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Chapter III. Galois Theory
The essence of Galois Theory: The systematically developed connection between
two seemingly unrelated subjects, the theory of fields and the theory of groups.
More specifically, but in the same line, is the idea of studying a mathematical
object by its group of automorphisms, an idea emphasized in Klein’s Erlanger Program, which has been accepted as a powerful tool in a great variety of mathematical
disciplines.
The Galois Theory of field extensions combines the esthetic appeal of a theory
of nearly perfect beauty with the technical development and difficulty that reveal the
depth of the theory and that make possible its great usefulness primarily in algebraic
number theory and related parts of algebraic geometry.
( From Roger Lyndon i:
Encyclopedia of Mathematics and Its Applications)
BASIC CONCEPTS AND THEOREMS.
An automorphism of a field L is a bijective mapping of L onto itself which is also
an isomorphism, i.e. a bijective mapping σ that sends a sum of elements in L into the
sum of the corresponding elements and a product of elements in L into the product
of the corresponding elements: σ(x + y) = σ(x) + σ(y) and σ(xy) = σ(x)σ(y) for all
x, y ∈ L. The set Aut(L) of all automorphisms of L forms a group where the product
of two automorphisms σ and τ is defined as the composition στ : στ (x) := σ(τ (x))
∀x ∈ L. The neutral element in this group is the identity mapping sending every
element in L into itself. This mapping is denoted 1L or e.
Example 3.1. For some fields the identity mapping is the only automorphism. For
instance the field Q of all rational numbers has no other automorphism than the
identity. (Why?)
If S is an arbitrary subset of Aut(L) we define the fixed set F (S) as
F (S) = {x ∈ L | σx = x ∀σ ∈ S}.
It is easily seen that F (S) is a subfield of L which is called the fixed field for S.
Example 3.2. Let L = C og σ = complex conjugation. It is straightforward to
check that F ({σ}) = R.
Let K be a subfield of L. We define
Aut(L/K) = {σ ∈ Aut(L) | σRes,K = 1K } = {σ ∈ Aut(L) | σk = k ∀k ∈ K}.
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Aut(L/K) is easily seen to be a subgroup of Aut(L). It is called the relative
automorphism group for L over K.
The following is an immediate consequence of the definition:
1) F (Aut(L/K)) K
2) Aut(L/F (S)) S
In general, ”=” does not hold in 1) and 2).
Remark 3.3. If ”=” holds in 2) then S must be a subgroup of Aut(L). We shall
later (Corollary 3.21) see that ”=” holds in 2) if S is a finite group.
For computations of relative automorphism groups the following is often quite
useful:
Lemma 3.4. Let L/K be an arbitrary field extension and f (x) = xn + a1 xn−1 +
· · · + an a (not necessarily irreducible) polynomial in K[x]. If α is an element in L
and is a root of f (x), then σ(α) is also a root of f (x) for every automorphism in
σ ∈ Aut(L/K).
Proof. Since α is a root of f (x) we get
αn + a1 αn−1 + · · · + an = 0
By applying the automorphism σ on each side of the above equation we obtain
(σ(α))n + a1 (σ(α))n−1 + · · · + an = 0 ()
where we have used that a1 , · · · an are fixed under the automorphism σ. But the
equation () just means that σ(α) is a root of f (x).
Example 3.5. L = C, K = R, Aut(L/K) = {1L , complex conjugation}. Here
F (Aut(L/K)) = K. Indeed, any number
in L can be written uniquely as k1 + k2 · i
√
where k1 and k2 lie in K and i = −1. An automorphism in Aut(L/K) is uniquely
determined by its value on i. Since i is a root of x2 + 1 the above lemma means
that such an automorphism can only send i into i or −i. On the other hand 1L
and complex conjugation are automorphisms in Aut(L/K) and a complex number is
invariant under complex conjugation if and only if it is real.
√
√
2)
,
K
=
Q,
Aut(L/K)
=
{1
,
σ}
where
σ(a
+
b
2) =
Example
3.6.
L
=
Q(
L
√
a−b 2; a, b ∈ Q. Here F (Aut(L/K)) = K. This can be verified by the same method
as in the previous example.
√
3
2) , K = Q, Aut(L/K)
= 1L . Indeed [L : K] =
Example 3.7.
L
=
Q(
√
√
3
3
3
2, Q).√ Hence every element in L
degree(Irr( 2, Q)) = 3 , since x − 2 = Irr(
√
3
has a unique representation of the form q0 + q1 2 + q2 ( 3 2)2 , where q0 , q1 and q2 are
rational numbers. An automorphism in Aut(L/K) is uniquely determined by its value
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Mat 3AL 3.3
√
√
on 3 2. Since L consists entirely of real numbers and 3 2 is the only real root of x3 −2
the identity 1L is the only automorphism in Aut(L/K). Hence F (Aut(L/K)) = L.
Example 3.8. L = R, K = Q. Here is Aut(L/K) = 1L hence: F (Aut(L/K)) = L.
(The proof is non-trivial.)
Example 3.9. L = C, K = Q, Aut(L/K) = Aut(C) has the cardinality 22
F (Aut(L/K)) = K. (The proof is non-trivial.)
ℵ
0
and
The above examples concern the relation 1).
Example 3.10. In 2) it can happen that Aut(L/F (S)) S even when S is a group.
Let L = C(x), K = C, S = all “translations”
f (x)
f (x + a)
},
S = {σ ∈ Aut C(x) | σ
=
g(x)
g(x + a)
where a runs through C. Here F (S) = C and Aut(L/F (S)) S since x →
an automorphism in Aut(C(X)/C) which does not belong to S.
1
x
induces
GALOIS THEORY STARTS.
Definition 3.11. L K is called a normal extension if K = F (Aut(L/K)). If
moreover [L : K] is finite, L is called a finite normal extension of K.
Theorem 3.12. If M is the splitting field over the field K for a separable polynomial
f (x) ∈ K[x] then M is a finite normal extension of K.
Proof. [M : K] < ∞ is clear. If all roots of f (x) lie in K then M = K and there
is nothing to prove. So let us assume that M K. We may then choose roots
α1 , ..., αt of f (x) such that M = K(α1 , . . . , αt ) and K K(α1 ) K(α1 , α2 ) · · · K(α1 , ..., αt) = M . We shall prove that ∀β ∈ M \ K ∃σ ∈ Aut(M/K) such that
σ(β) = β.
Assume β ∈ K(α1 , . . . , αi ) \ K(α1 , . . . , αi−1 ). To ease the notation we set L =
/ L. Thus:
K(α1 , . . . , αi−1 ), γ = αi . Therefore β ∈ L(γ), β ∈
Irr(γ, L) Irr(γ, K)f (x) .
Since f (x) is separable, Irr(γ, K) and thus in particular Irr(γ, L) has only simple
roots.
If n is the degree of Irr(γ, L), then [L(γ) : L] = n and Irr(γ, L) has exactly n
distinct roots γ1 , γ2 , . . . , γn where for instance γ = γ1 .
β can be written in the form
β = a0 + a1 γ + · · · + an−1 γ n−1 , a0 , . . . , an−1 ∈ L
and β ∈
/ L ⇒ ai = 0 for at least one i 1.
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Not all the elements a0 + a1 γj + · · · + an−1 γjn−1 , j = 1, 2, . . . , n can be = β.
Otherwise the non-zero polynomial β − a0 − a1 x − · · · − an−1 xn−1 of degree < n
would have n distinct roots γ1 , . . . , γn .
Assume for instance a0 + a1 γ2 + · · · + an−1 γ2n−1 = β.
According to the uniqueness theorem for adjunction of a root of an irreducible
polynomial (Theorem 2.57) there exists an isomorphism ϕ : L(γ) → L(γ2 ) for which
ϕRes,L = 1L and ϕ(γ) = γ2 .
M
L(γ2 )
L(γ)
β
ϕ
L
L
Here ϕ(β) = ϕ(a0 + a1 γ + · · · + an−1 γ n−1 ) = a0 + a1 γ2 + · · · + an−1 γ2n−1 , which we
just have seen is = β. Now M is splitting field for f (x) over L(γ) and M is splitting
field for f (x) over L(γ2 ).
Since f (x) ∈ K[x] L[x] the mapping of L(γ)[x] → L(γ2 )[x] induced by ϕ will
send f (x) into itself. By the theorem concerning the uniqueness of splitting fields
(Theorem 2.60 in chap.2) ϕ can be prolonged to an automorphism σ : M → M .
Clearly σ has the properties σ(β) = β and σRes,K = 1K . In other words: σ is an
automorphism in Aut(M/K) with the desired property.
−−−−−−−−−−−−−−−
Example 3.13. M = K(T ), f (x) = (x − T )(x − T1 ) = x2 − (T + T1 )x + 1. M is a
splitting field for f (x) over L = K(T + T1 ). Clearly f (x) is separable. Aut(M/L) =
{1L and σ} where σ(g(T )) = g( T1 ) for g(T ) ∈ K(T ). From this we get a theorem
(which can be formulated without use of splitting fields etc.) about rational functions:
a rational function g(T ) satisfies the ”functional equation” g(T ) = g( T1 ) if and only
if g = rational function of T + T1 .
Theorem 3.14. If Fpn as usual denotes the finite field with pn elements, the relative automorphism group Aut(Fpn /Zp ) is cyclic of order n and is generated by the
Frobenius automorphism sending every element into its p-th power.
Proof. By the theorem of Abel-Steinitz there exists an element α ∈ Fpn such that
Fpn = Zp (α). Let f (x) = Irr(α, Zp ). An automorphism σ ∈ Aut(Fpn /Zp ) is uniquely
determined by its value on α; thus (by Lemma 3.4) there are at most n possibilities for
σ since f (x) is of degree n. On the other hand as we have seen in Chap.2 the mapping
σ : a → ap (sending every element into its p-th power) is an automorphism. Now
σ n = the identity 1L on L and σ i = 1L for 0 < i < n. Indeed, otherwise all elements
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i
of Fpn would be roots of the polynomial xp − x. But this is impossible since Fpn has
pn elements. Therefore Aut(Fpn /Zp ) consists just of the powers σ, σ 2 , . . . , σ n = 1L .
Theorem 3.15. Every set of distinct automorphisms σ1 , . . . , σn of a field L is independent, i.e. if a1 , . . . , an ∈ L and a1 σ1 (x) + · · · + an σn (x) = 0 for all x ∈ L then
a1 = · · · = an = 0.
Proof. Induction on n.
The case n = 1 is clear: for x = 1 we get a1 σ1 (1) = a1 = 0.
Concerning n − 1 → n:
Assume that
a1 σ1 (x) + a2 σ2 (x) + · · · + an σn (x) = 0 ∀x ∈ L .
(∗)
For every b ∈ L we have a1 σ1 (bx) + a2 σ2 (bx) + · · · + an σn (bx) = 0 ∀x ∈ L or
a1 σ1 (b)σ1 (x) + a2 σ2 (b)σ2 (x) + · · · + an σn (b)σn (x) = 0 ∀x ∈ L .
By multiplication of (∗) by σ1 (b) we get
a1 σ1 (b)σ1 (x) + a2 σ1 (b)σ2 (x) + · · · + an σ1 (b)σn (x) = 0 ∀x ∈ L
and thus
a2 (σ2 (b) − σ1 (b))σ2 (x) + · · · + an (σn (b) − σ1 (b))σn (x) = 0 ∀x ∈ L .
If we choose b ∈ L such that σn (b) = σ1 (b) the inductive assumption yields an = 0.
We insert this in (∗) and applying again the inductive assumption we see that a1 =
· · · = an−1 = 0.
Theorem 3.16. Let σ1 , . . . , σn be distinct automorphisms of L and let K be a subfield of F ({σ1 , . . . , σn }). Then: [L : K] n.
Proof. By way of contradiction assume [L : K] = r < n. Let ω1 , . . . , ωr be a basis
for L viewed as a vector space over K. The system of homogeneous linear equations
x1 σ1 (ω1 ) + · · · + xn σn (ω1 ) = 0
...
(∗∗)
x1 σ1 (ωr ) + · · · + xn σn (ωr ) = 0
has a proper solution (i.e. not all the xi ’s are 0 ) since r < n (a well-known theorem
from linear algebra1 ). Any element β ∈ L kan be written β = a1 ω1 + · · · + ar ωr ,
where a1 , . . . , ar ∈ K. Because of (∗∗) we have
x1 σ1 (β) + · · · + xn σn (β) = 0.
This gives us the desired contradiction in view of Theorem 3.15.
1A
system of homogeneous linear equations has a proper solution if the number of unknowns is
bigger than the number of equations.
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Theorem 3.17. Let G be a group of automorphisms of L. Then [L : F (G)] = |G|
(in the sense that if one of the sides is finite so is the other and in that case the
cardinalities coincide).
Proof. In view of the preceding theorem it is enough to consider the case where G is
finite. Assume G has order n < ∞. Again by the preceding theorem it is enough to
show that [L : F (G)] n. To prove this we must show that any (n + 1) elements of
L are linearly dependent over F (G).
For this we need two lemmas
Lemma 3.18. For every x ∈ L “the trace” S(x) := σ∈G σ(x) is an element in
F (G).
Proof. For every σ ∈ G we get σ (S(x)) = σ∈G σ σ(x) = σ∈G σ(x) = S(x). Lemma 3.19. There exists an element x ∈ L for which S(x) = 0.
Proof. Apply Theorem 3.15 about the independence of distinct automorphisms. We now return to the proof of theorem 3.17. Let G = {σ1 , . . . , σn }.
We must show that any n + 1 elements ω1 , . . . , ωn+1 of L are linearly dependent
over K.
The system of homogeneous linear equations
x1 σ1−1 (ω1 ) + · · · + xn+1 σ1−1 (ωn+1 ) = 0
·
·
·
·
x1 σn−1 (ω1 ) + · · · + xn+1 σn−1 (ωn+1 ) = 0
has a proper (i.e. non-zero) solution (x1 , . . . , xn+1 ) in L. We may assume that x1 = 0
and by multiplication by a suitable element in L we can (because of Lemma 3.19)
obtain that S(x1 ) = 0.
By applying the automorphisms σ1 , . . . , σn on the above system of equations and
taking sums we get
S(x1 )ω1 + · · · + S(xn+1 )ωn+1 = 0
in other words: ω1 , . . . , ωn+1 are linearly dependent over F (G).
Remark 3.20. If G is infinite then [L : F (G)] is also infinite, but in general the
cardinality of G is bigger than than that of [L : F (G)]. For instance let Q be the
algebraic closure of Q in C and let G be Aut(Q/Q). It can be proved that G is
uncountable (actually has the cardinality 2ℵ0 ) while [Q : Q] = ℵ0 .
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Corollary 3.21. If G is a finite group of automorphisms of L then: Aut(L/F (G)) =
G.
Proof. It is clear that Aut(L/F (G)) G.
Let us now prove the inverse inclusion. Let G be a finite group of order n consisting
of the elements σ1 , . . . , σn . Then [L : F (G)] = n. If there were some automorphism σ of L, σ not lying in G, but fixing F (G) then F (G) would be equal to F (σ , σ1 , . . . σn ).
But this would contradict Theorem 3.16 since [L : F (G)] = n.
Corollary 3.22. Distinct finite automorphism groups of L have distinct fixed fields.
Proof. Let G1 and G2 be two finite automorphism groups of L. If F (G1 ) = F (G2 )
the above corollary implies that G1 = Aut(L/F (G1 )) = Aut(L/F (G2 )) = G2 .
We recall the notion ”finite normal extension”. L/K is a finite normal extension if
[L : K] < ∞ and K = F (Aut(L/K)). Aut(L/K) is called the Galois group for L/K
and is denoted Gal(L/K). In principle we could denote the Galois group of a normal
extension L/K by Aut(L/K). But it is customary to use the notation Gal(L/K) for
a normal extension.
In view of Theorem 3.17 the order of the Galois group Gal(L/K) equals the dimension [L : K].
Furthermore we note that a finite extension L/K is normal ⇔ K = fixed field for
a group of automorphisms of L.
Remark 3.23. From the above we can also conclude the following. Let L/K be a
finite extension. Then | Aut(L/K)| [L : K] and | Aut(L/K)| = [L : K] if and only
if L/K is normal.
√
√
Example 3.24. Q( 2)/Q is normal while Q( 3 2)/Q is not normal. (Cf. Examples
3.6 and 3.7).
√ √
√ √
Example 3.25. Q( 2, 3)/Q is normal. Indeed Q( 2, 3) is the splitting field of
the √separable
polynomial (x2 − 2)(x2 − 3), so
√
√ we
√ can apply Theorem 3.12. Since
[Q( 2, 3) : Q] = 4 the Galois group Gal(Q( 2, 3)/Q) has order 4. There are two
groups of order 4: Kleins 4-group and the cyclic group of order 4. To determine which
of these√groups
√ is isomorphic to the Galois group we note that
√ an automorphism
√
√ in
by its values on 2 and 3. Since 2 is
Gal(Q( 2, 3)/Q) is
√ uniquely determined
2
2
− 2 and √ 3 is a root
− 3. Lemma
implies that an automorphism
a root of x √
√ of x √
√ 3.4√
must send 2 into 2 or − 2 and 3 into 3 or - 3. Thus there are at most 4
possibilities. Since the Galois group has order 4 each of these possibilities can be
realized as an automorphism √
in exactly
√ one way. They can be listed in this
√ way:
2
and
3,
an
automorphism
σ
which
fixes
the identity
which
fixes
both
1
√
√
√
√2 and
√
sends 3 to − 3, an automorphism
√ σ2 which
√ fixes√ 3 and√sends 2 to − 2 and
an automorphism σ3 which sends 2 to − 2 and 3 to − 3. It is easily checked
that σ12 = σ22 = σ32 = the identity. Hence the Galois group is isomorphic to Kleins
4-group.
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We now give an extremely important characterization of the finite normal extensions.
Theorem 3.26. Let L/K be a finite extension. Then L/K is normal ⇔ L = splitting field for a separable polynomial f (x) over K.
Proof. “⇐” has already been proved in Theorem 3.12.
“⇒” here we need the following which is very important in its own right.
Theorem 3.27. Let L/K be a finite normal extension. If an irreducible polynomial
p(x) in K[x] has some root α ∈ L, then all roots of p(x) lie in L and all of them are
simple, and they are just the distinct elements among {σ(α) | σ ∈ Aut(L/K)}. In
particular p(x) is separable.
Proof. W.l.o.g we may asume that p(x) is monic, hence p(x) = Irr(α, K). Let
σ1 (α), . . . , σj (α) be the distinct elements in the set {σ(α) | σ ∈ Aut(L/K)}. For every
automorphism τ ∈ Aut(L/K) we have {τ σ1 (α), . . . , τ σj (α)} = {σ1 (α), . . . , σj (α)}.
Let τ̂ be automorphism of L[x] induced by τ . (τ̂ (a polynomial in L[x]) = the
polynomial obtained by replacing the coefficients by their images under τ ). For the
polynomial f (x) := [x − σ1 (α)] · · · [x − σj (α)] we get
τ̂ f (x) = [x − τ σ1 (α)] · · · [x − τ σj (α)] = f (x) .
This holds for all τ ∈ Gal(L/K), hence f (x) has coefficients in F (Gal(L/K)) = K.
f (x) is thus a polynomial in K[x] and f (α) = 0 and so p(x) = Irr(α, K)|f (x).
On the other hand (by Lemma 3.4) p(α) = 0 ⇒ p(σ(α)) = 0 ∀σ ∈ Gal(L/K).
Consequently each of the elements σ1 (α), . . . , σj (α) is a root of p(x) and this implies
that f (x) = [x−σ1 (α)] · · · [x−σj (α)] | p(x). But f (x) and p(x) are monic polynomials
for which f (x) | p(x) og p(x) | f (x). This means f (x) = p(x).
Now we return to the proof of “⇒” in Theorem 3.26.
Theorem 3.27 implies in particular that every finite normal extension is separable.
By Abel-Steinitz’s theorem there exists an α in L such that L = K(α). By Theorem
3.27 the polynomial p(x) = Irr(α, K) splits into linear factors inside L, and no proper
subfield of L containing K has this property; consequently L is the splitting field for
Irr(α, K) and Irr(α, K) is separable.
We give yet another characterization of finite normal extensions.
Theorem 3.28. A finite extension L/K is normal if and only if L/K is separable
and every irreducible polynomial in K[x] having some root in L has all its roots in
L.
Proof. “only if” has already been proved.
“if”: By Abel-Steinitz’s theorem L = K(α) for a suitable α ∈ L. Therefore L is
the splitting field for the separable polynomial p(x) = Irr(α, K).
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Remark 3.29. Some authors call a finite extension L/K normal if every irreducible
polynomial in K[x] having some root in L has all its roots in L, while an extension
wich is normal in the sense of these notes is called a Galois extension. For perfect
fields, in particular for fields of characteristic 0, the two definitions coincide.
Theorem 3.30. Assume M/K is a finite normal extension and L is an intermediate
extension, i.e. M ⊇ L ⊇ K, then M/L is a finite normal extension.
Proof. M/K finite normal ⇒ M = splitting field for a separable polynomial f (x)
over K ⇒ M = splitting field for f (x) over L ⇒ M/L is a finite normal extension Example 3.31. If M/K is a finite normal extension and L is an intermediate field,
i.e. M L K, then L may not be a√normal extension of K. A counterexample can
be gotten by taking K = Q, L = Q( 3 2) and M = the splitting field for x3 − 2 over
Q. By theorem 3.12 M/Q is normal, while L/Q (cf.Example 3.24) is not normal.
THE FUNDAMENTAL THEOREM OF GALOIS THEORY.
By means of the previous results we are now in a position to prove
Theorem 3.32. The Fundamental Theorem of Galois Theory. Let M/K be
a finite normal extension with Galois group G = Gal(M/K). (Here is [M : K] = |G|
(cf. Theorem 3.17)). There exists a (1 − 1) correspondence between the subgroups
of G and the intermediate fields of M/K (i.e. subfields of M containing K). The
(1 − 1) correspondence is obtained in the following way:
To an intermediate field L we associate T (L) = Gal(M/L) (which is a subgroup
of G); to a subgroup H of G we associate the fixed field F (H) (K ⊆ F (H) ⊆ M ).
Then:
1) F (T (L)) = L, T (F (H)) = H (this gives the (1−1) correspondence mentioned
above)
2) |T (L)| = [M : L]; [L : K] = [G : T (L)].
3) L1 L2 ⇔ T (L1 ) T (L2 ); H1 H1 ⇔ F (H1 ) F (H2 ).
4) For an intermediate field L holds: L/K is normal ⇔ T (L) G.
5) If L/K is normal [so by 4) T (L)
G] then Gal(L/K) ∼
= G/T (L) and every automorphism of Gal(L/K) can be extended to an automorphism of Gal(M/K).
Proof.
ad 1) F (T (L)) = L follows from Theorem 3.30 and T (F (H)) = H follows from
Corollary 3.21.
ad 2) This follows immediately from Theorem 3.17.
ad 3) The two arrows ⇒ from the left to the right are trivial. Let us now consider
the inverse implications. Assume T (L1 ) T (L2 ). Applying the (trivial)
implication ⇒ on this inclusion we get F (T (L1 )) F (T (L2 )), which by
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1) just means that L1 L2 . The other arrow ⇐ is obtained by a similar
argument.
ad 4) For this we need two sublemmas.
Sublemma 3.33. An intermediate field L is normal over K if and only if σL = L
for all σ ∈ G.
Proof. “only if”: It is enough to show that σL L for all σ ∈ G (why ?).
Let α ∈ L, p(x) = Irr(α, K) then σ(α) (cf. Lemma 3.4) is also a root of p(x) for
all σ ∈ G. Theorem 3.27 implies that σ(α) is an element in L i.e.: σ(α) ∈ L for all
α ∈ L and for all σ ∈ G.
“if”: Since M/K is normal, M/K is separable, in particular L/K is separable. By
Abel-Steinitz’s theorem L = K(α) for a suitable α ∈ L. The roots of p(x) = Irr(α, K)
are (by Theorem 3.27) exactly the elements σ(α), σ ∈ G.
Since σL = L all these roots lie in L, which therefore is the splitting field for the
separable polynomial p(x) over K, hence L/K is normal by Theorem 3.12.
Sublemma 3.34. T (σL) = σT (L)σ −1 for every intermediate field L and every
automorphism σ in G = Gal(M/K).
Proof. By the definitions we have
T (σL) = {τ ∈ G|τ σ = σ ∀ ∈ L} = {τ ∈ G|σ −1 τ σ = ∀ ∈ L}
= {τ ∈ G|σ −1 τ σ ∈ T (L)} = {τ |τ ∈ σ T (L)σ −1 } = σT (L)σ −1 .
We now apply the sublemmas to prove 4). By Sublemma 3.33 we know that
L/K is normal if and only if σL = L for all σ ∈ G. Since T is injective the latter
equality holds if and only if T (σL) = T (L) for all σ ∈ G. By Sublemma 3.34
T (σL) = σT (L)σ −1 . Hence T (σL) = T (L) for all σ ∈ G if and only if T (L) is a
normal subgroup of G.
ad 5) Assume that L/K is normal and thus T (L) G.
We define a mapping ϕ : G = Gal(M/K) → Gal(L/K) by ϕ(σ) = σRes,L . Sublemma 3.34 implies that σRes,L really is an automorphism for L (which is the identity
on K) i.e.: σRes,L ∈ Gal(L/K). Therefore ϕ is a well defined homomorphism from
G into Gal(L/K).
For the kernel of this homomorphism we find Ker ϕ = {σ ∈ G | σRes,L = 1L } =
Gal(M/L) = T (L). Therefore we get an isomorphism G/T (L) ϕG Gal(L/K).
Now by 2) we have |G/T (L)| = [L : K].
Theorem 3.17 says that | Gal(L/K)| = [L : K]; therefore ϕG is a subgroup of
Gal(L/K) having the same order as Gal(L/K). Consequently ϕG = Gal(L/K). Exercise 3.35. Let M/L and L/K be finite normal extensions. Show by an√example
4
that
√ M/K is not necessarily a normal extension. (Consider e.g. M = Q( 2), L =
Q( 2) and K = Q.)
Exercise 3.36. Let M/L and L/K be finite normal extensions and assume furthermore that every automorphism in Gal(L/K) can be extended to an automorphism
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of M . Prove that this implies that M/K is a normal extensione. (Compare this with
5) in the fundamental theorem of Galois theory.)
Exercise 3.37. Let M/Q be a finite normal extension. Show that complex conjugation induces an automorphism in Gal(M/Q) (which is the identity if M is contained
in the field of real numbers; otherwise it is an automorphism of order 2). (Hint: use
e.g. Theorem 3.27.)
−−−−−−−−−−−−−−−
Theorem 3.38. Supplement to the Fundamental Theorem of Galois Theory. Let M/K be a finite normal extension. For two intermediate fields L1 , L2 there
exists a smallest field (“the compositum”) containing L1 and L2 , denoted {L1 , L2 }
or just L1 L2 . Correspondingly for two subgroups H1 , H2 of a given group there is a
smallest group,(”the compositum”), denoted {H1 , H2 } containing H1 and H2 . With
the notations from the fundamental theorem of Galois theory the following holds:
T {L1 , L2 } = T (L1 ) ∩ T (L2 ),
T (L1 ∩ L2 ) = {T (L1 ), T (L2 )},
F (H1 ∩ H2 ) = {F H1 , F H2 },
F {H1 , H2 } = F H1 ∩ F H2 .
Proof. We just check the first one:
{L1 , L2 } L1 ⇒ T (L1 ) T ({L1 , L2 })
⇒ T ({L1 , L2 }) T (L1 ) ∩ T (L2 )
{L1 , L2 } L2 ⇒ T (L2 ) T ({L1 , L2 })
Furthermore
F (T (L1 ) ∩ T (L2 )) F (T (L1 )) = L1
⇒ F (T (L1 ) ∩ T (L2 )) {L1 , L2 }
F (T (L1 ) ∩ T (L2 )) F (T (L2 )) = L2
⇒ T (L1 ) ∩ T (L2 ) T ({L1 , L2 }).
Corollary 3.39. With the notations from above the following holds: L1 /K normal
and L2 /K normal ⇒ L1 ∩ L2 /K normal and {L1 , L2 }/K normal.
WHAT DO GALOIS GROUPS LOOK LIKE?
We shall now prove a theorem which is quite useful for the explicit computation
of the Galois group for a normal extension that is given as the splitting field of a
separable polynomial.
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Theorem 3.40. Let M be the splitting field over a field K for a polynomial p(x)
having degree n and no multiple roots. Then there is an isomorphism ϕ from the
Galois group Gal(M/K) onto a subgroup of the symmetric group Sn . Hence the
following holds: [M : K] n! and even [M : K] | n!. If p(x) is irreducible in K[x]
then ϕ Gal(M/K) is a transitive subgroup of Sn . Hence: n [M : K] n! and even
n | [M : K] | n!
Proof. We can write M = K(α1 , . . . , αn ) where α1 , . . . , αn are the distinct roots of
p(x). An automorphism σ ∈ Gal(M/K) is uniquely determined bu its values on
α1 , . . . , αn . According to Lemma 3.4 theelements σ(αi ) , 1 i n, are roots of
α 1 , . . . , αn
p(x). This means that
is a permutation of the roots α1 , . . . , αn .
σ(α1 ), . . . , σ(αn)
α
,
.
.
.
,
α
1
n
Furthermore the mapping Gal(M/K) ϕ- Sn , ϕσ =
is an
σ(α1 ), . . . , σ(αn)
homomorphism from Gal(M/K) into Sn ; since σ is uniquely determined by its values
on α1 , . . . , αn we conclude that ϕ is injective. If p(x) is assumed to be irreducible in
K[x] Theorem 3.27 implies that ϕ(Gal(M/K)) is a transitive subgroup of Sn . The
transitivity theorem (Theorem 2.47 in Chap.II) implies n | [M : K], since M contains
for instance K(α1 ) which has dimension n over K.
Theorem 3.41. Let M be the splitting field over a field K for a polynomial p(x)
having degree n and no multiple roots. Assume moreover that the characteristic of K
is = 2. If ϕ is the isomorphism introduced in Theorem 3.40 the image ϕ Gal(M/K)
is a subgroup of the alternating group An if and only if the discriminant discrim(p)
is the square of an element in K.
Proof. M = K(α1 , . . . , αn ) where α1 , . . . , αn are the distinct roots of p(x). We
consider the mapping ϕ : Gal(M/K) → Sn introduced in
Theorem 3.40.
(αi −αj ) is invariant under
Assume ϕ(Gal(M/K) ⊆ An . Then Δ(α1 , . . . , αn ) =
i>j
all automorphisms in Gal(M/K); therefore Δ(α1 , . . . , αn ) ∈ K. Hence discrim(p) =
(Δ(α1 , . . . , αn ))2 is the square of an element in K.
Next assume that ϕ Gal(M/K) An . Then there is an automorphism σ ∈
Gal(M/K) such that σΔ(α1 , . . . , αn ) = −Δ(α1 , . . . , αn ). Since the characteristic of
K is = 2 the element Δ(α1 , . . . , αn ) cannot lie in K; thus discrim(p) is not the square
of an element in K.
Example 3.42. Let M be the splitting field for x3 − 2 over Q. The polynomial
x3 − 2 is irreducible (by Eisenstein’s criterion) and separable (all polynomials over
Q are separable). By theorem 3.40 the Galois
group Gal(M/Q) is (isomorphic to)
√
3
a transitive √subgroup of S3 . Since M = Q( 2, ε) where ε is the 3rd root of unity
2πi
3
e 3 = −1+i
and ε is a root of the polynomial x2 +x+1, it is an easy consequence of
2
the transitivity theorem that the dimension [M : Q] is 6. Therefore Gal(M/Q) S3 .
It is an instructive exercise to write down all subgroups of Gal(M/Q) S3 and the
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corresponding fixed fields of M . By the fundamental theorem of Galois theory one
then gets a diagram of all subfields of M .
Exercise 3.43. Show that every normal extension M/Q with S3 as Galois group
can be obtained as the splitting field over Q for an irreducible cubic polynomial f (x)
in Q[x]. [Let L be the fixed field of a transposition in S3 . Notice that [L : Q] = 3;
by Abel-Steinitz’s theorem there is an α ∈ L so that L = Q(α). As f (x) one can use
Irr(α, Q).]
√
Example 3.44. Let M be the splitting field for x4 − 2 over Q. Here M = Q( 4 2, i)
which implies that [M : Q] = 8. We want to find Gal(M/Q). First notice that
without any computation we can predict that it must be the dihedral group D4 of
order 8. Indeed, by Theorem 3.40 Gal(M/Q) is a subgroup of S4 of order 8. But by
Sylow’s theorems any subgroup of S4 of order 8 is (isomorphic to) D4 .
We now study what the automorphisms in the Galois group look like, i.e. how
the automorphisms act explicitly on the numbers in M .
√
4
2
An automorphism in Gal(M/Q)
is
uniquely
determined
by
its
values
on
√
and √
i. For√the √
values √
on 4 2 there are (cf. Lemma 3.4) the following possibilities 4 2, − 4 2, i 4 2, −i 4 2. For the values on i there are the possibilties +i and −i.
Since Gal(M/Q) has order 8 any one of the above 8 combinations can be realized by
exactly one automorphism in Gal M/Q). The Galois group consists of the following
automorphisms:
σ , σ 2 , σ 3 , σ 4 = e , τ , τ σ , τ σ 2 , τ σ 3 , where
√
√
√
√
√
√
√
√
4
4
4
4
4
4
4
4
σ( 2) = 2i , σ 2 ( 2) = − 2 ; σ 3 ( 2) = −i 2 ; σ 4 ( 2) = 2
σ(i) = i ; σ 2 (i) = i ; σ 3 (i) = i ; σ 4 (i) = i
√
√
√
√
√
√
√
√
4
4
4
4
4
4
4
4
τ ( 2) = 2 τ σ( 2) = −i 2 τ σ 2 ( 2) = − 2 τ σ 3 ( 2) = i 2
τ (i) = −i τ σ(i) = −i τ σ 2 (i) = −i τ σ 3 (i) = −i .
Here we have τ 2 = e, στ = τ σ −1 .
By direct computation one finds all subgroups of Gal(M/Q):
{e} VLVVV
LLL VVVV
gggpgpgp
g
g
g
g
p
LLL VVVVVV
gg ppp
g
g
g
g
VVVV
g
LLL
p
g
g
p
g
VVVV
g
p
g
g
L
p
g
VV
p
gggg
2
{e,
τ
}
{e, τ σ}
{e,
σ
}
{e, τ σ 2 }
{e, τ σ 3 }
VVVV
gg
M
g
g
M
V
o
g
VVVV
o
gg
VVVV MMMM
ooo gggggggg
o
VVVV MMM
o
o
ggg
VVV M
ooo gggg
2
3
2
3
{e, σ 2 , τ, τ σ 2}
{e, σ , τ σ, τ σ }
{e, σ, σ , σ }
WWWWW
h
WWWWW
hhhh
hhhh
WWWWW
h
h
h
WWWWW
hhh
W
hhhh
Gal(M/K)
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From the fundamental theorem of Galois theory we then know that the mutual positions of the subfields of M are as in the above diagram. Explicit computations show
that the subfields are:
√
4
2, i) T
Q(
g
g
JJ TTTT
g
gg ppp
g
g
JJ TTT
g
g
g
p
g
JJ
g
TTTT
p
g
g
p
g
p
g
JJ
g
TTTT
p
g
g
p
g
J
p
gg
TT
g
p
g
√
√
√
√
√
4
4
4
Q( 2(1 + i))
Q( 2, i) T
Q( 2(1 − i))
Q( 2)
Q( 4 2i)
g
g
TTTT
II
gg
n
II
TTTT
ggggg
nnn
g
g
n
I
g
T
n
g
T
g
n
g
T
g
n
TTTT IIII
n ggggg
n
n
TTT
√
√ nggggg
Q(i)
Q(
2)
Q( 2i) WWWW
ii
WWWWW
i
i
i
WWWWW
ii
WWWWW
iiii
i
i
i
WWWWW
iii
WWWWW
iiii
i
Q
Let us for instance check that the fixed field of {e, σ, σ 2, σ 3 } is Q(i). It is clear that
Q(i) F ({e, σ, σ 2, σ 3 })
From 2) in the fundamental theorem of Galois theory we know that the dimension
of F ({e, σ, σ 2, σ 3 }) over Q is 2. The dimension [Q(i) : Q] is also 2. Therefore the
fixed field must be exactly Q(i). The determination of the other fixed fields can be
done similarly. The
hardest ones are the fixed fields of {e, τ σ} and {e, τ σ 3 }. It is
√
4
clear that L := Q( 2(1 − i)) F ({e, τ σ}). To show equality it suffices to verify that
the dimension of M over L is 2. For this one should just observe that
M = L(i).
√
4
3
An analogous argument shows that the fixed field of {e, τ σ } is Q( 2(1 + i)).
Which subfields of M are normal over Q?
How can the commutator subgroup of Gal(M/Q) be interpreted in terms of Galois
theory?
Exercise 3.45∗ . Let f (x) be an irreducible polynomial in Q[x] of degree n. Assume
that the number r of real roots of f (x) satisfies 0 < r < n. Let M be the splitting
field for f (x) over Q (viewed as a subfield of the complex number field C). Show that
L = M ∩ R is a non-normal extension of Q and that [L : Q] n.
1) Show that [M : Q] ≥ 2n.
Now let f (x) be x4 − 2x3 − 2x + 1.
2) Show that f (x) = x4 − 2x3 − 2x + 1 is irreducible over Q (consider f (x + 1)).
3) Show that f (x) has exactly 2 real roots.
4) Let M be the splitting field for f (x) over Q. Find [M : Q] and Gal(M/Q).
(Hint: Observe that a number α is a root of f (x) if and only if 1/α is a root
of f (x). Hence the roots of f (x) have the form α, 1/α, β, 1/β for suitable
numbers α and β.)
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THE TRANSLATION THEOREM AND APPLICATIONS.
Theorem 3.46. Translation Theorem. Let M and L be extension fields of a
field K, both of them being contained in a common extension field of K. Assume
M/K is a finite normal extension while L is an arbitrary extension of K. Then the
compositum M L is a normal extension of L and Gal(M L/L) Gal(M/M ∩ L). In
particular [M L : L] = [M : M ∩ L].
Furthermore by associating N → N L, (M ∩ L N M ) and Λ → Λ ∩ M ,
(L Λ M L) we get a (1 − 1) correspondence between the fields N between M ∩ L
and M and between the fields Λ between L and M L. The correspondence is etablished
by the following
N L ∩ M = N and (Λ ∩ M )L = Λ .
Proof. Since M/K is finite normal, M = K(α) for a suitable α ∈ M , and p(x) =
Irr(α, K) is separable. M L = L(α) is splitting field for p(x) over L; hence M L/L is
a finite normal extension.
The mutual positions of the fields can be illustrated this way:
ML
M
M ∩L
K
L
uu
u
uu
uu
u
uu
If σ is an automorphism in Gal(M L/L) the restriction σRes,M will be an automorphism in Gal(M/K) since σM M (use e.g. Theorem 3.27). The mapping
ϕ : Gal(M L/L) → Gal(M/K) defined by ϕσ = σRes,M is a homomorphism. Since σ
is uniquely determined by its value on α, the mapping ϕ is injective.
ϕ Gal(M L/L) is a subgroup of Gal(M/K) and is therefore by the fundamental
theorem of Galois theory determined by F (ϕ Gal(M L/L). According to the definitions we get
F (ϕ Gal(M L/L) = {m ∈ M |σ(m) = m ∀σ ∈ Gal(M L/L)} =
{m ∈ M |m ∈ F (Gal(M L/L) = L} = M ∩ L .
Therefore we get ϕ Gal(M L/L) = Gal(M/M ∩ L). The injectivity of ϕ yields the
isomorphism mentioned in the translation theorem and in particular that [M L : L] =
[M : M ∩ L].
To prove the second half of the translation theorem we have to show
1)
NL ∩ M = N
2)
(Λ ∩ M )L = Λ .
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ad 1)
M
ML
N
NL
M ∩L
K
L
uu
u
uu
uu
u
uu
It is clear that N L ∩ M N . Now M/N is normal, and since M L = M (N L) the
first half of the translation theorem yields: [M L : N L] = [M : N L ∩ M ]. Since
[M : M ∩ L] = [M L : L] we get: [N L ∩ M : M ∩ L] = [N L : L].
Let N = (M ∩ L)(β) for a suitable β in N ; then: [N : M ∩ L] =
degree(Irr(β, M ∩ L)) degree(Irr(β, L)) = [L(β) : L] = [N L : L]. From this we
conclude: [N : M ∩ L] [N L ∩ M : M ∩ L]. This taken together with N L ∩ M N
gives us [N : M ∩ L] = [N L ∩ M : M ∩ L] and thus N L ∩ M = N .
ad 2)
ML
M
Λ
M ∩ ΛL
LLL
LLL
LLL
(M ∩ Λ)L
M ∩L
K
L
qqq
q
q
qqq
q
q
qq
It is clear that Λ (M ∩ Λ)L. From the proof of 1) follows (with N = M ∩ Λ) that
[(M ∩ Λ)L : L] = [M ∩ Λ : M ∩ L]. According to the first half of the translation
theorem we get
[M : M ∩ Λ] = [M L : Λ] og [M : M ∩ L] = [M L : L],
and from theorem 2.47 in chap. II (the transitivity theorem) we then get
[M ∩ Λ : M ∩ L] = [Λ : L]
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Consequently [(M ∩ Λ)L : L] = [Λ : L]. Combining this with Λ (M ∩ Λ)L we
conclude Λ = (M ∩ Λ)L.
Theorem 3.47. Theorem about the compositum of finite normal extensions. Let L and M be finite normal extensions of a field K and assume that L
and M are contained in a common extension field. The compositum LM is a finite normal extension of K and the Galois group Gal(LM/K) is isomorphic to a
subgroup of the direct product Gal(L/K) × Gal(M/K). There is an isomorphism
Gal(LM/K) Gal(L/K) × Gal(M/K) if L ∩ M = K.
Proof. By Theorem 3.26 there exist separable polynomials f (x) and g(x) ∈ K[x] such
that L (resp. M ) is the splitting field over K for f (x) (resp. g(x)). Let α1 , . . . , αs
(resp. β1 , . . . , βt ) be the roots of f (x) (resp. g(x)). Then LM = K(α1 , . . . , αs , β1 , . . . , βt )
is the splitting field over K for f (x)g(x) and thus a finite normal extension of K,
(cf. Theorem 3.26). Lemma 3.4 implies that σL = L and σM = M for any σ in
Gal(LM/K). Hence there is a well defined homomorphic mapping Gal(LM/K) →
Gal(L/K) × Gal(M/K) determined by
σ ∈ Gal(LM/K) → (σRes,L , σRes,M ) .
Since σ is uniquely determined by its values on α1 , . . . , αs , β1 , . . . , βt the above mapping is injective.
If L ∩ M = K we conclude from the Translation theorem that
| Gal(LM/K)| = [LM : K] = [L : K][M : K] = | Gal(L/K) × Gal(M/K)|
so the injective mapping
Gal(LM/K) → Gal(L/K) × Gal(M/K)
becomes surjective and thus an isomorphism.
Corollary 3.48. Let M1 , . . . , Mt be finite normal extensions of K and assume that
for each i = 1, . . . , t the intersection of Mi and the compositum of the remaining
fields Mj , 1 t, i = j, is K. Then the compositum of the fields M1 , . . . , Mt is a
normal extension of K whose Galois group is Gal(M1 /K) × · · · × Gal(Mt /K).
Exercise 3.49. Construct a finite normal extension M/Q, whose Galois group is
Z/4Z × Z/2Z.
√
Example 3.50. Let b be a real algebraic number and i = −1. Applying the
translation theorem (with K = Q, M = Q(i) and L = Q(b)) we conclude that
[Q(i, b) : Q(b)] = [Q(i) : Q] = 2, hence [Q(i, b) : Q] = 2[Q(b) : Q]. To find the relation
between [Q(ib) : Q] and [Q(b) : Q] we have to distinguish between two cases:
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i) i ∈
/ Q(ib)
and
ii) i ∈ Q(ib).
In the case i) we use the translation theorem (K = Q, M = Q(i) and L = Q(ib)).
Since i ∈
/ Q(ib) the intersection Q(i) ∩ Q(ib) is Q; therefore [Q(i, b) : Q(ib)] =
[Q(i) : Q] = 2 and thus [Q(i, b) : Q(i)] = [Q(ib) : Q]. Hence [Q(b) : Q] = [Q(ib) : Q].
In the case ii) Q(i, b) = Q(ib) since i ∈ Q(ib). Therefore we get [Q(ib) : Q] =
2[Q(b) : Q].
√
Exercise 3.51. Let α be a complex number written as α = a + ib, where i = −1
and a and b are real numbers. Prove that α is an algebraic number if and only if a
and b are algebraic numbers.
Let n be the degree of α over Q.
Show that the degree of a is a most n(n − 1)/2 and that the degree of ib is at
most n(n − 1).
If i ∈ Q(ib) then the degree of b is at most n(n − 1)/2.
Exercise 3.52. Let α and β be two algebraic numbers. Is it true that Q(α) = Q(β)
implies that Q(iα) = Q(iβ)?