Definitions etc

Question 1

If E is a field and F is a subfield of E, we say that…

Answer 1

E is an extension of F.

Notation: E : F.

Question 2

An extension E : F is called finite if…

Answer 2

E is a finite dimensional vector space over F. In that case the degree of the extension is the dimension of E as a vector space over F. Notation: (E : F) = dimF E.

Question 3

State the Tower Law

Answer 3

Let F, B, E be fields with F ⊂B ⊂ E such that B : F and E :B are finite extensions.

Then E : F is a finite extension and (E : F) = (B : F)(E : B).

Question 4

If F ⊂ F1 ⊂ F2 ⊂ · · · ⊂ Fn is a tower of finite field

extensions, then

Answer 4

Fn : F is a finite extension and (Fn : F) = (F1 : F)(F2 : F1). . .(Fn : Fn−1).

Question 5

State Gauss’ Lemma

Answer 5

A polynomial with integer coefficients is irreducible

over Q if and only if it is irreducible over Z

Question 6

State Eisenstein’s Criterion

Answer 6

Let f = a0 + a1x + · · · + anx n ∈ Z[x]
with n ≥ 1 and an ̸= 0, and let p be a prime such that
(i) p /| an,
(ii) p | ai (i = 0, 1, . . . , n − 1,)
(iii) p^2 /| a0.
Then f is irreducible over Q

Question 7

An element α ∈ E is called algebraic over F, if…

Answer 7

if α is a zero of a polynomial f ∈ F[x].

Question 8

An monomorphism from a field E into a field E’ is

Answer 8

is an injective map σ : E → E′ such that for all α, β ∈ E
σ(α + β) = σ(α) + σ(β)
and
σ(αβ) = σ(α)σ(β).
If σ is bijective, it is called an isomorphism. An isomorphism from E
to E is called an automorphism of E.

Question 9

Theorem 7 (L. Kronecker). Let F be a field. For any non-constant polynomial f ∈ F[x] there exists…

Answer 9

there exists an extension E of F in which f has a root.

Question 10

Theorem 8. Let σ : F → F′ be an isomorphism of fields, let f be an irreducible polynomial in F[x] and let f′ be the corresponding polynomial in F′[x] (obtained by applying σ to the coefficients of f).
If E = F(β) and E′ = F′(β′), where f(β) = 0 in E and f′(β′) = 0 in E′, then…

Answer 10

then σ can be extended to an isomorphism between E and E′ such that σ(β) = β′.

Question 11

If F, B and E are fields with F ⊆ B ⊆ E, then we call B …

Answer 11

an intermediate field of the extension E : F.

Question 12

Let p ∈ F[x] with deg p ≥ 1. An extension E of F in

which p can be factored into linear factors is called a splitting field for p over F if…

Answer 12

if such factorization cannot be carried out in any intermediate
field.

Question 13

Theorem 9. Let F be a field, p ∈ F[x] with deg p = n ≥ 1. Then there exists …

Answer 13

a splitting field of p over F.

Question 14

What is Γ(E)?

Answer 14

Γ(E) = the group of automorphisms of E

Question 15

Define Galois group Γ(E : F)

Answer 15

If E is an extension field of F, then the Galois group Γ(E : F) is the set of all automorphisms of E that leave F fixed (it is also a subgroup of Γ(E) )

An F-automorphism the field E is a bijective map σ : E→ E such that for all x, y ∈ E one has
σ(x + y) = σ(x) + σ(y) and σ(xy) = σ(x)σ(y),
and such that σ(x) = x for all x ∈ F. The F-automorphisms of E form a group under composition of maps, and this is the Galois group Γ(E : F)

Question 16

Define the fixed field Φ(H)

Answer 16

The field of fixed points for H in E is called the fixed field for H.

The fixed field Φ(H) of H in E is defined as
Φ(H) = {x ∈ E ; σ(x) = x for all σ ∈ H}

Question 17

Define a solvable group

Answer 17

A group G is called solvable (or soluble) if there is a finite chain of subgroups

{1} = Hr ≤ Hr−1 ≤ · · · ≤ H2 ≤ H1 ≤ H0 = G

such that each Hi is normal in Hi−1 and the factor group Hi−1/Hi is abelian (i=1,2,. . . ,r).

Question 18

Explain what is meant when saying “solvable by radicals”

Answer 18

A polynomial equation f(x) = 0 with f ∈ F[x] is said to be solvable by radicals if the splitting field for f is contained in some radical extension of F.

Question 19

Define a radical extension

Answer 19

An extension field E of F is called an extension by radicals or radical
extension if there exist intermediate fields

F = C0 ⊆ C1 ⊆ C2 ⊆ · · · ⊆ Cr = E

such that Ci = Ci−1(αi) (i = 1, . . . , r) where αi
is a root of a polynomial x^ni − ai with ai ∈ Ci−1.

Question 20

State Galois’ Theorem

Answer 20

Let F be a field of characteristic zero and let f ∈ F[x]. Then the polynomial equation f(x) = 0 is solvable by radicals if and only if the group Γ(f) is solvable.

Question 21

What are the primitive nth roots of unity?

Answer 21

The generators of ⟨ϵ⟩ are called the primitive n-th roots of unity. They are the elements of order n in this cyclic group, i.e. the powers ϵ^k with 1 ≤ k ≤ n − 1 such that (k, n) = 1.

Question 22

Explain what is a Kummer extension of F

Answer 22

If a field F contains a primitive n-th root of unity, then any splitting field of a polynomial

(xn − a1)(xn − a2)· · ·(xn − ar), (a1, . . . , ar ∈ F \ {0})

is called a Kummer extension of F (or Kummer field for short).

Question 23

Define the Galois group Γ(f) of the polynomial f ∈ Q[x]

Answer 23

The Galois group of the polynomial f ∈ Q[x] is defined as
Γ(f) = Γ(E : Q)
where E is the splitting field of f over Q.

Question 24

The Fundamental Theorem of Galois Theory

Answer 24

Let E be a normal extension of a field F with Galois group G = Γ(E : F).

Then

(i) There is a one-to-one correspondence between the set H of all subgroups of G and the set B of all intermediate fields of E : F given by H → Φ(H) (H ∈ H) with inverse B → Γ(E : B) (B ∈ B).
(ii) For each intermediate field B ∈ B we have (E : B) = |Γ(E : B)| and (B : F) = [G : Γ(E : B)]. In particular, |G| = (E : F).
(iii) An intermediate field B is a normal extension of F if and only if Γ(E : B) is a normal subgroup of G. In this case we have Γ(B : F) ∼= G/Γ(E : B).