Galois Theory:
On infinite extensions
We assume that Zorn’s Lemma (stated below) is true. Note that Zorn’s Lemma
is equivalent to the Axiom of Choice, which is controversial among some mathematicians. However, most mathematicians assume Zorn’s Lemma.
Zorn’s Lemma: Suppose X is a nonempty, partially ordered set. A chain C of
X is a collection of elements {ai }i∈I of X so that for every i, j ∈ I, either ai ≤ aj
or aj ≤ ai . Suppose that every nonempty chain C of X has an upper bound in X;
then X has a maximal element.
Definitions. A field K is called algebraically closed if every f ∈ K[t] r K has a
root in K (and hence splits over K). An algebraic closure of a field K is a field K
which is algebraic over K and is algebraically closed.
Lemma. There is an algebraic extension E : K which contains a root of every
irreducible f ∈ K[t] (and hence for every g ∈ K[t] r K).
Proof. Let {qi }i∈I be the set of all irreducible polynomials over K (I some indexing
set). Consider R = K[{ti }i∈I ]. Let A be the ideal of R generated by {qi }i∈I . We
claim A 6= R: For the sake of contradiction, suppose A = R. So 1 ∈ A, and hence
1=
X
uj qj (tj )
j∈J
for some finite set J, J ⊆ I, with uj ∈ R. We can construct an extension E : K
so that for all j ∈ J, qj has a root αj ∈ E. Then we can define a homomorphism
ϕ : R → E so that ϕ is the identity map on K, ϕ(tj ) = αj for all j ∈ J, and
ϕ(ti ) = 0 for all i ∈ I r J. Then
1 = ϕ(1) =
X
ϕ(uj )ϕ(qj )(αj ) =
j∈J
X
ϕ(uj )qj (αj ) = 0,
j∈J
a contradiction. So A 6= R.
Let B be a maximal ideal of R so that A ⊆ B (such an ideal exists by Zorn’s
Lemma). Set E = R/B. So E : K is a field extension. Define ψ : K → E by
ψ(c) = c + B, and identify c with ψ(c). Let αi = ti + B for all i ∈ I. Define
σ : R → E by σ(u) = u + B. So σ is a surjective homomorphism, and σ(ti ) = αi
for all i ∈ I. Hence for all i ∈ I,
qi (αi ) = σ(qi (ti )) = qi (ti ) + B = 0 + B
since qi (ti ) ∈ A ⊆ B. Therefore every qi has a root in E. Also, each αi is algebraic
over K, so E = K[{αi }i∈I ] is an algebraic extension of K. 1
2
Theorem. For K a field, there is an algebraic extension K of K so that K is
algebraically closed.
Proof. We construct a sequence of fields K = E0 ⊆ E1 ⊆ · · · ⊆ En−1 ⊆ En ⊆ · · ·
inductively: For n ∈ Z+ , En is an algebraic extension of En−1 containing a root of
every f ∈ En−1 [t] r En−1 . So each En is algebraic over K. Hence K = ∪n∈Z+ En
is algebraic over K. Suppose f ∈ K[t] r K. Since f has finitely many nonzero
coefficients, f ∈ En−1 [t] for some n ∈ Z+ . Therefore f has a root in En ⊆ K. So
K is algebraically closed. Theorem. Let E be an algebraic extension of K, and suppose K ⊆ E ⊆ K. Given
a homomorphism ϕ : K → K, ϕ can be extended to a homomorphism from E into
K.
Proof. Let S be the set of all pairs (F, ψ) where F is a field with K ⊆ F ⊆ E, and
ψ : F → K is a homomorphism extending ϕ. Since (K, ϕ) ∈ S, we have S 6= ∅.
We partially order S by defining (F1 , ψ1 ) ≤ (F2 , ψ2 ) if F1 ⊆ F2 and ψ2 extends
ψ1 . Suppose {(Fi , ψ)}i∈I is a chain in S. Set F = ∪i∈I Fi . So F is a subfield of E
(check!). Define ψ : F → K by ψ(α) = ψj (α) where j ∈ I so that α ∈ Fj . Then ψ
is a homomorphism extending ψi for all i ∈ I (check!). So every nonempty chain in
S has an upper bound in S. Thus by Zorn’s Lemma, S contains a maximal element
(M, µ). Suppose M ( E. Take α ∈ E rM . Then α is algebraic over K and hence α
is algebraic over M , so we can extend µ to a homomorphism ν : M (α) → K, giving
us (M (α), ν) ∈ S, and thereby contradicting that (M, µ) is a maximal element of
S. For your amusement (and your well-rounded education), here are some more
results on extensions that are possibly infinite extensions.
Proposition. An extension L : K is algebraic and normal if and only if it is a
splitting field extension for some S ⊆ K[t].
Proof. First, for S ⊆ K[t], let A ⊆ K be the set of all roots of the polynomials in
S. Then K(A) denotes the smallest subfield of K containing K and A.
Suppose L : K is algebraic and normal. Let S = {mα (K) : α ∈ L }. Then L : K
is a splitting field extension for S.
Now suppose L : K is a splitting field extension for S ⊆ K[t]. So L : K is an
algebraic extension. Suppose f ∈ K[t] is irreducible over K and has a root in α in
L. With A the set of roots of S, we have that
Q L = K(A) is algebraic, and α ∈ K(B)
where B is a finite subset of A. Let h = β∈B mβ (K). Therefore K(B) : K is a
splitting field extension for h, and hence K(B) : K is normal. Since α ∈ K(B),
we have that f splits over K(B). We have K(B) ⊆ K(A) = L, so f splits over L.
Thus L : K is normal. Definition. An extension L : K is a Galois extension if it is an algebraic extension
and K = LGal(L:K) .
3
Proposition. Suppose L : K is an algebraic, normal, separable extension. The
LGal(L:K) = K.
Proof. Assume K ⊆ L.
First suppose that L : K is an algebraic, normal, separable extension. Since
L : K is algebraic and normal, it is a splitting field extension for some S ⊆ K[t].
Thus L lies in an algebraic closure of K, so we can assume K ⊆ L ⊆ K. [The
argument now proceeds just as in the case that [L : K] < ∞.] Set G = Gal(L : K).
So K ⊆ LG . Take α ∈ LG . Let β ∈ L be a root of mα (K). We know there is some
ϕ ∈ G so that ϕ(α) = β. But α ∈ LG , so ϕ(α) = α. Since L : K is normal, mα (K)
splits over L; thus α is the only root of mα (K). Hence mα (K) = (t − α)r for some
r ∈ Z+ . But L : K is separable, so mα (K) has no multiple roots. Thus r = 1 and
α ∈ K. Hence LG ⊆ K, and consequently LG = K. Proposition. Suppose L : K is an algebraic extension with K = LGal(L:K) . Then
L : K is normal and separable.
Proof. [This proof is almost identical to the proof in the case that [L : K] < ∞.]
Let G = Gal(L : K). For α ∈ L, let α, α2 , . . . , αr be the distinct elements in the
G-orbit of α. Set
fα = (t − α)(t − α2 ) · · · (t − αr ).
Then fα is fixed by G (since G permutes the roots of fα ), so fα ∈ LG [t] = K[t]. By
construction, fα is separable. Since fα ∈ K[t] and α is a root of fα , we know that
mα (K) divides fα , and hence mα (K) is separable. [In fact, mα (K) = fα , since for
i = 2, . . . , r, there is some ϕ ∈ G so that ϕ(α) = αi , so αi is a root of mαi (K) for
i = 2, . . . , r.] Therefore S = {mα (K) : α ∈ L } is a set of separable polynomials
from K[t], and L : K is a splitting field extension for S. Hence L : K is normal and
separable. Corollary. Suppose L : K is an algebraic extension. Then K = LGal(L:K) if and
only if L : K is a normal, separable extension.
Proposition. Let K be a field. The following are equivalent:
(1) Every f ∈ K[t] r K has a root in K.
(2) Every irreducible element of K[t] has degree 1.
(3) The only aglebraic extension of K is itself.
Proof. To show (1) implies (2): Suppose f ∈ K[t] is irreducible over K. Then there
is some α ∈ K so that f (α) = 0. Thus f = (t − α)h for some h ∈ K[t]. Since f is
irreducible over K, h must be a unit in K[t], meaning h ∈ K × . So deg f = 1.
To show (2) implies (3): Let α be algebraic over K (so α ∈ L for some extension
L of K). Then mα (K) is monic and irreducible over K, hence deg f = 1. Thus
mα (K) = t − α, and hence α ∈ K.
To show (3) implies (2): Let f ∈ K[t] be irreducible over K, and let L = K[t]/(f ).
So L : K is a field extension with [L : K] = deg f < ∞. Hence L : K is an algebraic
extention, so [L : K] = 1. Hence deg f = 1.
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To show (2) implies (1): Take f ∈ K[t] r K, and let h ∈ K[t] be an irreducible
factor of f . So deg h = 1, and hence h = λt − β, λ, β ∈ K with λ =
6 0. Thus
−1
λ β ∈ K is a root of h and hence a root of f . Corollary. Let K be a field. Then K is a maximal algebraic extension of K.
Propostion. Suppose L, M are algebraic closures of K. Then L ' M .
Proof. Identify K with its isomorphic image in L (so we assume K ⊆ L). By the
previous theorem, the inclusion map ϕ : K → M extends to a K-homomorphism
ψ : L → M . Since L is a field, we know ψ must be injective. Also, M is algebraic
over K. Since ψ(L) ' L and L is algebraically closed, we must have that ψ(L)
is algebraically closed (check!). By the preceding proposition, the only algebraic
extension of ψ(L) is ψ(L). But M : ψ(L) is an algebraic extension as M : K is an
algebraic extension, so we must have M = ψ(L).