Any Sub-extension of a Radical Extension is Solvable?












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I have seen the following theorem stated without proof. (I assume that all fields are characteristic zero.)



Theorem: Let $Fsubset K$ be a radical extension of fields. That is, suppose that $K$ can be obtained from $F$ by successive adjunction of radicals. Then for any intermediate field $Fsubseteq Lsubseteq K$ the group $mathrm{Aut}_F(L)$ (i.e., automorphisms of $L$ that fix $F$) is solvable.



Can someone please point me to a proof? The application I have in mind is when $L$ is the splitting field of a polynomial.










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    I have seen the following theorem stated without proof. (I assume that all fields are characteristic zero.)



    Theorem: Let $Fsubset K$ be a radical extension of fields. That is, suppose that $K$ can be obtained from $F$ by successive adjunction of radicals. Then for any intermediate field $Fsubseteq Lsubseteq K$ the group $mathrm{Aut}_F(L)$ (i.e., automorphisms of $L$ that fix $F$) is solvable.



    Can someone please point me to a proof? The application I have in mind is when $L$ is the splitting field of a polynomial.










    share|cite|improve this question









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      0





      $begingroup$


      I have seen the following theorem stated without proof. (I assume that all fields are characteristic zero.)



      Theorem: Let $Fsubset K$ be a radical extension of fields. That is, suppose that $K$ can be obtained from $F$ by successive adjunction of radicals. Then for any intermediate field $Fsubseteq Lsubseteq K$ the group $mathrm{Aut}_F(L)$ (i.e., automorphisms of $L$ that fix $F$) is solvable.



      Can someone please point me to a proof? The application I have in mind is when $L$ is the splitting field of a polynomial.










      share|cite|improve this question









      $endgroup$




      I have seen the following theorem stated without proof. (I assume that all fields are characteristic zero.)



      Theorem: Let $Fsubset K$ be a radical extension of fields. That is, suppose that $K$ can be obtained from $F$ by successive adjunction of radicals. Then for any intermediate field $Fsubseteq Lsubseteq K$ the group $mathrm{Aut}_F(L)$ (i.e., automorphisms of $L$ that fix $F$) is solvable.



      Can someone please point me to a proof? The application I have in mind is when $L$ is the splitting field of a polynomial.







      galois-theory galois-extensions






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      asked Jan 15 at 17:17









      Drew ArmstrongDrew Armstrong

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          I found a reference. This is Theorem V.9.4 in Hungerford.






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            I found a reference. This is Theorem V.9.4 in Hungerford.






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              I found a reference. This is Theorem V.9.4 in Hungerford.






              share|cite|improve this answer









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                I found a reference. This is Theorem V.9.4 in Hungerford.






                share|cite|improve this answer









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                I found a reference. This is Theorem V.9.4 in Hungerford.







                share|cite|improve this answer












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                answered Jan 15 at 21:28









                Drew ArmstrongDrew Armstrong

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