Galois Theory
UW-Madison
Galois Group
Definition. If E/F is an extension of fields, let Gal(E/F ) be the set of isomorphisms φ : E → E
that fix F .
Remark. Note that if K is a field between F and E then Gal(E/K) is a subgroup of Gal(E/F ).
Definition. If E/F is an extension of fields, let
Fix(Gal(E/F )) = {α ∈ E : σ(α) = α ∀σ ∈ Gal(E/F )}.
It is not hard to see that Fix(Gal(E/F )) is a subfield of E. We say that E/F is Galois if |E : F | < ∞
and F = Fix(Gal(E/F )).
Lemma. Let E/F be a finite field extension and let K = Fix(Gal(E/F )). Then, Gal(E/K) =
Gal(E/F ). In particular E/K is a Galois extension.
Proof. Gal(E/K) ⊆ Gal(E/F ) since F ⊆ K. On the other hand Gal(E/F ) ⊆ Gal(E/K) by definition
of K.
Galois extensions are the “nice” extensions. It is for these extensions that we have the most information.
Definition. An extension E/F is normal if for all α ∈ E, α is algebraic over F and minF (α) splits
in E.
In simple words an extension is normal if it is closed for conjugates.
Theorem (Isaacs, 18.4). Let E/F be a field extension with |E : F | < ∞. Then, E is normal over
F if and only if E is the splitting field of some polynomial g ∈ F [x].
Definition. An extension E/F is separable if for all α ∈ E algebraic over F , minF (α) has distinct
roots in some splitting field.
One simple criteria to determine if a polynomial has distinct roots is the following.
Theorem (Isaacs, 19.4). If f ∈ F [x] is a non-zero polynomial, then f (x) has distinct roots (in a
splitting field) if and only if f (x) and f 0 (x) are relatively prime polynomials.
It is always important to deal with the irreducible cases. The following two corollaries are really useful
when dealing with problems over fields of positive characteristic.
0
Corollary (Isaacs, 19.5). Let f ∈ F [x] irreducible. Then f has distinct roots iff f is a non zero
polynomial.
Corollary (Isaacs, 19.9). Let F have prime characteristic p and let f ∈ F [x] be irreducible. Then
e
f (x) = g(xp ) for some integer e ≥ 0 and some irreducible, separable polynomial g ∈ F [x].
Inseparable extensions are “rare,” and the following theorem makes this more precise.
Theorem (Isaacs, 18.11). If E/F is not separable, then F is infinite and has prime characteristic.
One nice thing about separable extensions it that they are generated by a single element.
Theorem (Primitive Element. Isaacs, 18.17). If E/F is a separable and |E : F | < ∞ then there
is some α ∈ E such that E = F [α].
Let
Finally, this theorem gives a useful characterization of Galois extensions.
Theorem (Isaacs, 18.13). Let F ⊆ E with |E : F | < ∞. Then the following are equivalent.
1. E is a Galois extension of F .
2. E is both separable and normal over F .
3. E is the splitting field over F for some polynomial f ∈ F [x] with distinct roots.
Remark. From this it follows that if E/F is Galois and F ⊆ K ⊆ E is an intermediate field, then
E/K is Galois.
Finally, there is one more useful fact I will give about Galois extensions.
Lemma (Isaacs, 18.3). Let E/F be a field extension. Let f ∈ F [x] be an irreducible polynomial
whose splitting field is E and let Ω = {α ∈ E : f (α) = 0}. Then, Gal(E/F ) acts transitively on Ω
with trivial kernel.
Sometimes it is useful to know what are the transitive subgroups of Sn so we can decide what is the
Galois group of an extension.
The examples along this lines to have in mind are S3 and S4 . The only transitive subgroups of S4 , up
to isomorphism, are: S4 , A4 , D8 and any group of order 4. For S3 we have just A3 and S3 .
Example: Let f (x) = x3 − 5. If K is the splitting field of f over Q, then G = Gal(K/Q) ∼
= S3 .
Short proof: By Eisenstein’s and
√ the above lemma we know that G is a transitive subgroup of S3 .
One the other hand K = Q(ω, 3 5) where ω is a primitive cubic root of 1. From this we see that K/Q
is an extension of degree six, so G can’t be A3 hence it is S3 .
A detailed proof of this describing the action of generators of Gal(K/Q) on K is the following.
Proof. From Eisenstein’s criterion (with the prime 5), f (x) is irreducible over Q. Let G = Gal(K/Q).
Then, from Lemma 18.3 applied to f (x), G acts transitively on the roots with trivial kernel.
√
√
√
√
The roots of f (x) are the cube roots of 5 in C. These are 3 5, α 3 5 and α2 3 5 where 3 5 is a real cube
root of 5, and α is a primitive third root of unity. Now, α3 = 1 so α is a root of x3 −1 = (x−1)(x2 +x+1).
This latter quadratic factor is irreducible. Since α 6= 1, it follows that minQ (α) = x2 + x + 1 and hence
√
−1 + −3
α=
2
(for√some choice
of the square root). Thus, Q[α] is a degree two subfield of Q. Further, since α 6∈ R,
√
3
3
Q[ 5, α] > Q[ 5]. Thus,
√
3
K = Q[ 5, α]
and
√
√
√
3
3
3
|K : Q| = |Q[ 5, α] : Q[ 5]||Q[ 5] : Q| = 3 · 2 = 6.
√
√
Since G acts transitively on the√roots of x3 − 5 there is a ρ ∈ G with ρ( 3 5) = α 3 5. The action of ρ
is determined by its action on 3 5 and α. We don’t know yet whether ρ(α) = α or its conjugate.
√
Let L = Q[ 3 5]. As remarked above, Gal(K/L) is a subgroup
of Gal(K/Q). As shown above,
√
3
|K : L| = 2 and hence x2 + x + 1 is irreducible
in
Q[
5][x].
Again,
by Lemma 18.3, there is
√
−1− −3
2
=α .
τ ∈ Gal(K/L) ⊆ Gal(K/Q) so that τ (α) =
2
√
√
If ρ(α) = α, let σ = ρ. Then, σ fixes α and maps 3 5 to α 3 5.
Otherwise, let σ = τ ρ. Then,
√
√
3
3
( 5)σ = ( 5)τ ρ
√
√
3
3
= ( 5)ρ = α 5
(α)σ = (α)τ ρ
= (α2 )ρ = (α)ρ2 = α4 = α.
Thus, σ ∈ Gal(K/Q) maps
√
3
√
5 to α 3 5 and fixes α.
√
√
3 − 5 as x = 3 5, x = α 3 5 and
From this
information,
we
can
see
that
if
we
label
the
roots
of
x
1
2
√
x3 = α2 3 5 then σ induces the permutation (1, 2, 3) and τ induces the permutation (2, 3). It is easy to
see that these generate the group S3 . From Lemma 18.3, we know that if g ∈ Gal(K/Q) fixes x1 , x2
and x3 , then g = 1 and hence it follows that Gal(K/Q) ∼
= S3 .
Fundamental Theorem of Galois Theory
Now I will state the fundamental theorem of Galois theory.
Theorem (FTGT. Isaacs, 18.21). Let E/F be a Galois extension. Let
F = {K|F ⊆ K ⊆ E}
be the set of intermediate fields K between F and E. Let
G = {H|H ⊆ G}
be the set of subgroups of G = Gal(E/F ). Define f : G → F by f (H) = Fix(H) and g : F → G by
g(K) = Gal(E/K). Then, the following are true.
1. The maps f and g are inverse bijections. These maps reverse containment.
2. If g(K) = H then |E : K| = |H| and |K : F | = |G : H|. In particular, |E : F | = |G|.
3. If g(K) = H and σ ∈ G, then g(σ(K)) = H σ , the conjugate of H by σ. Also, H E G if and
only if K is Galois over F . In this case, Gal(K/F ) ∼
= G/H.
Let us consider the example from last time of K the splitting field of x3 − 5 over Q.
Here is a table of subgroups of G and the fixed fields.
Subgroup
Order
Fixed Field
1
hτ i
hστ i
hσ 2 τ i
hσi
hσ, τ i
1
2
2
2
3
6
√
3
K = Q[
5, α]
√
Q[ 3√5]
3
Q[α √
5]
Q[α2 3 5]
Q[α]
Q.
[Draw lattice of subgroups of S3 and lattice of subfields of K].
1
Example I
Before I give the next example, here is a definition.
Definition. Suppose that E/F is an extension of fields and K1 and K2 are fields with F ⊆ K1 ⊆ E
and F ⊆ K2 ⊆ E. The compositum of K1 and K2 , denoted hK1 , K2 i, is the smallest subfield of E
that contains K1 and K2 .
Here is a more general result about Galois extensions, and is the sort of thing one can prove fairly
easily using the fundamental theorem.
Proposition. Suppose that E/F is a Galois extension and K1 and K2 are subfields of E that are
Galois over F . Suppose that K1 ∩ K2 = F and hK1 , K2 i = E. Then,
Gal(E/F ) ∼
= Gal(K1 /F ) × Gal(K2 /F ).
Proof. Let G = Gal(E/F ). Let N1 = Gal(E/K1 ) and N2 = Gal(E/K2 ). Since K1 and K2 are Galois
over F , by the fundamental theorem, N1 and N2 are normal in G. Moreover, G/N1 ∼
= Gal(K1 /F )
and G/N2 ∼
= Gal(K2 /F ).
Since N1 and N2 are normal, K = N1 N2 is a subgroup of G. Then, Fix(K) ⊆ Fix(N1 ) and Fix(K) ⊆
Fix(N2 ). Thus, Fix(K) ⊆ Fix(N1 ) ∩ Fix(N2 ) = K1 ∩ K2 = F . Thus, Fix(K) = F and by the
correspondence, it follows that K = G.
Let H = N1 ∩ N2 . Then, since H ⊆ N1 , Fix(H) ⊇ Fix(N1 ) = K1 and since H ⊆ N2 , Fix(H) ⊇
Fix(N2 ) = K2 . Thus, Fix(H) ⊇ hK1 , K2 i = E. Thus, Fix(H) = E and H = 1.
Thus, G = N1 N2 and N1 ∩ N2 = 1. Hence, G ∼
= N1 × N2 . Note that G/N1 ∼
= N2 and G/N2 ∼
= N1 .
Thus,
G∼
= G/N1 × G/N2 ∼
= Gal(K1 /F ) × Gal(K2 /F ),
as desired.
2
Example II
Lemma 18.3 that we have used frequently states that if K is the splitting field of a separable, irreducible
polynomial f (x) of degree n over F , then Gal(K/F ) acts transitively on the roots of f (x), and is hence
isomorphic to a transitive subgroup of Sn . We will use the fundamental theorem to prove the converse.
Theorem. Suppose that E/F is a Galois extension and Gal(E/F ) is isomorphic to a transitive
subgroup of Sn . Then, E is the splitting field of some irreducible f (x) ∈ F [x] with deg f = n.
Proof. Let G = Gal(E/F ) and consider G as a subgroup of Sn that is transitive. Let H ⊆ G be the
stabilizer of 1 in G. We will show that |G : H| = n and coreG (H) = 1.
Since the size of the orbit of 1 is n, the orbit-stabilizer theorem shows that |G : H| = n.
Suppose that k ∈ {1, . . . , n} and gk ∈ G is an element that maps k to 1. Then, g ∈ G fixes k if and
only if ggk maps k to 1. This occurs if and only if gk−1 ggk fixes 1. Hence, the stabilizer of k is gk Hgk−1 .
Thus, we have
coreG (H) =
\
g∈G
g
−1
Hg ⊆
k
\
gi Hgi−1 .
i=1
Now, gi Hgi−1 fixes i, and hence an element of coreG (H) must fix all of {1, 2, . . . , n} pointwise. Thus,
coreG (H) = 1.
Now, let K = Fix(H). Since K/F is separable, the primitive element theorem applies and there is an
element α so that K = F [α]. This α ∈ K has the property that minF (α) has distinct roots. Note that
by the fundamental theorem, |K : F | = |G : H| = n and hence if f = minF (α) then f is irreducible,
and deg f = n. Let L be a splitting field of f contained in E. Then, L/F is Galois.
By the fundamental theorem, since L/F is Galois, N = Gal(E/L) is normal in G. Also, since L ⊇ K,
we have N = Gal(E/L) ⊆ Gal(E/K) = H. Now,
\
\
N=
g −1 N g ⊆
g −1 Hg = coreG (H) = 1.
g∈G
g∈G
Thus, N = 1 and hence L = E. Hence, E is the splitting field of an irreducible polynomial f ∈ F [x]
of degree n.
Cyclotomic Extensions
Cyclotomic extensions are special types of Galois extensions. They show up quite frequently [on the
exam, and in mathematics in general] and have many nice properties.
Definition. Suppose that F is a field, α ∈ F . Then, α is called a root of unity if there is an integer
n such that αn = 1. If αk 6= 1 for 1 ≤ k < n, we say that α is a primitive nth root of unity.
Lemma. Suppose that ∈ F is a primitive nth root of unity.
1. The elements k for 0 ≤ k < n are distinct and are all the nth roots of unity in F .
2. The elements k for 0 ≤ k < n and gcd(k, n) = 1 are all of the primitive nth roots of unity in
F.
3. F contains precisely φ(n) primitive nth roots of unity, where φ(n) is Euler’s function.
If α is a primitive nth root of unity, then α generates a cyclic subgroup of order n of the abelian group
F × . Conversely, every finite subgroup is of this form.
Lemma (Isaacs, 17.12). Suppose that G ⊆ F × is a finite subgroup. Then, G is cyclic.
Using this lemma, if α ∈ F × generates G, and |G| = n then α ∈ F × is an element of order n. This
means that α is a primitive nth root of unity.
Lemma (20.3). A field F has an extension that contains a primitive nth root of unity if and only
if char F does not divide n. If E/F is an extension and ∈ E is a primitive nth root of unity, then
F [] is a splitting field for xn − 1 over F .
Definition. Let
Y
Φn (x) =
(x − ),
where runs over all primitive nth roots of unity in C. The polynomial Φn (x) is called the nth
cyclotomic polynomial.
Note that if p is a prime Φp (x) =
i
0≤i<p x .
P
Lemma (20.4). For n ≥ 1,
Y
xn − 1 =
Φd (x).
d|n
This identity is useful in computing cyclotomic polynomials. Using it one can prove the following.
Corollary (20.5). We have that
Φn (x) ∈ Z[x].
Also with a little more of work we have that
Φn (x) =
Y
(xd − 1)µ(n/d)
d|n
where µ : N → Z is the möbius function, defined by µ(1) = 1, µ(n) = 0 if n has a square factor and
µ(n) = (−1)r if n is a product of r distinct primes.
We will now work toward describing Gal(F []/F ) where is a root of unity.
Definition. Let Un = (Z/nZ)× , the group of units of the ring Z/nZ. Note that Un is an abelian
group.
Lemma (Isaacs, 20.6). We have that Aut(Zn ) ∼
= Un . If n is prime, then Un is cyclic of order n − 1.
Now, we will describe the Galois group of F []/F .
Lemma (Isaacs, 20.7). Let F ⊆ E and suppose that E = F [], where is a root of unity. Then
E is Galois over F and Gal(E/F ) is isomorphic to a subgroup of Aut(G) where G = hi ⊆ E × . In
particular, Gal(E/F ) is abelian.
In the situation F = Q we can obtain more accurate information.
Theorem (Isaacs, 20.8). The cyclotomic polynomial Φn (x) ∈ Z[x] is irreducible for all n ≥ 1.
Remark. This implies that if Q ⊆ E and ∈ E is a primitive nth root of unity, then |Q[] : Q| = φ(n)
and Gal(Q[]/Q) ∼
= Un .
Many applications involve extensions that contains cyclotomic extensions. Here is a sample application.
Proposition. Suppose that F ⊆ K is an extension of fields. Suppose K = F [α, ], where is a
primitive nth root of unity for n > 1 and αn ∈ F × . Then the following hold.
1. K is Galois over F .
2. Let k be the smallest positive integer such that αk ∈ F []. Then, Gal(K/F []) ∼
= Z/kZ.
Proof.
Let f (x) = xn − αn . Since αn ∈ F , f (x) ∈ F [x]. Further, k α for 0 ≤ k < n are n distinct roots of
f (x) in K. Thus, f (x) is a separable polynomial. Note that a splitting field for f (x) in K contains
α, =
α
α
and hence K is a splitting field for a separable polynomial. Thus, K/F is Galois.
Define φ : Gal(K/F []) → hi ⊆ F [] by
(σ)φ =
(α)σ
.
α
Since (α)σ = l α for some l, it follows that (σ)φ ∈ hi.
Suppose that (α)σ = l α and (α)τ = m α. Then,
(α)στ = (l α)τ
= l (α)τ
= l (m α)
= l+m α.
It follows that φ is a homomorphism.
Also, if (σ)φ = 1 then (α)σ = α and hence σ fixes α so σ = 1. Thus, φ is injective. Thus, Gal(K/F [])
is isomorphic to a subgroup of hi which is cyclic of order n. Thus, Gal(K/F []) is cyclic. It suffices
to determine its order.
I claim that minF [] (α) = xk − αk . This then implies that |K : F []| = k and the desired result follows.
It suffices to show therefore that xk − αk ∈ F [][x] is irreducible.
We will show that k|n. Note that αk ∈ F [] by assumption and αn ∈ F ⊆ F []. Let d = gcd(k, n).
Then, there are integers x and y so that d = kx + ny. Then,
αd = (αk )x · (αn )y ∈ F [].
It follows that d = k and so k|n.
For this part of the argument we will use the “constant term of the minimal polynomial trick.” Clearly,
xk − αk ∈ F [][x].
Suppose that g(x) = minF [] (α). Then, g(x)|xk − αk and so the roots of g(x) in the splitting field are
of the form α times some kth root of unity. Since k|n, this kth root of unity is some power of . It
follows that the constant term of g(x) is the product of these roots. Thus,
g(0) = (−1)deg g αdeg g m .
Since m ∈ F [], it follows that αdeg g ∈ F []. By assumption, k is the smallest integer so that αk ∈ F []
and hence deg g = k. Thus, xk − αk is irreducible and and so F [, α]/F [] is a degree k extension.
Problems
1. IMPORTANT Read Isaacs pg 320-327, Finite fields.
2. Let G be a finite abelian group. Show that there exists a finite Galois extension F/Q of the
rational numbers, such that Gal(F/Q) ∼
= G. (Hint: Dirichlet Theorem on arithmetic progressions.)
3. Let f (x) = x4 − 2. If K is the splitting field of f over Q, then G = Gal(K/Q) ∼
= D8 . (Hint:
Why is it enough to show that [K : Q] = 8?)
4. A field F is called perfect if every finite extension K/F is separable. Show that a field F of
characteristic p > 0 is perfect iff the Frobenious homomorphism φp : F → F , sending x to xp is
an isomorphism.
5. Suppose that F is an algebraic closure of F and suppose that E/F is a normal extension with
E ⊆ F . Suppose that σ : E → F is an injective ring homomorphism and that it is also a F -linear
transformation. Prove that σ(E) = E.
6. Suppose that E/F is an extension of fields, char F 6= 2 and |E : F | = 2. Prove that E/F is
Galois.
7. Suppose that K/F is a Galois extension of fields with |K : F | = n. If p is a prime so that pa |n
but pa+1 - n, prove there is a field F ⊆ E ⊆ K so that |K : E| = pa .
8. If K/F is a Galois extension and F ⊆ E ⊆ K is a subfield, prove that there is a unique minimal
subfield L with F ⊆ L ⊆ E so that E/L is Galois.
9. Suppose that f (x) ∈ Q[x] is an irreducible cubic polynomial. If f (x) has a non-real root and K
is the splitting field of f over Q, prove that Gal(K/Q) ∼
= S3 .