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Characterizing Splitting Fields Normal Extensions Size of the Galois Group Normal Field Extensions Bernd Schr oder logo1 Bernd Schr oder Louisiana Tech University, College of Engineering and Science Normal Field Extensions


  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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

  26. 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

  27. 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

  28. 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

  29. 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

  30. 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

  31. 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

  32. 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

  33. 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

  34. 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

  35. 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

  36. 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

  37. 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

  38. 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

  39. 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

  40. 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

  41. 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

  42. 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

  43. 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

  44. 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

  45. 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

  46. 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

  47. 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

  48. 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

  49. 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

  50. 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

  51. 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

  52. 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

  53. 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

  54. 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

  55. 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

  56. 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

  57. 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

  58. 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

  59. 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

  60. 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

  61. 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

  62. 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

  63. 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

  64. 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

  65. 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

  66. 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

  67. 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

  68. 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

  69. 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

  70. 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

  71. 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

  72. 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

  73. 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

  74. 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

  75. 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

  76. 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

  77. 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

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