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.
galois-theory galois-extensions
asked Jan 15 at 17:17
Drew ArmstrongDrew Armstrong
427210
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add a comment |
$begingroup$
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.
galois-theory galois-extensions
asked Jan 15 at 17:17
Drew ArmstrongDrew Armstrong
427210
$endgroup$
add a comment |
$begingroup$
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.
galois-theory galois-extensions
asked Jan 15 at 17:17
Drew ArmstrongDrew Armstrong
427210
$endgroup$
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
galois-theory galois-extensions
asked Jan 15 at 17:17
Drew ArmstrongDrew Armstrong
427210
asked Jan 15 at 17:17
Drew ArmstrongDrew Armstrong
427210
asked Jan 15 at 17:17
Drew ArmstrongDrew Armstrong
427210
asked Jan 15 at 17:17
Drew ArmstrongDrew Armstrong
427210
asked Jan 15 at 17:17
Drew ArmstrongDrew Armstrong
427210
427210
add a comment |
add a comment |
1 Answer
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I found a reference. This is Theorem V.9.4 in Hungerford.
answered Jan 15 at 21:28
Drew ArmstrongDrew Armstrong
427210
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add a comment |
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$begingroup$
I found a reference. This is Theorem V.9.4 in Hungerford.
answered Jan 15 at 21:28
Drew ArmstrongDrew Armstrong
427210
$endgroup$
add a comment |
$begingroup$
I found a reference. This is Theorem V.9.4 in Hungerford.
answered Jan 15 at 21:28
Drew ArmstrongDrew Armstrong
427210
$endgroup$
add a comment |
$begingroup$
I found a reference. This is Theorem V.9.4 in Hungerford.
answered Jan 15 at 21:28
Drew ArmstrongDrew Armstrong
427210
$endgroup$
I found a reference. This is Theorem V.9.4 in Hungerford.
answered Jan 15 at 21:28
Drew ArmstrongDrew Armstrong
427210
answered Jan 15 at 21:28
Drew ArmstrongDrew Armstrong
427210
answered Jan 15 at 21:28
Drew ArmstrongDrew Armstrong
427210
answered Jan 15 at 21:28
Drew ArmstrongDrew Armstrong
427210
427210
add a comment |
add a comment |
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