10.2 The Main Theorem of Galois Theory
1.
a. 
is the splitting field of x5 − 1 over 
, and 
is separable as char 
. By Example 4 §10.1, 
where σ(u) = u2 (because 
. The lattices are:
is the splitting field of x5 − 1 over , and is separable as char . By Example 4 §10.1, where σ(u) = u2 (because . The lattices are:
Thus H
is the only intermediate field, 
is Galois (as H 
G), and 
. We have σ2(u) = u4 = u−1 so σ2(u + u4) = (u + u4). Hence 
so, since 
, 
. Of course 
is the only intermediate field, is Galois (as H G), and . We have σ2(u) = u4 = u−1 so σ2(u + u4) = (u + u4). Hence so, since , . Of course
c. 
splits (x2 + 1)(x2 − 3) over 
, so it is a Galois extension. Clearly x2 + 1 is the minimal polynomial of i, and has roots {i, − i} in E ; and x2 − 3 is the minimal polynomial of 
and has roots 
It follows by Theorem 1 that 
where G = gal(E : Q) . Construct 
such that 
then extend σ0 to 
where σ(i) = − i. Hence σ 
G and 
Similarly, construct 
such that τ0(i) = i ; then extend τ0 to 
where 
Hence τ G and τ(i) = i. It follows easily that σ2 = ε and τ2 = ε, so 
Now consider 
Since 
and 
it follows that 
If H0 = 
σ 
, H1 = τ and H2 = στ , the lattices are
splits (x2 + 1)(x2 − 3) over , so it is a Galois extension. Clearly x2 + 1 is the minimal polynomial of i, and has roots {i, − i} in E ; and x2 − 3 is the minimal polynomial of and has roots It follows by Theorem 1 that where G = gal(E : Q) . Construct such that then extend σ0 to where σ(i) = − i. Hence σ G and Similarly, construct such that τ0(i) = i ; then extend τ0 to where Hence τ G and τ(i) = i. It follows easily that σ2 = ε and τ2 = ε, so Now consider Since and it follows that If H0 = σ , H1 = τ and H2 = στ , the lattices are
Here each 
is Galois (because Hi G) and 
Hence 
means 
; then 
because 
. Similarly 
Finally, 
so 
is Galois (because Hi G) and Hence means ; then because . Similarly Finally, so
e. 
, 
; so 
by Theorem 6 §10.1. Then E is the splitting field of x4 − 2 = (x − u)(x + u)(x − iu)(x + iu) so 
is Galois. By Exercise 13 §10.1, 
where o(σ) = 4, o(τ) = 2, στσ = τ and 
The lattice diagrams are:
, ; so by Theorem 6 §10.1. Then E is the splitting field of x4 − 2 = (x − u)(x + u)(x − iu)(x + iu) so is Galois. By Exercise 13 §10.1, where o(σ) = 4, o(τ) = 2, στσ = τ and The lattice diagrams are:
Here σ2 , σ2, τ , σ and σ2, τσ are normal in G (because σ2 = Z(G)), so 
, 
, 
and 
are Galois.
σ2 , σ2, τ , σ and σ2, τσ are normal in G (because σ2 = Z(G)), so , , and are Galois.
1. 
: Hence σ2(i) = i; σ(u2) = − u2 so
: Hence σ2(i) = i; σ(u2) = − u2 so
Thus 
. Also 
and so we are done because 
. Now 
by Theorem 6 §10.1.
. Also and so we are done because . Now by Theorem 6 §10.1.
2. 
: We have 
; and 
: We have ; and
3. 
: We have τ(u2) = u2 and σ2(u2) = u2 (in (1)) so 
. But 
and 
: We have τ(u2) = u2 and σ2(u2) = u2 (in (1)) so . But and
4. 
: Check 
and 
: Check and
5. 
: Clearly 
and 
as x4 − 2 is irreducible.
: Clearly and as x4 − 2 is irreducible.
6. 
: We have 
as τσ(iu) = − u and τσ(u) = − iu. This is equality as x4 + 8 is the minimum polynomial of 
: We have as τσ(iu) = − u and τσ(u) = − iu. This is equality as x4 + 8 is the minimum polynomial of
7. 
: Compute 
; 
has minimum polynomial x4 − 2.
: Compute ; has minimum polynomial x4 − 2.
8. 
: We have 
because τσ3(u) = iu and τσ3(iu) = u. The minimum polynomial of 
is x4 + 8.
: We have because τσ3(u) = iu and τσ3(iu) = u. The minimum polynomial of is x4 + 8.
2.
a. If G = gal(E : F) then |G| = p2 implies G is abelian, so 
or G ≅ Cp × Cp by Theorem 7, §8.2. If 
, the lattices are:
or G ≅ Cp × Cp by Theorem 7, §8.2. If , the lattices are:
Here σ2 ⊇ F is Galois because σ2 G, and [ σ2 : F] = 2 . If G≅ Cp × Cp = σ, τ the lattices are:
σ2 ⊇ F is Galois because σ2 G, and [ σ2 : F] = 2 . If G≅ Cp × Cp = σ, τ the lattices are:
Each of σ ⊇ F and τ ⊇ F are Galois because [ σ : F] = 
and [ τ : F].
σ ⊇ F and τ ⊇ F are Galois because [ σ : F] = and [ τ : F].
3. By Example 6 §10.1, 
where σ : E → F is the Frobenius automorphism given by σ(u) = up. By the Dedekind-Artin theorem [E : G] = |G| = n. But 
so, since 
, 
. Thus 
is a Galois extension by Lemma 4.
where σ : E → F is the Frobenius automorphism given by σ(u) = up. By the Dedekind-Artin theorem [E : G] = |G| = n. But so, since , . Thus is a Galois extension by Lemma 4.
Now the (inverted) lattice of subgroups of G = σ , when o(σ) = 12, is shown below at the right. Write Ek = σk = {u E 
σk(u) = u} for each divisor k of n. Then the subfield lattice is shown below on the left.
σ , when o(σ) = 12, is shown below at the right. Write Ek = σk = {u E σk(u) = u} for each divisor k of n. Then the subfield lattice is shown below on the left.
Note that
so |Ek| = pk, and Ek ≅ GF(pk).
4. a. 
, σi X, 
. Thus u H is fixed by all τ H if and only if σ(u) = u for all σ X.
, σi X, . Thus u H is fixed by all τ H if and only if σ(u) = u for all σ X.
5. a. Let 
. Then
. Then
If char F = 2, this always holds and K = EG = E. Thus t K so m = x − t. If char F ≠ 2, write h(t) = f(t)g(− t). Then h(t) = h(− t) so h(t) = k(t2) for some polynomial k (because char F ≠ 2) . Thus f(t)g(− t) = k(t2) . Similarly and g(t)g(− t) = l(t2) for some polynomial l. Thus 
, from which K = F(t2) . It follows that m = x2 − t2. Note that this shows [E : K] = 2 = |G| in this case, as the Dedekind-Artin theorem guarantees.
K so m = x − t. If char F ≠ 2, write h(t) = f(t)g(− t). Then h(t) = h(− t) so h(t) = k(t2) for some polynomial k (because char F ≠ 2) . Thus f(t)g(− t) = k(t2) . Similarly and g(t)g(− t) = l(t2) for some polynomial l. Thus , from which K = F(t2) . It follows that m = x2 − t2. Note that this shows [E : K] = 2 = |G| in this case, as the Dedekind-Artin theorem guarantees.
7. a. K′ is a subgroup of the abelian group G, so it is normal. Then the main theorem applies.
9.
a. If H → H is onto, and K is an intermediate field, then K = H for a subgroup H, so K′ = H′ = H′ = K. Thus K is closed. Conversely, if all intermediate fields K are closed then K = K′ is the image of K′, and the map is onto.
is onto, and K is an intermediate field, then K = H for a subgroup H, so K′ = H′ = H′ = K. Thus K is closed. Conversely, if all intermediate fields K are closed then K = K′ is the image of K′, and the map is onto.
11. Write G = gal(E : F) = {σ autE σ fixes F}. Since K ⊇ F we have
autE σ fixes F}. Since K ⊇ F we have
Thus:
| E ⊇ K is Galois |  |
K is the set of elements of E fixed by gal(E : K) |
|
K = {u E σ(u) = u for all σ K′} |
|
|
K = K′ |
|
|
K is closed in E ⊇ F . |
13. If [E : K] = 6 then 
, so K′ is a subgroup of A4 of order 6. There is no such subgroup (Exercise 34 §2.6).
, so K′ is a subgroup of A4 of order 6. There is no such subgroup (Exercise 34 §2.6).
15. a. By its definition, K 
L is the smallest intermediate field containing both K and L. Since the Galois connection K → K′ is order reversing, (K L)′ is the largest subgroup of G = gal(E : F) contained in both K′ and L′; that is (K L)′ = K′ ∩ L′.
L is the smallest intermediate field containing both K and L. Since the Galois connection K → K′ is order reversing, (K L)′ is the largest subgroup of G = gal(E : F) contained in both K′ and L′; that is (K L)′ = K′ ∩ L′.
17. If K = σ(K1) let 
. If 
write 
, u K1. Then
. If write , u K1. Then
so σK′σ−1 ⊆ K′. On the other hand, if μ K′ then 
for all 
, so μ[σ(u)] = σ(u) for all u K1. Hence 
, so 
. This proves 
K′ then for all , so μ[σ(u)] = σ(u) for all u K1. Hence , so . This proves
Conversely, assume 
, that is 
. If u K1 then μ(u) = u for all 
, so σ−1λσ(u) = u for all λ K′, that is λ[σ(u)] = σ(u) for all λ K′, that is σ(u) K′. Since E ⊇ F is Galois, K′ = K, so this shows σ(u) K, u σ−1(K). Hence K1 ⊆ σ−1(K), so σ(K1) ⊆ K. Similarly K ⊆ σ(K1).
, that is . If u K1 then μ(u) = u for all , so σ−1λσ(u) = u for all λ K′, that is λ[σ(u)] = σ(u) for all λ K′, that is σ(u) K′. Since E ⊇ F is Galois, K′ = K, so this shows σ(u) K, u σ−1(K). Hence K1 ⊆ σ−1(K), so σ(K1) ⊆ K. Similarly K ⊆ σ(K1).
19. a. Let E = F(u1, u2, . . ., um) where X = {u1, u2, . . ., um} is the set of distinct roots of f in E. If σ G then σ(ui) X for all i so σ : X → X is one-to-one (hence onto). Thus σ induces a permutation of these roots:
G then σ(ui) X for all i so σ : X → X is one-to-one (hence onto). Thus σ induces a permutation of these roots:
Now the map 
is a group homomorphism G → Sm because
is a group homomorphism G → Sm because
for all σ, τ in G. Moreover 
means σ(ui) = ui for all i, so σ = ε in G by Theorem 3 §10.1. Thus 
is an embedding.
means σ(ui) = ui for all i, so σ = ε in G by Theorem 3 §10.1. Thus is an embedding.
20.
a. If 
then, given
then, given
But τσ traces all of G as σ does, so 
. This is true for all τ G, so 
. The proof that T(u) is in F is analogous.
. This is true for all τ G, so . The proof that T(u) is in F is analogous.
c. If K ⊇ F is Galois and H = gal(K : F), then |H| = n by Theorem 3, Corollary 3. Now p splits in K and has distinct roots u1, u2, . . ., um (Theorem 3). Hence p = (x − u1)(x − u2) 
(x − un). For each i there exists σi H such that σi(u) = ui, so p = ∏ σG(x − σ(u)). Thus the coefficients of xn−1 and 1 are, respectfully
(x − un). For each i there exists σi H such that σi(u) = ui, so p = ∏ σG(x − σ(u)). Thus the coefficients of xn−1 and 1 are, respectfully
21. If 
define 
for τ G. Thus fτ = ∏ σG[x − τσ(u)] = f because τσ runs through G as σ does. It follows that 
for all τ G, whence 
for each j. Thus f F[x]. Since f(u) = 0, p|f where p is the minimal polynomial of u over F. Write f = pmg where p and g are relatively prime in F[x]. If g ≠ 1, let q be an irreducible factor of g. Since q|f, 0 = q[σ(u)] = σ[q(u)] for some σ G, so q(u) = 0. Since p(u) = 0 this is a contradiction because gcd(p, q) = 1 in F[x]. In fact, 1 = pr + qs gives 1 = 0 when we substitute x = u.
define for τ G. Thus fτ = ∏ σG[x − τσ(u)] = f because τσ runs through G as σ does. It follows that for all τ G, whence for each j. Thus f F[x]. Since f(u) = 0, p|f where p is the minimal polynomial of u over F. Write f = pmg where p and g are relatively prime in F[x]. If g ≠ 1, let q be an irreducible factor of g. Since q|f, 0 = q[σ(u)] = σ[q(u)] for some σ G, so q(u) = 0. Since p(u) = 0 this is a contradiction because gcd(p, q) = 1 in F[x]. In fact, 1 = pr + qs gives 1 = 0 when we substitute x = u.
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