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logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Normal Field Extensions

Bernd Schr¨

• der

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Introduction

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Introduction

• 1. Ultimately, we want to compute Galois groups of splitting

fields.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Introduction

• 1. Ultimately, we want to compute Galois groups of splitting

fields.

• 2. To do so, we need some more properties of splitting fields

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Introduction

• 1. Ultimately, we want to compute Galois groups of splitting

fields.

• 2. To do so, we need some more properties of splitting fields,

and we are especially interested in properties that connect to Galois groups.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Introduction

• 1. Ultimately, we want to compute Galois groups of splitting

fields.

• 2. To do so, we need some more properties of splitting fields,

and we are especially interested in properties that connect to Galois groups.

• 3. Demanding characteristic 0 will make some proofs a bit

easier.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a finite extension of F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a finite extension of F. Then the following are equivalent.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a finite extension of F. Then the following are equivalent.

• 1. E is the splitting field for a polynomial f of positive degree

in F[x].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a finite extension of F. Then the following are equivalent.

• 1. E is the splitting field for a polynomial f of positive degree

in F[x].

• 2. Every irreducible polynomial p ∈ F[x] that has one zero in

E actually splits in E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a finite extension of F. Then the following are equivalent.

• 1. E is the splitting field for a polynomial f of positive degree

in F[x].

• 2. Every irreducible polynomial p ∈ F[x] that has one zero in

E actually splits in E.

• 3. F is the fixed field of G(E/F).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a finite extension of F. Then the following are equivalent.

• 1. E is the splitting field for a polynomial f of positive degree

in F[x].

• 2. Every irreducible polynomial p ∈ F[x] that has one zero in

E actually splits in E.

• 3. F is the fixed field of G(E/F).

Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a finite extension of F. Then the following are equivalent.

• 1. E is the splitting field for a polynomial f of positive degree

in F[x].

• 2. Every irreducible polynomial p ∈ F[x] that has one zero in

E actually splits in E.

• 3. F is the fixed field of G(E/F).
• Proof. We will prove “1⇒2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a finite extension of F. Then the following are equivalent.

• 1. E is the splitting field for a polynomial f of positive degree

in F[x].

• 2. Every irreducible polynomial p ∈ F[x] that has one zero in

E actually splits in E.

• 3. F is the fixed field of G(E/F).
• Proof. We will prove “1⇒2 ⇒3

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a finite extension of F. Then the following are equivalent.

• 1. E is the splitting field for a polynomial f of positive degree

in F[x].

• 2. Every irreducible polynomial p ∈ F[x] that has one zero in

E actually splits in E.

• 3. F is the fixed field of G(E/F).
• Proof. We will prove “1⇒2 ⇒3 ⇒1”.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn). Thus E = E(θ1) is the splitting field of f over F(θ1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn). Thus E = E(θ1) is the splitting field of f over F(θ1) (the splitting field must contain E and θ1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn). Thus E = E(θ1) is the splitting field of f over F(θ1) (the splitting field must contain E and θ1), and E(θ2) is the splitting field of f over F(θ2)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn). Thus E = E(θ1) is the splitting field of f over F(θ1) (the splitting field must contain E and θ1), and E(θ2) is the splitting field of f over F(θ2) (the splitting field must contain E and θ2).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn). Thus E = E(θ1) is the splitting field of f over F(θ1) (the splitting field must contain E and θ1), and E(θ2) is the splitting field of f over F(θ2) (the splitting field must contain E and θ2). There is an isomorphism Φ : E → E(θ2) that continues ϕ.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn). Thus E = E(θ1) is the splitting field of f over F(θ1) (the splitting field must contain E and θ1), and E(θ2) is the splitting field of f over F(θ2) (the splitting field must contain E and θ2). There is an isomorphism Φ : E → E(θ2) that continues ϕ. In particular, Φ fixes F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn). Thus E = E(θ1) is the splitting field of f over F(θ1) (the splitting field must contain E and θ1), and E(θ2) is the splitting field of f over F(θ2) (the splitting field must contain E and θ2). There is an isomorphism Φ : E → E(θ2) that continues ϕ. In particular, Φ fixes F. But then [E : F] = [E(θ2) : F]

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn). Thus E = E(θ1) is the splitting field of f over F(θ1) (the splitting field must contain E and θ1), and E(θ2) is the splitting field of f over F(θ2) (the splitting field must contain E and θ2). There is an isomorphism Φ : E → E(θ2) that continues ϕ. In particular, Φ fixes F. But then [E : F] = [E(θ2) : F], which implies that E = E(θ2).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn). Thus E = E(θ1) is the splitting field of f over F(θ1) (the splitting field must contain E and θ1), and E(θ2) is the splitting field of f over F(θ2) (the splitting field must contain E and θ2). There is an isomorphism Φ : E → E(θ2) that continues ϕ. In particular, Φ fixes F. But then [E : F] = [E(θ2) : F], which implies that E = E(θ2). Because θ2 was an arbitrary root of p, we conclude that all roots of p are in E

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“1⇒2”). Let E be the splitting field for a polynomial f ∈ F[x] of positive degree. Let µ1,...,µn ∈ E\F be the roots

• f f that are not in F. Then E = F(µ1,...,µn).

Let p ∈ F[x] be an arbitrary irreducible polynomial with a root θ1 ∈ E. Let θ2 = θ1 be another arbitrary root of p. Then θ2 lies in some extension of F, but, at this stage, we do not know if it lies in E. There is an isomorphism ϕ : F(θ1) → F(θ2) so that ϕ(θ1) = θ2 and ϕ fixes F. Moreover, any field in which f splits must contain µ1,...,µn, so it must contain E = F(µ1,...,µn). Thus E = E(θ1) is the splitting field of f over F(θ1) (the splitting field must contain E and θ1), and E(θ2) is the splitting field of f over F(θ2) (the splitting field must contain E and θ2). There is an isomorphism Φ : E → E(θ2) that continues ϕ. In particular, Φ fixes F. But then [E : F] = [E(θ2) : F], which implies that E = E(θ2). Because θ2 was an arbitrary root of p, we conclude that all roots of p are in E, that is p splits in E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF}

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E. There is a

θ ∈ E = F(µ1,...,µn) so that E = F(µ1,...,µn) = F(θ)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E. There is a

θ ∈ E = F(µ1,...,µn) so that E = F(µ1,...,µn) = F(θ) and θ is a zero of an irreducible polynomial p ∈ F[x].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E. There is a

θ ∈ E = F(µ1,...,µn) so that E = F(µ1,...,µn) = F(θ) and θ is a zero of an irreducible polynomial p ∈ F[x]. (We used that F has characteristic 0 here.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E. There is a

θ ∈ E = F(µ1,...,µn) so that E = F(µ1,...,µn) = F(θ) and θ is a zero of an irreducible polynomial p ∈ F[x]. (We used that F has characteristic 0 here.) Then

• 1,θ,...,θ deg(p)−1

is a basis for E = F(θ) over F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E. There is a

θ ∈ E = F(µ1,...,µn) so that E = F(µ1,...,µn) = F(θ) and θ is a zero of an irreducible polynomial p ∈ F[x]. (We used that F has characteristic 0 here.) Then

• 1,θ,...,θ deg(p)−1

is a basis for E = F(θ) over F. Moreover deg(p) = [E : F] = n.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E. There is a

θ ∈ E = F(µ1,...,µn) so that E = F(µ1,...,µn) = F(θ) and θ is a zero of an irreducible polynomial p ∈ F[x]. (We used that F has characteristic 0 here.) Then

• 1,θ,...,θ deg(p)−1

is a basis for E = F(θ) over F. Moreover deg(p) = [E : F] = n. By assumption, p splits in E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E. There is a

θ ∈ E = F(µ1,...,µn) so that E = F(µ1,...,µn) = F(θ) and θ is a zero of an irreducible polynomial p ∈ F[x]. (We used that F has characteristic 0 here.) Then

• 1,θ,...,θ deg(p)−1

is a basis for E = F(θ) over F. Moreover deg(p) = [E : F] = n. By assumption, p splits in E. Let θ1 := θ and let θ2,...,θn be the

• ther zeros of p

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E. There is a

θ ∈ E = F(µ1,...,µn) so that E = F(µ1,...,µn) = F(θ) and θ is a zero of an irreducible polynomial p ∈ F[x]. (We used that F has characteristic 0 here.) Then

• 1,θ,...,θ deg(p)−1

is a basis for E = F(θ) over F. Moreover deg(p) = [E : F] = n. By assumption, p splits in E. Let θ1 := θ and let θ2,...,θn be the

• ther zeros of p (which must all be distinct).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E. There is a

θ ∈ E = F(µ1,...,µn) so that E = F(µ1,...,µn) = F(θ) and θ is a zero of an irreducible polynomial p ∈ F[x]. (We used that F has characteristic 0 here.) Then

• 1,θ,...,θ deg(p)−1

is a basis for E = F(θ) over F. Moreover deg(p) = [E : F] = n. By assumption, p splits in E. Let θ1 := θ and let θ2,...,θn be the

• ther zeros of p (which must all be distinct).

Let j ∈ {2,...,n}.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3”). In case E = F, we have G(F/F) = {idF} and clearly E = F is the fixed field of G(F/F) = {idF}. So for the remainder of this part, we can assume that E = F. Let {µ1,...,µn} be a basis of E over F. Every µk must be a zero of a polynomial in F[x], because otherwise

• µj

k : j ∈ N0

• would be an infinite F-linearly independent set in E. There is a

θ ∈ E = F(µ1,...,µn) so that E = F(µ1,...,µn) = F(θ) and θ is a zero of an irreducible polynomial p ∈ F[x]. (We used that F has characteristic 0 here.) Then

• 1,θ,...,θ deg(p)−1

is a basis for E = F(θ) over F. Moreover deg(p) = [E : F] = n. By assumption, p splits in E. Let θ1 := θ and let θ2,...,θn be the

• ther zeros of p (which must all be distinct).

Let j ∈ {2,...,n}. There is an isomorphism ϕ : F(θ1) → F(θj) so that ϕ(θ1) = θj and ϕ fixes F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• =

n−1

∑

i=0

fiΦ(θ)i

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• =

n−1

∑

i=0

fiΦ(θ)i =

n−1

∑

i=0

fiθ i

j .

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• =

n−1

∑

i=0

fiΦ(θ)i =

n−1

∑

i=0

fiθ i

j .

Let f(x) := −e+

n−1

∑

i=0

fixi.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• =

n−1

∑

i=0

fiΦ(θ)i =

n−1

∑

i=0

fiθ i

j .

Let f(x) := −e+

n−1

∑

i=0

• fixi. Then f ∈ E[x] and deg(f) ≤ n−1.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• =

n−1

∑

i=0

fiΦ(θ)i =

n−1

∑

i=0

fiθ i

j .

Let f(x) := −e+

n−1

∑

i=0

• fixi. Then f ∈ E[x] and deg(f) ≤ n−1. But

the above shows that f has the n distinct zeros θ1,...,θn.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• =

n−1

∑

i=0

fiΦ(θ)i =

n−1

∑

i=0

fiθ i

j .

Let f(x) := −e+

n−1

∑

i=0

• fixi. Then f ∈ E[x] and deg(f) ≤ n−1. But

the above shows that f has the n distinct zeros θ1,...,θn. This is

• nly possible if f = 0.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• =

n−1

∑

i=0

fiΦ(θ)i =

n−1

∑

i=0

fiθ i

j .

Let f(x) := −e+

n−1

∑

i=0

• fixi. Then f ∈ E[x] and deg(f) ≤ n−1. But

the above shows that f has the n distinct zeros θ1,...,θn. This is

• nly possible if f = 0. But then −e+f0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• =

n−1

∑

i=0

fiΦ(θ)i =

n−1

∑

i=0

fiθ i

j .

Let f(x) := −e+

n−1

∑

i=0

• fixi. Then f ∈ E[x] and deg(f) ≤ n−1. But

the above shows that f has the n distinct zeros θ1,...,θn. This is

• nly possible if f = 0. But then −e+f0 = f(0)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• =

n−1

∑

i=0

fiΦ(θ)i =

n−1

∑

i=0

fiθ i

j .

Let f(x) := −e+

n−1

∑

i=0

• fixi. Then f ∈ E[x] and deg(f) ≤ n−1. But

the above shows that f has the n distinct zeros θ1,...,θn. This is

• nly possible if f = 0. But then −e+f0 = f(0) = 0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“2⇒3” concl.). Moreover, (note that E = F(θ) is the splitting field of p) there is an isomorphism Φ : E → E that continues ϕ. In particular, Φ fixes F and Φ(θ1) = θj. Now let e ∈ E be fixed by G(E/F). Because E = F(θ) we have e =

n−1

∑

i=0

fiθ i, and because there is a Φ ∈ G(E/F) with Φ(θ) = θj, we also have e = Φ(e) = Φ

• n−1

∑

i=0

fiθ i

• =

n−1

∑

i=0

fiΦ(θ)i =

n−1

∑

i=0

fiθ i

j .

Let f(x) := −e+

n−1

∑

i=0

• fixi. Then f ∈ E[x] and deg(f) ≤ n−1. But

the above shows that f has the n distinct zeros θ1,...,θn. This is

• nly possible if f = 0. But then −e+f0 = f(0) = 0, that is,

e = f0 ∈ F, which was to be proved.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”). In case E = F, E is the splitting field for f(x) = x over F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”). In case E = F, E is the splitting field for f(x) = x over F. So for the remainder of this part, we can assume that E = F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”). In case E = F, E is the splitting field for f(x) = x over F. So for the remainder of this part, we can assume that E = F. For each a ∈ E let a1,...,an be the elements of

• σ(a) : σ ∈ G(E/F)
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”). In case E = F, E is the splitting field for f(x) = x over F. So for the remainder of this part, we can assume that E = F. For each a ∈ E let a1,...,an be the elements of

• σ(a) : σ ∈ G(E/F)
• and let

fa(x) :=

n

∏

j=1

(x−aj)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”). In case E = F, E is the splitting field for f(x) = x over F. So for the remainder of this part, we can assume that E = F. For each a ∈ E let a1,...,an be the elements of

• σ(a) : σ ∈ G(E/F)
• and let

fa(x) :=

n

∏

j=1

(x−aj) =

n

∑

k=0

bkxk ∈ E[x].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”). In case E = F, E is the splitting field for f(x) = x over F. So for the remainder of this part, we can assume that E = F. For each a ∈ E let a1,...,an be the elements of

• σ(a) : σ ∈ G(E/F)
• and let

fa(x) :=

n

∏

j=1

(x−aj) =

n

∑

k=0

bkxk ∈ E[x]. Let σ ∈ G(E/F).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”). In case E = F, E is the splitting field for f(x) = x over F. So for the remainder of this part, we can assume that E = F. For each a ∈ E let a1,...,an be the elements of

• σ(a) : σ ∈ G(E/F)
• and let

fa(x) :=

n

∏

j=1

(x−aj) =

n

∑

k=0

bkxk ∈ E[x]. Let σ ∈ G(E/F). The function ˜ σ : E[x] → E[x] defined by ˜ σ

• n

∑

k=0

ckxk

• :=

n

∑

k=0

σ(ck)xk is an isomorphism

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”). In case E = F, E is the splitting field for f(x) = x over F. So for the remainder of this part, we can assume that E = F. For each a ∈ E let a1,...,an be the elements of

• σ(a) : σ ∈ G(E/F)
• and let

fa(x) :=

n

∏

j=1

(x−aj) =

n

∑

k=0

bkxk ∈ E[x]. Let σ ∈ G(E/F). The function ˜ σ : E[x] → E[x] defined by ˜ σ

• n

∑

k=0

ckxk

• :=

n

∑

k=0

σ(ck)xk is an isomorphism (good exercise).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”). In case E = F, E is the splitting field for f(x) = x over F. So for the remainder of this part, we can assume that E = F. For each a ∈ E let a1,...,an be the elements of

• σ(a) : σ ∈ G(E/F)
• and let

fa(x) :=

n

∏

j=1

(x−aj) =

n

∑

k=0

bkxk ∈ E[x]. Let σ ∈ G(E/F). The function ˜ σ : E[x] → E[x] defined by ˜ σ

• n

∑

k=0

ckxk

• :=

n

∑

k=0

σ(ck)xk is an isomorphism (good exercise). Let a ∈ E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1”). In case E = F, E is the splitting field for f(x) = x over F. So for the remainder of this part, we can assume that E = F. For each a ∈ E let a1,...,an be the elements of

• σ(a) : σ ∈ G(E/F)
• and let

fa(x) :=

n

∏

j=1

(x−aj) =

n

∑

k=0

bkxk ∈ E[x]. Let σ ∈ G(E/F). The function ˜ σ : E[x] → E[x] defined by ˜ σ

• n

∑

k=0

ckxk

• :=

n

∑

k=0

σ(ck)xk is an isomorphism (good exercise). Let a ∈ E. Because σ permutes the a1,...,an we infer

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k. (We’re “sort of” using that F has characteristic 0 here.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k. (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G(E/F) was arbitrary, all bk are in F, the fixed field of G(E/F).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k. (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G(E/F) was arbitrary, all bk are in F, the fixed field of G(E/F). By definition, fa splits in E

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k. (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G(E/F) was arbitrary, all bk are in F, the fixed field of G(E/F). By definition, fa splits in E and, because a = id(a) ∈ {a1,...,an}, we have fa(a) = 0.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k. (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G(E/F) was arbitrary, all bk are in F, the fixed field of G(E/F). By definition, fa splits in E and, because a = id(a) ∈ {a1,...,an}, we have fa(a) = 0. Let v1,...,vm be a basis of E over F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k. (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G(E/F) was arbitrary, all bk are in F, the fixed field of G(E/F). By definition, fa splits in E and, because a = id(a) ∈ {a1,...,an}, we have fa(a) = 0. Let v1,...,vm be a basis of E over F. Then f :=

m

∏

j=1

fvk ∈ F[x] splits in E

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k. (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G(E/F) was arbitrary, all bk are in F, the fixed field of G(E/F). By definition, fa splits in E and, because a = id(a) ∈ {a1,...,an}, we have fa(a) = 0. Let v1,...,vm be a basis of E over F. Then f :=

m

∏

j=1

fvk ∈ F[x] splits in E and f(vk) = 0 for all k.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k. (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G(E/F) was arbitrary, all bk are in F, the fixed field of G(E/F). By definition, fa splits in E and, because a = id(a) ∈ {a1,...,an}, we have fa(a) = 0. Let v1,...,vm be a basis of E over F. Then f :=

m

∏

j=1

fvk ∈ F[x] splits in E and f(vk) = 0 for all k. Hence the splitting field of f must contain E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k. (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G(E/F) was arbitrary, all bk are in F, the fixed field of G(E/F). By definition, fa splits in E and, because a = id(a) ∈ {a1,...,an}, we have fa(a) = 0. Let v1,...,vm be a basis of E over F. Then f :=

m

∏

j=1

fvk ∈ F[x] splits in E and f(vk) = 0 for all k. Hence the splitting field of f must contain E. So E is the splitting field of f.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (“3⇒1” concl.).

n

∑

k=0

σ(bk)xk = ˜ σ(fa) = ˜ σ

• n

∏

j=1

(x−aj)

• =

n

∏

j=1

• x− ˜

σ(aj)

• =

n

∏

j=1

(x−aj) = fa =

n

∑

k=0

bkxk, so σ(bk) = bk for all k. (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G(E/F) was arbitrary, all bk are in F, the fixed field of G(E/F). By definition, fa splits in E and, because a = id(a) ∈ {a1,...,an}, we have fa(a) = 0. Let v1,...,vm be a basis of E over F. Then f :=

m

∏

j=1

fvk ∈ F[x] splits in E and f(vk) = 0 for all k. Hence the splitting field of f must contain E. So E is the splitting field of f.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Definition.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Definition. Let (F,+,·) be a field and let E be a finite extension
• f F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Definition. Let (F,+,·) be a field and let E be a finite extension
• f F. Then E is called a normal extension of F iff every

irreducible polynomial p ∈ F[x] that has one zero in E actually splits in E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Definition. Let (F,+,·) be a field and let E be a finite extension
• f F. Then E is called a normal extension of F iff every

irreducible polynomial p ∈ F[x] that has one zero in E actually splits in E. Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Definition. Let (F,+,·) be a field and let E be a finite extension
• f F. Then E is called a normal extension of F iff every

irreducible polynomial p ∈ F[x] that has one zero in E actually splits in E.

• Example. R is not a normal extension of Q.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Definition. Let (F,+,·) be a field and let E be a finite extension
• f F. Then E is called a normal extension of F iff every

irreducible polynomial p ∈ F[x] that has one zero in E actually splits in E.

• Example. R is not a normal extension of Q.

The polynomial p(x) = x5 −3x4 −3x3 +3x2 +3x+3 from the Eisenstein presentation is irreducible in Q[x]

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Definition. Let (F,+,·) be a field and let E be a finite extension
• f F. Then E is called a normal extension of F iff every

irreducible polynomial p ∈ F[x] that has one zero in E actually splits in E.

• Example. R is not a normal extension of Q.

The polynomial p(x) = x5 −3x4 −3x3 +3x2 +3x+3 from the Eisenstein presentation is irreducible in Q[x], one of its zeros is real

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Definition. Let (F,+,·) be a field and let E be a finite extension
• f F. Then E is called a normal extension of F iff every

irreducible polynomial p ∈ F[x] that has one zero in E actually splits in E.

• Example. R is not a normal extension of Q.

The polynomial p(x) = x5 −3x4 −3x3 +3x2 +3x+3 from the Eisenstein presentation is irreducible in Q[x], one of its zeros is real (in fact, three of them are)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Definition. Let (F,+,·) be a field and let E be a finite extension
• f F. Then E is called a normal extension of F iff every

irreducible polynomial p ∈ F[x] that has one zero in E actually splits in E.

• Example. R is not a normal extension of Q.

The polynomial p(x) = x5 −3x4 −3x3 +3x2 +3x+3 from the Eisenstein presentation is irreducible in Q[x], one of its zeros is real (in fact, three of them are), but not all of them are real.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Definition. Let (F,+,·) be a field and let E be a finite extension
• f F. Then E is called a normal extension of F iff every

irreducible polynomial p ∈ F[x] that has one zero in E actually splits in E.

• Example. R is not a normal extension of Q.

The polynomial p(x) = x5 −3x4 −3x3 +3x2 +3x+3 from the Eisenstein presentation is irreducible in Q[x], one of its zeros is real (in fact, three of them are), but not all of them are real.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Theorem.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a normal extension of F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a normal extension of F. Then

• G(E/F)
• = [E : F].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a normal extension of F. Then

• G(E/F)
• = [E : F].

Proof.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a normal extension of F. Then

• G(E/F)
• = [E : F].
• Proof. Because E is a normal extension, it is the splitting field
• f some polynomial f ∈ F[x].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a normal extension of F. Then

• G(E/F)
• = [E : F].
• Proof. Because E is a normal extension, it is the splitting field
• f some polynomial f ∈ F[x]. Hence E = F(γ1,...,γm), where

γ1,...,γm are the zeros of f in E\F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a normal extension of F. Then

• G(E/F)
• = [E : F].
• Proof. Because E is a normal extension, it is the splitting field
• f some polynomial f ∈ F[x]. Hence E = F(γ1,...,γm), where

γ1,...,γm are the zeros of f in E\F. There is a θ ∈ E so that E = F(γ1,...,γm) = F(θ) and θ is a zero of a polynomial in F[x].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a normal extension of F. Then

• G(E/F)
• = [E : F].
• Proof. Because E is a normal extension, it is the splitting field
• f some polynomial f ∈ F[x]. Hence E = F(γ1,...,γm), where

γ1,...,γm are the zeros of f in E\F. There is a θ ∈ E so that E = F(γ1,...,γm) = F(θ) and θ is a zero of a polynomial in F[x]. (We’re using that F has characteristic 0 here.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a normal extension of F. Then

• G(E/F)
• = [E : F].
• Proof. Because E is a normal extension, it is the splitting field
• f some polynomial f ∈ F[x]. Hence E = F(γ1,...,γm), where

γ1,...,γm are the zeros of f in E\F. There is a θ ∈ E so that E = F(γ1,...,γm) = F(θ) and θ is a zero of a polynomial in F[x]. (We’re using that F has characteristic 0 here.) But then θ also is a zero of an irreducible polynomial of degree n in F[x].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a normal extension of F. Then

• G(E/F)
• = [E : F].
• Proof. Because E is a normal extension, it is the splitting field
• f some polynomial f ∈ F[x]. Hence E = F(γ1,...,γm), where

γ1,...,γm are the zeros of f in E\F. There is a θ ∈ E so that E = F(γ1,...,γm) = F(θ) and θ is a zero of a polynomial in F[x]. (We’re using that F has characteristic 0 here.) But then θ also is a zero of an irreducible polynomial of degree n in F[x]. Hence the set

• 1,θ,...,θ n−1

is a basis for E over F.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

• Theorem. Let (F,+,·) be a field of characteristic 0 and let E

be a normal extension of F. Then

• G(E/F)
• = [E : F].
• Proof. Because E is a normal extension, it is the splitting field
• f some polynomial f ∈ F[x]. Hence E = F(γ1,...,γm), where

γ1,...,γm are the zeros of f in E\F. There is a θ ∈ E so that E = F(γ1,...,γm) = F(θ) and θ is a zero of a polynomial in F[x]. (We’re using that F has characteristic 0 here.) But then θ also is a zero of an irreducible polynomial of degree n in F[x]. Hence the set

• 1,θ,...,θ n−1

is a basis for E over F. In particular, n = [E : F].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (concl.).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (concl.). Now

• G(E/F)
• =
• G(F(θ)/F)
• ≤ n.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (concl.). Now

• G(E/F)
• =
• G(F(θ)/F)
• ≤ n. Moreover,

p, being irreducible, has n distinct zeros θ1 = θ,θ2,...,θn.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (concl.). Now

• G(E/F)
• =
• G(F(θ)/F)
• ≤ n. Moreover,

p, being irreducible, has n distinct zeros θ1 = θ,θ2,...,θn. Because E is a normal extension of F, all zeros of p are in E.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (concl.). Now

• G(E/F)
• =
• G(F(θ)/F)
• ≤ n. Moreover,

p, being irreducible, has n distinct zeros θ1 = θ,θ2,...,θn. Because E is a normal extension of F, all zeros of p are in E. For every j ∈ {1,...,n} there is an automorphism of E that fixes F and that maps θ to θj.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (concl.). Now

• G(E/F)
• =
• G(F(θ)/F)
• ≤ n. Moreover,

p, being irreducible, has n distinct zeros θ1 = θ,θ2,...,θn. Because E is a normal extension of F, all zeros of p are in E. For every j ∈ {1,...,n} there is an automorphism of E that fixes F and that maps θ to θj. No two of these n automorphisms are equal to each other.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (concl.). Now

• G(E/F)
• =
• G(F(θ)/F)
• ≤ n. Moreover,

p, being irreducible, has n distinct zeros θ1 = θ,θ2,...,θn. Because E is a normal extension of F, all zeros of p are in E. For every j ∈ {1,...,n} there is an automorphism of E that fixes F and that maps θ to θj. No two of these n automorphisms are equal to each other. So

• G(E/F)
• ≥ n.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (concl.). Now

• G(E/F)
• =
• G(F(θ)/F)
• ≤ n. Moreover,

p, being irreducible, has n distinct zeros θ1 = θ,θ2,...,θn. Because E is a normal extension of F, all zeros of p are in E. For every j ∈ {1,...,n} there is an automorphism of E that fixes F and that maps θ to θj. No two of these n automorphisms are equal to each other. So

• G(E/F)
• ≥ n. Hence
• G(E/F)
• = n = [E : F].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions

logo1 Characterizing Splitting Fields Normal Extensions Size of the Galois Group

Proof (concl.). Now

• G(E/F)
• =
• G(F(θ)/F)
• ≤ n. Moreover,

p, being irreducible, has n distinct zeros θ1 = θ,θ2,...,θn. Because E is a normal extension of F, all zeros of p are in E. For every j ∈ {1,...,n} there is an automorphism of E that fixes F and that maps θ to θj. No two of these n automorphisms are equal to each other. So

• G(E/F)
• ≥ n. Hence
• G(E/F)
• = n = [E : F].

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Normal Field Extensions