2Field extensions
II Galois Theory
2.5 Algebraic closures
The splitting field gives us the field with the root of one particular polynomial.
We could be greedy and ask for the roots for all polynomials, and get the
algebraic closure. The algebraic closure will not be of much use in this course,
but is a nice thing to know about. The major theorems would be the existence
and uniqueness of algebraic closures.
Definition (Algebraically closed field). A field
L
is algebraically closed if for all
f ∈ L[t], we have
f = a(t − α
1
)(t − α
2
) ···(t − α
n
)
for some a, α
i
∈ L. In other words, L contains all roots of its polynomials.
Let L/K be a field extension. We say L is an algebraic closure of K if
– L is algebraic over K
– L is algebraically closed.
Example.
L
is an algebraically closed field iff (
L ⊆ E
is a finite extension
implies E = L).
This is since if
L ⊆ E
is finite, then
E
is algebraic over
L
, and hence must
be L.
Example.
C
is algebraically closed by the fundamental theorem of algebra, and
is the algebraic closure of R (but not Q).
Before we prove our next theorem, we need the following technical lemma:
Lemma. If
R
is a commutative ring, then it has a maximal ideal. In particular,
if I is an ideal of R, then there is a maximal ideal that contains I.
Proof. Let
P = {I : I is an ideal of R, I = R}.
If
I
1
⊆ I
2
⊆ ···
is any chain of
I
i
∈ P
, then
I
=
S
I
i
∈ P
. By Zorn’s lemma,
there is a maximal element of
P
(containing
I
). So
R
has at least one maximal
ideal (containing I).
Theorem (Existence of algebraic closure). Any field
K
has an algebraic closure.
Proof. Let
A = {λ = (f, j) : f ∈ K[t] irreducible monic, 1 ≤ j ≤ deg f }.
We can think of
j
as labelling which root of
f
we want. For each
λ ∈ A
, we
assign a variable t
λ
. We take
R = K[t
λ
: λ ∈ A]
to be the polynomial ring over
K
with variables
t
λ
. This
R
contains all the
“roots” of the polynomials in
K
. However, we’ve got a bit too much. For example,
(if
K
=
Q
), in
R
,
√
3
and
√
3
+ 1 would be put down as separate, unrelated
variables. So we want to quotient this R by something.
For every monic and irreducible f ∈ K[t], we define
˜
f = f −
deg f
Y
j=1
(t − t
(f,j)
) ∈ R[t].
If we want the
t
(f,j)
to be roots of
f
, then
˜
f
should vanish for all
f
. Denote the
coefficient of t
ℓ
in
˜
f by b
(f,ℓ)
. Then we want b
(f,ℓ)
= 0 for all f, ℓ.
To do so, let
I ⊆ R
be the ideal generated by all such coefficients. We now
want to quotient R by I. We first have to check that I = R.
Suppose not. So there are
b
(f
1
,ℓ
1
)
, ··· , b
(f
r
,ℓ
r
)
with
g
1
, ··· , g
r
∈ R
such that
g
1
b
(f
1
,ℓ
1
)
+ ··· + g
r
b
(f
r
,ℓ
r
)
= 1. (∗)
We will attempt to reach a contradiction by constructing a homomorphism
ϕ
that sends each b
(f
i
,ℓ
i
)
to 0.
Let E be a splitting field of f
1
f
2
···f
r
. So in E[t], for each i, we can write
f
i
=
deg f
i
Y
j=1
(t − α
i,j
).
Then we define a homomorphism ϕ : R → E by
(
ϕ(t
(f
i
,j)
) = α
i,j
ϕ(t
λ
) = 0 otherwise
This induces a homomorphism
˜
ϕ : R[t] → E[t].
Now apply
˜
ϕ(
˜
f
i
) =
˜
ϕ(f
i
) −
deg f
i
Y
j=1
˜
ϕ(t − t
(f
i
,j)
)
= f
i
−
deg f
i
Y
j=1
(t − α
i,j
)
= 0
So ϕ(b
(f
i
,ℓ
i
)
) = 0 as b
(f
i
,ℓ
i
)
is a coefficient of f
i
.
Now we apply ϕ to (∗) to obtain
ϕ(g
1
b
(f
1
,ℓ
1
)
+ ··· + g
r
b
(f
r
,ℓ
r
)
) = ϕ(1).
But this is a contradiction since the left had side is 0 while the right is 1. Hence
we must have I = R.
We would like to quotient by
I
, but we have to be a bit more careful, since
the quotient need not be a field. Instead, pick a maximal ideal
M
containing
I
, and consider
L
=
R/M
. Then
L
is a field. Moreover, since we couldn’t
have quotiented out anything in
K
(any ideal containing anything in
K
would
automatically contain all of
R
), this is a field extension
L/K
. We want to show
that L is an algebraic closure.
Now we show that
L
is algebraic over
K
. This should all work out smoothly,
since that’s how we constructed
L
. First we pick
α ∈ L
. Since
L
=
R/M
and
R
is generated by the terms t
λ
, there is some (f
1
, j
1
), ··· , (f
r
, j
r
) such that
α ∈ K(
¯
t
(f
i
,j
i
)
, ··· ,
¯
t
(f
r
,j
r
)
).
So
α
is algebraic over
K
if each
¯
t
(f
i
,j
i
)
is algebraic over
K
. To show this, note
that
˜
f
i
= 0, since we’ve quotiented out each of its coefficients. So by definition,
0 = f
i
(t) −
deg f
i
Y
j=1
(t −
¯
t
(f
i
,j)
).
So f
i
(
¯
t
(f
i
,j
i
)
) = 0. So done.
Finally, we have to show that
L
is algebraically closed. Suppose
L ⊆ E
is a
finite (and hence algebraic) extension. We want to show that L = E.
Consider arbitrary
β ∈ E
. Then
β
is algebraic over
L
, say a root of
f ∈ L
[
t
]. Since every coefficient of
f
can be found in some finite extension
K
(
¯
t
(f
i
,j
i
)
, ··· ,
¯
t
(f
r
,j
r
)
), there is a finite extension
F
of
K
that contains all coeffi-
cients of
f
. Since
F
(
β
) is a finite extension of
F
, we know
F
(
β
) is a finite and
hence algebraic extension of K. In particular, β is algebraic in K.
Let
P
β
be the minimal polynomial of
β
over
K
. Since all polynomials in
K
split over
L
by construction (
f
(
t
) =
Q
(
t −
¯
t
(f,j)
)), its roots must in
L
. In
particular, β ∈ L. So L = E.
Theorem (Uniqueness of algebraic closure). Any field
K
has a unique algebraic
closure up to K-isomorphism.
This is the same proof as the proof that the splitting field is unique — given
two algebraic closures, we take the largest subfield of the algebraic closures that
biject with each other. However, since there could be infinitely many subfields,
we have to apply Zorn’s lemma to obtain the maximal such subfield.
Proof. (sketch) Suppose L, L
′
are both algebraic closures of K. Let
H = {(F, ψ) : K ⊆ F ⊆ L, ψ ∈ Hom
K
(F, L
′
)}.
We define a partial order on
H
by (
F
1
, ψ
1
)
≤
(
F
2
, ψ
2
) if
F
1
≤ F
2
and
ψ
1
=
ψ
2
|
F
1
.
We have to show that chains have upper bounds. Given a chain
{
(
F
α
, ψ
α
)
}
,
we define
F =
[
F
α
, ψ(x) = ψ
α
(x) for x ∈ F
α
.
Then (
F, ψ
)
∈ H
. Then applying Zorn’s lemma, there is a maximal element of
H, say (F, ψ).
Finally, we have to prove that
F
=
L
, and that
ψ
(
L
) =
L
′
. Suppose
F
=
L
.
Then we attempt to produce a larger
˜
F
and a
K
-isomorphism
˜
F →
˜
F
′
⊆ L
′
.
Since
F
=
L
, there is some
α ∈ L \ F
. Since
L
is an algebraic extension of
K
,
there is some irreducible g ∈ K[t] such that deg g > 0 and g(α) = 0.
Now there is an isomorphism
F
[
t
]
/⟨g⟩ → F
(
α
) defined by
¯
t 7→ α
. The
isomorphism
ψ
:
F → F
′
then extends to an isomorphism
µ
:
F
[
t
]
→ F
′
[
t
]
and thus to
F
[
t
]
/⟨g⟩ → F
′
[
t
]
/⟨µ
(
g
)
⟩
. Then if
α
′
is a root of
µ
(
g
), then we have
F
′
[
t
]
//⟨µ
(
g
)
⟩
∼
=
F
′
(
α
′
). So this gives an isomorphism
F
(
α
)
→ F
(
α
′
). This
contradicts the maximality of ϕ.
By doing the argument the other way round, we must have
ψ
(
L
) =
L
′
. So
done.