
α
i
such that
F
i
=
F
i−1
(
α
i
). Then by
our previous lemma, we have
|Hom
K
(L, E)| ≤ [F
n
: F
n−1
]|Hom
K
(F
n−1
, E)|
≤ [F
n
: F
n−1
][F
n−1
: F
n−2
]|Hom
K
(F
n−2
, E)|
.
.
.
≤ [F
n
: F
n−1
][F
n−1
: F
n−2
] ···[F
1
: F
0
]|Hom
K
(F
0
, E)|
= [F
n
: F
0
]
= [L : K]
(ii) (a)
If equality holds in (i), then every inequality in the proof above has
to an equality. Instead of directly decomposing
K ⊆ L
as a chain
above, we can first decompose
K ⊆ F
, then
F ⊆ L
, then join them
together. Then we can assume that F = F
i
for some i. Then we get
|Hom
K
(L, E)| = [L : F ]|Hom
K
(F, E)| = [L : K].
Then the tower law says
|Hom
K
(F, E)| = [F : K].
(b)
By the proof of the lemma, for each
ψ ∈ Hom
K
(
F, E
), we know that
{ϕ : Hom
K
(L, E) : ϕ|
F
= ψ} ≤ [L : F ]. (∗)
As we know that
|Hom
K
(F, E)| = [F : K], |Hom
K
(L, E)| = [L : K]
we must have had equality in (
∗
), or else we won’t have enough
elements. So in particular
{ϕ
:
Hom
K
(
L, E
) :
ϕ|
F
=
ψ} ≥
1. So the
map is surjective.
With this result, we can prove prove the following result characterizing
separable extensions.
Theorem. Let
L/K
be a finite field extension. Then the following are equivalent:
(i) There is some extension E of K such that |Hom
K
(L, E)| = [L : K].
(ii) L/K is separable.
(iii) L
=
K
(
α
1
, ··· , α
n
) such that
P
α
i
, the minimal polynomial of
α
i
over
K
,
is separable for all i.
(iv) L
=
K
(
α
1
, ··· , α
n
) such that
R
α
i
, the minimal polynomial of
α
i
over
K(α
1
, ··· , α
i−1
) is separable for all i.
Proof.
–
(i)
⇒
(ii): For all
α ∈ L
, if
P
α
is the minimal polynomial of
α
over
K
,
then since K(α) is a subfield of L, by our previous theorem, we have
|Hom
K
(K(α), E)| = [K(α) : K].
We also know that
|Root
P
α
(
E
)
|
=
|Hom
K
(
K
(
α
)
, E
)
|
, and that [
K
(
α
) :
K
] =
deg P
α
. So we know that
P
α
has no repeated roots in any splitting
field. So P
α
is a separable. So L/K is a separable extension.
– (ii) ⇒ (iii): Obvious from definition
–
(iii)
⇒
(iv): Since
R
α
i
is a minimal polynomial in
K
(
α
1
, ··· , α
i−1
), we
know that R
α
i
| P
α
i
. So R
α
i
is separable as P
α
i
is separable.
–
(iv)
⇒
(i): Let
E
be the splitting field of
P
α
1
, ··· , P
α
n
. We do induction
on
n
to show that this satisfies the properties we want. If
n
= 1, then
L = K(α
1
). Then we have
|Hom
K
(L, E)| = |Root
P
α
i
(E)| = deg P
α
1
= [K(α
1
) : K] = [L : K].
We now induct on
n
. So we can assume that (iv)
⇒
(i) holds for smaller
number of generators. For convenience, we write
K
i
=
K
(
α
1
, ··· , α
i
).
Then we have
|Hom
K
(K
n−1
, E)| = [K
n−1
: K].
We also know that
|Hom
K
(K
n
, E)| ≤ [K
n
: K
n−1
]|Hom
K
(K
n−1
, E)|.
What we actually want is equality. We now re-do (parts of) the proof of
this result, and see that separability guarantees that equality holds. If
we pick
ψ ∈ Hom
K
(
K
n−1
, E
), then there is a one-to-one correspondence
between
{ϕ ∈ Hom
K
(
K
n
, E
) :
ϕ|
K
n−1
=
ψ}
and
Root
q
(
E
), where
q ∈ M
[
t
]
is defined as the image of
R
α
n
under
K
n−1
[
t
]
→ M
[
t
], and
M
is the image
of ψ.
Since
P
α
n
∈ K
[
t
] and
R
α
n
| P
α
n
, then
q | P
α
n
. So
q
splits over
E
. By
separability assumption , we get that
|Root
q
(E)| = deg q = deg R
α
n
= [K
n
: K
n−1
].
Hence we know that
|Hom
K
(L, E)| = [K
n
: K
n−1
]|Hom
K
(K
n−1
, E)|
= [K
n
: K
n−1
][K
n−1
: K]
= [K
n
: K].
So done.
Before we finally get to the primitive element theorem, we prove the following
lemma. This will enable us to prove the trivial case of the primitive element
theorem, and will also be very useful later on.
Lemma. Let
L
be a field,
L
∗
=
L \{
0
}
be the multiplicative group of
L
. If
G
is a finite subgroup of L
∗
, then G is cyclic.
Proof.
Since
L
∗
is abelian,
G
is also abelian. Then by the structure theorem on
finite abelian groups,
G
∼
=
Z
⟨n
1
⟩
× ··· ×
Z
⟨n
r
⟩
,
for some
n
i
∈ N
. Let
m
be the least common multiple of
n
1
, ··· , n
r
, and let
f = t
m
− 1.
If α ∈ G, then α
m
= 1. So f(α) = 0 for all α ∈ G. Therefore
|G| = n
1
···n
r
≤ |Root
f
(L)| ≤ deg f = m.
Since
m
is the least common multiple of
n
1
, ··· , n
r
, we must have
m
=
n
1
···n
r
and thus (
n
i
, n
j
) = 1 for all
i
=
j
. Then by the Chinese remainder theorem, we
have
G
∼
=
Z
⟨n
1
⟩
× ··· ×
Z
⟨n
r
⟩
=
Z
⟨n
1
···n
r
⟩
.
So G is cyclic.
We now come to the main theorem of the lecture:
Theorem (Primitive element theorem). Assume
L/K
is a finite and separable
extension. Then L/K is simple, i.e. there is some α ∈ L such that L = K(α).
Proof.
At some point in our proof, we will require that
L
is infinite. So we
first do the finite case first. If
K
is finite, then
L
is also finite, which in turns
implies
L
∗
is finite too. So by the lemma,
L
∗
is a cyclic group (since it is a finite
subgroup of itself). So there is some
α ∈ L
∗
such that every element in
L
∗
is a
power of α. So L = K(α).
So focus on the case where
K
is infinite. Also, assume
K
=
L
. Then since
L/K
is a finite extension, there is some intermediate field
K ⊆ F ⊊ L
such that
L
=
F
(
β
) for some
β
. Now
L/K
is separable. So
F/K
is also separable, and
[
F
:
K
]
<
[
L
:
K
]. Then by induction on degree of extension, we can assume
F/K
is simple. In other words, there is some
λ ∈ F
such that
F
=
K
(
λ
). Now
L = K(λ, β). In the rest of the proof, we will try to replace the two generators
λ, β with just a single generator.
Unsurprisingly, the generator of
L
will be chosen to be a linear combination
of β and λ. We set
α = β + aλ
for some
a ∈ K
to be chosen later. We will show that
K
(
α
) =
L
. Actually,
almost any choice of
a
will do, but at the end of the proof, we will see which
ones are the bad ones.
Let
P
β
and
P
λ
be the minimal polynomial of
β
and
λ
over
K
respectively.
Consider the polynomial f = P
β
(α −at) ∈ K(α)[t]. Then we have
f(λ) = P
β
(α −aλ) = P
β
(β) = 0.
On the other hand, P
λ
(λ) = 0. So λ is a common root of P
λ
and f.
We now want to pick an
a
such that
λ
is the only common root of
f
and
P
λ
(in
E
). If so, then the gcd of
f
and
P
α
in
K
(
α
) must only have
λ
as a root.
But since
P
λ
is separable, it has no double roots. So the gcd must be
t −λ
. In
particular, we must have
λ ∈ K
(
α
). Since
α
=
β
+
aλ
, it follows that
β ∈ K
(
α
)
as well, and so K(α) = L.
Thus, it remains to choose an
a
such that there are no other common roots.
We work in a splitting field of P
β
P
λ
, and write
P
β
= (t −β
1
) ···(t −β
m
)
P
λ
= (t −λ
1
) ···(t −λ
n
).
We wlog β
1
= β and λ
1
= λ.
Now suppose θ is a common root of f and P
λ
. Then
(
f(θ) = 0
P
λ
(θ) = 0
⇒
(
P
β
(α −aθ) = 0
P
λ
(θ) = 0
⇒
(
α −aθ = β
i
θ = λ
j
for some i, j. Then we know that
α = β
i
+ aλ
j
.
However, by definition, we also know that
α = β + aλ
Now we see how we need to choose
a
. We need to choose
a
such that the elements
β + aλ = β
i
+ aλ
j
for all i, j. But if they were equal, then we have
a =
λ −λ
j
β
i
− β
,
and there are only finitely many elements of this form. So we just have to pick
an a not in this list.
Corollary. Any finite extension
L/K
of field of characteristic 0 is simple, i.e.
L = K(α) for some α ∈ L.
Proof.
This follows from the fact that all extensions of fields of characteristic
zero are separable.
We have previously seen that
Q
(
√
2,
√
3
)
/Q
is a simple extension, but that
is of course true from this theorem. A more interesting example would be one in
which this fails. We will need a field with non-zero characteristic.
Example. Let
L
=
F
p
(
s, u
), the fraction field of
F
p
[
s, u
]. Let
K
=
F
p
(
s
p
, u
p
).
We have L/K. We want to show this is not simple.
If
α ∈ L
, then
α
p
∈ K
. So
α
is a root of
t
p
− α
p
∈ K
[
t
]. Thus the minimal
polynomial
P
α
has degree at most
p
. So [
K
(
α
) :
K
] =
deg P
α
≤ p
. On the other
hand, we have [
L
:
K
] =
p
2
, since
{s
i
u
j
: 0
≤ i, j < p}
is a basis. So for any
α
,
we have
K
(
α
)
=
L
. So
L/K
is not a simple extension. This then implies
L/K
is not separable.
At this point, one might suspect that all fields with positive characteristic
are not separable. This is not true by considering a rather silly example.
Example. Consider
K
=
F
2
and
L
=
F
2
[
s
]
/⟨s
2
+
s
+ 1
⟩
. We can check manually
that
s
2
+
s
+ 1 has no roots and hence irreducible. So
L
is a field. So
L/F
2
is a
finite extension. Note that L only has 4 elements.
Now if
α ∈ L \ F
2
, and
P
α
is the minimal polynomial of
α
over
F
2
, then
P
α
| t
2
+ t + 1. So P
α
is separable as a polynomial. So L/F
2
is separable.
In fact, we have
Proposition. Let
L/K
be an extension of finite fields. Then the extension is
separable.
Proof.
Let the characteristic of the fields be
p
. Suppose the extension were not
separable. Then there is some non-separable element
α ∈ L
. Then its minimal
polynomial must be of the form P
α
=
P
a
i
t
pi
.
Now note that the map
K → K
given by
x 7→ x
p
is injective, hence surjective.
So we can write a
i
= b
p
i
for all i. Then we have
P
α
=
X
a
i
t
pi
=
X
b
i
t
i
p
,
and so P
α
is not irreducible, which is a contradiction.