A finite normal extension is also a splitting field












1












$begingroup$


Let $E/K $ be a finite normal extension, then there exists $p(x)in K[x] $ s.t. $E$ is the splitting field of $p(x) $.



The proof goes as follows:
$E=K(a_1,...,a_n) $ for some $a_1,...,a_nin Omega $ where $Omega $ is the algebraic closure of $K $.
Let $mu_i $ be the minimal polynomial of $a_i $ for $i=1,...,n $
We can send any $a_i $ onto another root of $mu _i $ through an homomorphism that fixes $K $.



Then the thesis follows.



I don't understand why we can send any $a_i $ onto another root of $mu _i $.



Note: the definition that we use of normal extension is:
$L/K$ is a normal extension if every homomorphism $phi :Krightarrow Omega $ that fixes $K $ sends $L $ onto itself.






field-theory galois-theory extension-field







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$endgroup$

















    1












    $begingroup$


    Let $E/K $ be a finite normal extension, then there exists $p(x)in K[x] $ s.t. $E$ is the splitting field of $p(x) $.



    The proof goes as follows:
    $E=K(a_1,...,a_n) $ for some $a_1,...,a_nin Omega $ where $Omega $ is the algebraic closure of $K $.
    Let $mu_i $ be the minimal polynomial of $a_i $ for $i=1,...,n $
    We can send any $a_i $ onto another root of $mu _i $ through an homomorphism that fixes $K $.



    Then the thesis follows.



    I don't understand why we can send any $a_i $ onto another root of $mu _i $.



    Note: the definition that we use of normal extension is:
    $L/K$ is a normal extension if every homomorphism $phi :Krightarrow Omega $ that fixes $K $ sends $L $ onto itself.






    field-theory galois-theory extension-field







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $E/K $ be a finite normal extension, then there exists $p(x)in K[x] $ s.t. $E$ is the splitting field of $p(x) $.



      The proof goes as follows:
      $E=K(a_1,...,a_n) $ for some $a_1,...,a_nin Omega $ where $Omega $ is the algebraic closure of $K $.
      Let $mu_i $ be the minimal polynomial of $a_i $ for $i=1,...,n $
      We can send any $a_i $ onto another root of $mu _i $ through an homomorphism that fixes $K $.



      Then the thesis follows.



      I don't understand why we can send any $a_i $ onto another root of $mu _i $.



      Note: the definition that we use of normal extension is:
      $L/K$ is a normal extension if every homomorphism $phi :Krightarrow Omega $ that fixes $K $ sends $L $ onto itself.






      field-theory galois-theory extension-field







      share|cite|improve this question











      $endgroup$




      Let $E/K $ be a finite normal extension, then there exists $p(x)in K[x] $ s.t. $E$ is the splitting field of $p(x) $.



      The proof goes as follows:
      $E=K(a_1,...,a_n) $ for some $a_1,...,a_nin Omega $ where $Omega $ is the algebraic closure of $K $.
      Let $mu_i $ be the minimal polynomial of $a_i $ for $i=1,...,n $
      We can send any $a_i $ onto another root of $mu _i $ through an homomorphism that fixes $K $.



      Then the thesis follows.



      I don't understand why we can send any $a_i $ onto another root of $mu _i $.



      Note: the definition that we use of normal extension is:
      $L/K$ is a normal extension if every homomorphism $phi :Krightarrow Omega $ that fixes $K $ sends $L $ onto itself.






      field-theory galois-theory extension-field




      field-theory galois-theory extension-field






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 6 at 11:22









      Jyrki Lahtonen

      110k13172390




      110k13172390










      asked Jan 5 at 17:31









      Lucio TanziniLucio Tanzini

      351114




      351114






















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          $begingroup$

          Here are the steps of a proof.




          • The $mu_{i}$ are irreducible.


          • Hence, given any two roots $a, b$ of $mu_{i}$, there is a $K$-isomorphism (meaning that it restricts to the identity on $K$) $$K(a) to K(b)$$ taking $a$ to $b$.


          • This isomorphism can be extended to an automorphism $phi$ of $Omega$.


          • Now since $L/K$ is normal, $phi$ has to send $L$ to $L$.







          share|cite|improve this answer









          $endgroup$














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            $begingroup$

            Here are the steps of a proof.




            • The $mu_{i}$ are irreducible.


            • Hence, given any two roots $a, b$ of $mu_{i}$, there is a $K$-isomorphism (meaning that it restricts to the identity on $K$) $$K(a) to K(b)$$ taking $a$ to $b$.


            • This isomorphism can be extended to an automorphism $phi$ of $Omega$.


            • Now since $L/K$ is normal, $phi$ has to send $L$ to $L$.







            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Here are the steps of a proof.




              • The $mu_{i}$ are irreducible.


              • Hence, given any two roots $a, b$ of $mu_{i}$, there is a $K$-isomorphism (meaning that it restricts to the identity on $K$) $$K(a) to K(b)$$ taking $a$ to $b$.


              • This isomorphism can be extended to an automorphism $phi$ of $Omega$.


              • Now since $L/K$ is normal, $phi$ has to send $L$ to $L$.







              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Here are the steps of a proof.




                • The $mu_{i}$ are irreducible.


                • Hence, given any two roots $a, b$ of $mu_{i}$, there is a $K$-isomorphism (meaning that it restricts to the identity on $K$) $$K(a) to K(b)$$ taking $a$ to $b$.


                • This isomorphism can be extended to an automorphism $phi$ of $Omega$.


                • Now since $L/K$ is normal, $phi$ has to send $L$ to $L$.







                share|cite|improve this answer









                $endgroup$



                Here are the steps of a proof.




                • The $mu_{i}$ are irreducible.


                • Hence, given any two roots $a, b$ of $mu_{i}$, there is a $K$-isomorphism (meaning that it restricts to the identity on $K$) $$K(a) to K(b)$$ taking $a$ to $b$.


                • This isomorphism can be extended to an automorphism $phi$ of $Omega$.


                • Now since $L/K$ is normal, $phi$ has to send $L$ to $L$.








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 at 17:39









                Andreas CarantiAndreas Caranti

                57k34397




                57k34397






























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