Radical field extension is soluble, **counter-example?**
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There is a theorem that says that if $E/F$ is a normal and radical extension with $text{char}(K)=0$ then then $text{Aut}(E/F)$ is soluble. But why do we need normal and $text{char}(K)=0?$ What would happen if $E/F$ is not normal? For example $mathbb{Q}(sqrt[3]{2})/mathbb{Q}$ is not normal but its Galois group ${Id}$ is solvable? What would be a counter-example if $text{char}(K) neq 0$?
galois-theory examples-counterexamples
edited Dec 28 '18 at 15:35
Yanko
7,8801830
asked Dec 28 '18 at 15:34
roi_saumonroi_saumon
61138
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add a comment |
$begingroup$
There is a theorem that says that if $E/F$ is a normal and radical extension with $text{char}(K)=0$ then then $text{Aut}(E/F)$ is soluble. But why do we need normal and $text{char}(K)=0?$ What would happen if $E/F$ is not normal? For example $mathbb{Q}(sqrt[3]{2})/mathbb{Q}$ is not normal but its Galois group ${Id}$ is solvable? What would be a counter-example if $text{char}(K) neq 0$?
galois-theory examples-counterexamples
edited Dec 28 '18 at 15:35
Yanko
7,8801830
asked Dec 28 '18 at 15:34
roi_saumonroi_saumon
61138
$endgroup$
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Perhaps you mean solvable in the title?
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– Yanko
Dec 28 '18 at 15:37
2
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@Yanko solvable = soluble (British vs American English, don't know which one is which)
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– Kenny Lau
Dec 28 '18 at 15:47
add a comment |
$begingroup$
There is a theorem that says that if $E/F$ is a normal and radical extension with $text{char}(K)=0$ then then $text{Aut}(E/F)$ is soluble. But why do we need normal and $text{char}(K)=0?$ What would happen if $E/F$ is not normal? For example $mathbb{Q}(sqrt[3]{2})/mathbb{Q}$ is not normal but its Galois group ${Id}$ is solvable? What would be a counter-example if $text{char}(K) neq 0$?
galois-theory examples-counterexamples
edited Dec 28 '18 at 15:35
Yanko
7,8801830
asked Dec 28 '18 at 15:34
roi_saumonroi_saumon
61138
$endgroup$
There is a theorem that says that if $E/F$ is a normal and radical extension with $text{char}(K)=0$ then then $text{Aut}(E/F)$ is soluble. But why do we need normal and $text{char}(K)=0?$ What would happen if $E/F$ is not normal? For example $mathbb{Q}(sqrt[3]{2})/mathbb{Q}$ is not normal but its Galois group ${Id}$ is solvable? What would be a counter-example if $text{char}(K) neq 0$?
galois-theory examples-counterexamples
galois-theory examples-counterexamples
edited Dec 28 '18 at 15:35
Yanko
7,8801830
asked Dec 28 '18 at 15:34
roi_saumonroi_saumon
61138
edited Dec 28 '18 at 15:35
Yanko
7,8801830
asked Dec 28 '18 at 15:34
roi_saumonroi_saumon
61138
edited Dec 28 '18 at 15:35
Yanko
7,8801830
edited Dec 28 '18 at 15:35
Yanko
7,8801830
edited Dec 28 '18 at 15:35
Yanko
7,8801830
7,8801830
asked Dec 28 '18 at 15:34
roi_saumonroi_saumon
61138
asked Dec 28 '18 at 15:34
roi_saumonroi_saumon
61138
asked Dec 28 '18 at 15:34
roi_saumonroi_saumon
61138
61138
$begingroup$
Perhaps you mean solvable in the title?
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– Yanko
Dec 28 '18 at 15:37
2
$begingroup$
@Yanko solvable = soluble (British vs American English, don't know which one is which)
$endgroup$
– Kenny Lau
Dec 28 '18 at 15:47
add a comment |
$begingroup$
Perhaps you mean solvable in the title?
$endgroup$
– Yanko
Dec 28 '18 at 15:37
2
$begingroup$
@Yanko solvable = soluble (British vs American English, don't know which one is which)
$endgroup$
– Kenny Lau
Dec 28 '18 at 15:47
$begingroup$
Perhaps you mean solvable in the title?
$endgroup$
– Yanko
Dec 28 '18 at 15:37
$begingroup$
Perhaps you mean solvable in the title?
$endgroup$
– Yanko
Dec 28 '18 at 15:37
2
2
$begingroup$
@Yanko solvable = soluble (British vs American English, don't know which one is which)
$endgroup$
– Kenny Lau
Dec 28 '18 at 15:47
$begingroup$
@Yanko solvable = soluble (British vs American English, don't know which one is which)
$endgroup$
– Kenny Lau
Dec 28 '18 at 15:47
add a comment |
1 Answer
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See e.g. here for a proof of the more general theorem that requires neither normality nor zero characteristic.
answered Dec 28 '18 at 15:53
Kenny LauKenny Lau
20k2160
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add a comment |
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$begingroup$
See e.g. here for a proof of the more general theorem that requires neither normality nor zero characteristic.
answered Dec 28 '18 at 15:53
Kenny LauKenny Lau
20k2160
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add a comment |
$begingroup$
See e.g. here for a proof of the more general theorem that requires neither normality nor zero characteristic.
answered Dec 28 '18 at 15:53
Kenny LauKenny Lau
20k2160
$endgroup$
add a comment |
$begingroup$
See e.g. here for a proof of the more general theorem that requires neither normality nor zero characteristic.
answered Dec 28 '18 at 15:53
Kenny LauKenny Lau
20k2160
$endgroup$
See e.g. here for a proof of the more general theorem that requires neither normality nor zero characteristic.
answered Dec 28 '18 at 15:53
Kenny LauKenny Lau
20k2160
answered Dec 28 '18 at 15:53
Kenny LauKenny Lau
20k2160
answered Dec 28 '18 at 15:53
Kenny LauKenny Lau
20k2160
answered Dec 28 '18 at 15:53
Kenny LauKenny Lau
20k2160
20k2160
add a comment |
add a comment |
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$begingroup$
Perhaps you mean solvable in the title?
$endgroup$
– Yanko
Dec 28 '18 at 15:37
2
$begingroup$
@Yanko solvable = soluble (British vs American English, don't know which one is which)
$endgroup$
– Kenny Lau
Dec 28 '18 at 15:47