Normal Field Extensions Bernd Schr oder logo1 Bernd Schr oder - PowerPoint PPT Presentation

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 lies in some extension of F , but, at this stage, we do not know if it lies in E . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . Thus E = E ( θ 1 ) is the splitting field of f over F ( θ 1 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . Thus E = E ( θ 1 ) is the splitting field of f over F ( θ 1 ) (the splitting field must contain E and θ 1 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . Thus E = E ( θ 1 ) is the splitting field of f over F ( θ 1 ) (the splitting field must contain E and θ 1 ), and E ( θ 2 ) is the splitting field of f over F ( θ 2 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . Thus E = E ( θ 1 ) is the splitting field of f over F ( θ 1 ) (the splitting field must contain E and θ 1 ), and E ( θ 2 ) is the splitting field of f over F ( θ 2 ) (the splitting field must contain E and θ 2 ). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . Thus E = E ( θ 1 ) is the splitting field of f over F ( θ 1 ) (the splitting field must contain E and θ 1 ), and E ( θ 2 ) is the splitting field of f over F ( θ 2 ) (the splitting field must contain E and θ 2 ). There is an isomorphism Φ : E → E ( θ 2 ) that continues ϕ . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . Thus E = E ( θ 1 ) is the splitting field of f over F ( θ 1 ) (the splitting field must contain E and θ 1 ), and E ( θ 2 ) is the splitting field of f over F ( θ 2 ) (the splitting field must contain E and θ 2 ). There is an isomorphism Φ : E → E ( θ 2 ) that continues ϕ . In particular, Φ fixes F . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . Thus E = E ( θ 1 ) is the splitting field of f over F ( θ 1 ) (the splitting field must contain E and θ 1 ), and E ( θ 2 ) is the splitting field of f over F ( θ 2 ) (the splitting field must contain E and θ 2 ). There is an isomorphism Φ : E → E ( θ 2 ) that continues ϕ . In particular, Φ fixes F . But then [ E : F ] = [ E ( θ 2 ) : F ] logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . Thus E = E ( θ 1 ) is the splitting field of f over F ( θ 1 ) (the splitting field must contain E and θ 1 ), and E ( θ 2 ) is the splitting field of f over F ( θ 2 ) (the splitting field must contain E and θ 2 ). There is an isomorphism Φ : E → E ( θ 2 ) that continues ϕ . In particular, Φ fixes F . But then [ E : F ] = [ E ( θ 2 ) : F ] , which implies that E = E ( θ 2 ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . Thus E = E ( θ 1 ) is the splitting field of f over F ( θ 1 ) (the splitting field must contain E and θ 1 ), and E ( θ 2 ) is the splitting field of f over F ( θ 2 ) (the splitting field must contain E and θ 2 ). There is an isomorphism Φ : E → E ( θ 2 ) that continues ϕ . In particular, Φ fixes F . But then [ E : F ] = [ E ( θ 2 ) : F ] , which implies that E = E ( θ 2 ) . Because θ 2 was an arbitrary root of p , we conclude that all roots of p are in E logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 µ 1 ,..., µ n ∈ E \ F be the roots of f that are not in F . Then E = F ( µ 1 ,..., µ n ) . Let p ∈ F [ x ] be an arbitrary irreducible polynomial with a root θ 1 ∈ E . Let θ 2 � = θ 1 be another arbitrary root of p . Then θ 2 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 ( θ 1 ) → F ( θ 2 ) so that ϕ ( θ 1 ) = θ 2 and ϕ fixes F . Moreover, any field in which f splits must contain µ 1 ,..., µ n , so it must contain E = F ( µ 1 ,..., µ n ) . Thus E = E ( θ 1 ) is the splitting field of f over F ( θ 1 ) (the splitting field must contain E and θ 1 ), and E ( θ 2 ) is the splitting field of f over F ( θ 2 ) (the splitting field must contain E and θ 2 ). There is an isomorphism Φ : E → E ( θ 2 ) that continues ϕ . In particular, Φ fixes F . But then [ E : F ] = [ E ( θ 2 ) : F ] , which implies that E = E ( θ 2 ) . Because θ 2 was an arbitrary root of p , we conclude that all roots of p are in E , that is p splits in E . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . There is a θ ∈ E = F ( µ 1 ,..., µ n ) so that E = F ( µ 1 ,..., µ n ) = F ( θ ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . There is a θ ∈ E = F ( µ 1 ,..., µ n ) so that E = F ( µ 1 ,..., µ n ) = F ( θ ) and θ is a zero of an irreducible polynomial p ∈ F [ x ] . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . There is a θ ∈ E = F ( µ 1 ,..., µ n ) so that E = F ( µ 1 ,..., µ n ) = F ( θ ) and θ is a zero of an irreducible polynomial p ∈ F [ x ] . (We used that F has characteristic 0 here.) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . There is a θ ∈ E = F ( µ 1 ,..., µ n ) so that E = F ( µ 1 ,..., µ n ) = F ( θ ) and θ is a zero of an irreducible polynomial p ∈ F [ x ] . (We used that F � 1 , θ ,..., θ deg ( p ) − 1 � has characteristic 0 here.) Then is a basis for E = F ( θ ) over F . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . There is a θ ∈ E = F ( µ 1 ,..., µ n ) so that E = F ( µ 1 ,..., µ n ) = F ( θ ) and θ is a zero of an irreducible polynomial p ∈ F [ x ] . (We used that F � 1 , θ ,..., θ deg ( p ) − 1 � has characteristic 0 here.) Then is a basis for E = F ( θ ) over F . Moreover deg ( p ) = [ E : F ] = n . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . There is a θ ∈ E = F ( µ 1 ,..., µ n ) so that E = F ( µ 1 ,..., µ n ) = F ( θ ) and θ is a zero of an irreducible polynomial p ∈ F [ x ] . (We used that F � 1 , θ ,..., θ deg ( p ) − 1 � has characteristic 0 here.) Then is a basis for E = F ( θ ) over F . Moreover deg ( p ) = [ E : F ] = n . By assumption, p splits in E . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . There is a θ ∈ E = F ( µ 1 ,..., µ n ) so that E = F ( µ 1 ,..., µ n ) = F ( θ ) and θ is a zero of an irreducible polynomial p ∈ F [ x ] . (We used that F � 1 , θ ,..., θ deg ( p ) − 1 � has characteristic 0 here.) Then is a basis for E = F ( θ ) over F . Moreover deg ( p ) = [ E : F ] = n . By assumption, p splits in E . Let θ 1 : = θ and let θ 2 ,..., θ n be the other zeros of p logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . There is a θ ∈ E = F ( µ 1 ,..., µ n ) so that E = F ( µ 1 ,..., µ n ) = F ( θ ) and θ is a zero of an irreducible polynomial p ∈ F [ x ] . (We used that F � 1 , θ ,..., θ deg ( p ) − 1 � has characteristic 0 here.) Then is a basis for E = F ( θ ) over F . Moreover deg ( p ) = [ E : F ] = n . By assumption, p splits in E . Let θ 1 : = θ and let θ 2 ,..., θ n be the other zeros of p (which must all be distinct). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . There is a θ ∈ E = F ( µ 1 ,..., µ n ) so that E = F ( µ 1 ,..., µ n ) = F ( θ ) and θ is a zero of an irreducible polynomial p ∈ F [ x ] . (We used that F � 1 , θ ,..., θ deg ( p ) − 1 � has characteristic 0 here.) Then is a basis for E = F ( θ ) over F . Moreover deg ( p ) = [ E : F ] = n . By assumption, p splits in E . Let θ 1 : = θ and let θ 2 ,..., θ n be the other zeros of p (which must all be distinct). Let j ∈ { 2 ,..., n } . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3”). In case E = F , we have G ( F / F ) = { id F } and clearly E = F is the fixed field of G ( F / F ) = { id F } . So for the remainder of this part, we can assume that E � = F . Let { µ 1 ,..., µ n } be a basis of E over F . Every µ k must be a � � µ j zero of a polynomial in F [ x ] , because otherwise k : j ∈ N 0 would be an infinite F -linearly independent set in E . There is a θ ∈ E = F ( µ 1 ,..., µ n ) so that E = F ( µ 1 ,..., µ n ) = F ( θ ) and θ is a zero of an irreducible polynomial p ∈ F [ x ] . (We used that F � 1 , θ ,..., θ deg ( p ) − 1 � has characteristic 0 here.) Then is a basis for E = F ( θ ) over F . Moreover deg ( p ) = [ E : F ] = n . By assumption, p splits in E . Let θ 1 : = θ and let θ 2 ,..., θ n be the other zeros of p (which must all be distinct). Let j ∈ { 2 ,..., n } . There is an isomorphism ϕ : F ( θ 1 ) → F ( θ j ) so that ϕ ( θ 1 ) = θ j and ϕ fixes F . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“2 ⇒ 3” concl.). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i e = i = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 we also have e logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 we also have e = Φ ( e ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 f i θ i ∑ we also have e = Φ ( e ) = Φ i = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 n − 1 f i θ i f i Φ ( θ ) i ∑ ∑ we also have e = Φ ( e ) = Φ = i = 0 i = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 n − 1 n − 1 f i Φ ( θ ) i = f i θ i f i θ i ∑ ∑ ∑ we also have e = Φ ( e ) = Φ = j . i = 0 i = 0 i = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 n − 1 n − 1 f i Φ ( θ ) i = f i θ i f i θ i ∑ ∑ ∑ we also have e = Φ ( e ) = Φ = j . i = 0 i = 0 i = 0 n − 1 f i x i . ∑ Let f ( x ) : = − e + i = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 n − 1 n − 1 f i Φ ( θ ) i = f i θ i f i θ i ∑ ∑ ∑ we also have e = Φ ( e ) = Φ = j . i = 0 i = 0 i = 0 n − 1 f i x i . Then f ∈ E [ x ] and deg ( f ) ≤ n − 1. ∑ Let f ( x ) : = − e + i = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 n − 1 n − 1 f i Φ ( θ ) i = f i θ i f i θ i ∑ ∑ ∑ we also have e = Φ ( e ) = Φ = j . i = 0 i = 0 i = 0 n − 1 f i x i . Then f ∈ E [ x ] and deg ( f ) ≤ n − 1. But ∑ Let f ( x ) : = − e + i = 0 the above shows that f has the n distinct zeros θ 1 ,..., θ n . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 n − 1 n − 1 f i Φ ( θ ) i = f i θ i f i θ i ∑ ∑ ∑ we also have e = Φ ( e ) = Φ = j . i = 0 i = 0 i = 0 n − 1 f i x i . Then f ∈ E [ x ] and deg ( f ) ≤ n − 1. But ∑ Let f ( x ) : = − e + i = 0 the above shows that f has the n distinct zeros θ 1 ,..., θ n . This is only possible if f = 0. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 n − 1 n − 1 f i Φ ( θ ) i = f i θ i f i θ i ∑ ∑ ∑ we also have e = Φ ( e ) = Φ = j . i = 0 i = 0 i = 0 n − 1 f i x i . Then f ∈ E [ x ] and deg ( f ) ≤ n − 1. But ∑ Let f ( x ) : = − e + i = 0 the above shows that f has the n distinct zeros θ 1 ,..., θ n . This is only possible if f = 0. But then − e + f 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 n − 1 n − 1 f i Φ ( θ ) i = f i θ i f i θ i ∑ ∑ ∑ we also have e = Φ ( e ) = Φ = j . i = 0 i = 0 i = 0 n − 1 f i x i . Then f ∈ E [ x ] and deg ( f ) ≤ n − 1. But ∑ Let f ( x ) : = − e + i = 0 the above shows that f has the n distinct zeros θ 1 ,..., θ n . This is only possible if f = 0. But then − e + f 0 = f ( 0 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 n − 1 n − 1 f i Φ ( θ ) i = f i θ i f i θ i ∑ ∑ ∑ we also have e = Φ ( e ) = Φ = j . i = 0 i = 0 i = 0 n − 1 f i x i . Then f ∈ E [ x ] and deg ( f ) ≤ n − 1. But ∑ Let f ( x ) : = − e + i = 0 the above shows that f has the n distinct zeros θ 1 ,..., θ n . This is only possible if f = 0. But then − e + f 0 = f ( 0 ) = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 ( θ 1 ) = θ j . Now let e ∈ E be fixed by G ( E / F ) . Because E = F ( θ ) we have n − 1 ∑ f i θ i , and because there is a Φ ∈ G ( E / F ) with Φ ( θ ) = θ j , e = i = 0 � � n − 1 n − 1 n − 1 f i Φ ( θ ) i = f i θ i f i θ i ∑ ∑ ∑ we also have e = Φ ( e ) = Φ = j . i = 0 i = 0 i = 0 n − 1 f i x i . Then f ∈ E [ x ] and deg ( f ) ≤ n − 1. But ∑ Let f ( x ) : = − e + i = 0 the above shows that f has the n distinct zeros θ 1 ,..., θ n . This is only possible if f = 0. But then − e + f 0 = f ( 0 ) = 0, that is, e = f 0 ∈ F , which was to be proved. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1”). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 a 1 ,..., a n be the elements of � � σ ( a ) : σ ∈ G ( E / F ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 a 1 ,..., a n be the elements of � � σ ( a ) : σ ∈ G ( E / F ) and let n ∏ f a ( x ) : = ( x − a j ) j = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 a 1 ,..., a n be the elements of � � σ ( a ) : σ ∈ G ( E / F ) and let n n b k x k ∈ E [ x ] . ∏ ∑ f a ( x ) : = ( x − a j ) = j = 1 k = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 a 1 ,..., a n be the elements of � � σ ( a ) : σ ∈ G ( E / F ) and let n n b k x k ∈ E [ x ] . ∏ ∑ f a ( x ) : = ( x − a j ) = j = 1 k = 0 Let σ ∈ G ( E / F ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 a 1 ,..., a n be the elements of � � σ ( a ) : σ ∈ G ( E / F ) and let n n b k x k ∈ E [ x ] . ∏ ∑ f a ( x ) : = ( x − a j ) = j = 1 k = 0 Let σ ∈ G ( E / F ) . The function ˜ σ : E [ x ] → E [ x ] defined by � � n n σ ( c k ) x k is an isomorphism c k x k ∑ ∑ : = σ ˜ k = 0 k = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 a 1 ,..., a n be the elements of � � σ ( a ) : σ ∈ G ( E / F ) and let n n b k x k ∈ E [ x ] . ∏ ∑ f a ( x ) : = ( x − a j ) = j = 1 k = 0 Let σ ∈ G ( E / F ) . The function ˜ σ : E [ x ] → E [ x ] defined by � � n n σ ( c k ) x k is an isomorphism (good c k x k ∑ ∑ : = σ ˜ k = 0 k = 0 exercise). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 a 1 ,..., a n be the elements of � � σ ( a ) : σ ∈ G ( E / F ) and let n n b k x k ∈ E [ x ] . ∏ ∑ f a ( x ) : = ( x − a j ) = j = 1 k = 0 Let σ ∈ G ( E / F ) . The function ˜ σ : E [ x ] → E [ x ] defined by � � n n σ ( c k ) x k is an isomorphism (good c k x k ∑ ∑ : = σ ˜ k = 0 k = 0 exercise). Let a ∈ E . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

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 a 1 ,..., a n be the elements of � � σ ( a ) : σ ∈ G ( E / F ) and let n n b k x k ∈ E [ x ] . ∏ ∑ f a ( x ) : = ( x − a j ) = j = 1 k = 0 Let σ ∈ G ( E / F ) . The function ˜ σ : E [ x ] → E [ x ] defined by � � n n σ ( c k ) x k is an isomorphism (good c k x k ∑ ∑ : = σ ˜ k = 0 k = 0 exercise). Let a ∈ E . Because σ permutes the a 1 ,..., a n we infer logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). n σ ( b k ) x k ∑ k = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). n σ ( b k ) x k ∑ = σ ( f a ) ˜ k = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n σ ( b k ) x k ∑ ∏ = σ ( f a ) = ˜ ˜ ( x − a j ) σ k = 0 j = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n ∏ = ( x − a j ) j = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n ∏ = ( x − a j ) = f a j = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k , ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 so σ ( b k ) = b k for all k . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k , ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 so σ ( b k ) = b k for all k . (We’re “sort of” using that F has characteristic 0 here.) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k , ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 so σ ( b k ) = b k for all k . (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G ( E / F ) was arbitrary, all b k are in F , the fixed field of G ( E / F ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k , ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 so σ ( b k ) = b k for all k . (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G ( E / F ) was arbitrary, all b k are in F , the fixed field of G ( E / F ) . By definition, f a splits in E logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k , ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 so σ ( b k ) = b k for all k . (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G ( E / F ) was arbitrary, all b k are in F , the fixed field of G ( E / F ) . By definition, f a splits in E and, because a = id ( a ) ∈ { a 1 ,..., a n } , we have f a ( a ) = 0. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k , ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 so σ ( b k ) = b k for all k . (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G ( E / F ) was arbitrary, all b k are in F , the fixed field of G ( E / F ) . By definition, f a splits in E and, because a = id ( a ) ∈ { a 1 ,..., a n } , we have f a ( a ) = 0. Let v 1 ,..., v m be a basis of E over F . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k , ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 so σ ( b k ) = b k for all k . (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G ( E / F ) was arbitrary, all b k are in F , the fixed field of G ( E / F ) . By definition, f a splits in E and, because a = id ( a ) ∈ { a 1 ,..., a n } , we have f a ( a ) = 0. m ∏ Let v 1 ,..., v m be a basis of E over F . Then f : = f v k ∈ F [ x ] j = 1 splits in E logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k , ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 so σ ( b k ) = b k for all k . (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G ( E / F ) was arbitrary, all b k are in F , the fixed field of G ( E / F ) . By definition, f a splits in E and, because a = id ( a ) ∈ { a 1 ,..., a n } , we have f a ( a ) = 0. m ∏ Let v 1 ,..., v m be a basis of E over F . Then f : = f v k ∈ F [ x ] j = 1 splits in E and f ( v k ) = 0 for all k . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k , ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 so σ ( b k ) = b k for all k . (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G ( E / F ) was arbitrary, all b k are in F , the fixed field of G ( E / F ) . By definition, f a splits in E and, because a = id ( a ) ∈ { a 1 ,..., a n } , we have f a ( a ) = 0. m ∏ Let v 1 ,..., v m be a basis of E over F . Then f : = f v k ∈ F [ x ] j = 1 splits in E and f ( v k ) = 0 for all k . Hence the splitting field of f must contain E . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions

Characterizing Splitting Fields Normal Extensions Size of the Galois Group Proof (“3 ⇒ 1” concl.). � � n n n σ ( b k ) x k ∑ ∏ ∏ � � = σ ( f a ) = ˜ ˜ ( x − a j ) = x − ˜ σ ( a j ) σ k = 0 j = 1 j = 1 n n b k x k , ∏ ∑ = ( x − a j ) = f a = j = 1 k = 0 so σ ( b k ) = b k for all k . (We’re “sort of” using that F has characteristic 0 here.) Because σ ∈ G ( E / F ) was arbitrary, all b k are in F , the fixed field of G ( E / F ) . By definition, f a splits in E and, because a = id ( a ) ∈ { a 1 ,..., a n } , we have f a ( a ) = 0. m ∏ Let v 1 ,..., v m be a basis of E over F . Then f : = f v k ∈ F [ x ] j = 1 splits in E and f ( v k ) = 0 for all k . Hence the splitting field of f must contain E . So E is the splitting field of f . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions