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Finding the Galois group of $mathbb{Q}(2^{1/p},zeta_{2n})$
0
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How to find the Galois group of the extension $mathbb{Q}(2^{1/p},zeta_{2n})$ of $mathbb{Q}$? Here $n$ is a positive integer, and $p$ is an odd prime factor of $n$.
galois-theory
asked Jan 12 at 4:18
saisai
1376
$endgroup$
add a comment |
0
$begingroup$
How to find the Galois group of the extension $mathbb{Q}(2^{1/p},zeta_{2n})$ of $mathbb{Q}$? Here $n$ is a positive integer, and $p$ is an odd prime factor of $n$.
galois-theory
asked Jan 12 at 4:18
saisai
1376
$endgroup$
$begingroup$
Did you show that $Gal(mathbb{Q}(zeta_{2n})/mathbb{Q}) = { zeta_{2n}^b mapsto zeta_{2n}^{ab}, a in mathbb{Z}/2nmathbb{Z}^times}$ ? Then assume $pnmid varphi(2n)$, what can you say of $[mathbb{Q}(2^{1/p}):mathbb{Q}] ,[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})],[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}] $ ? So what is $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q}(zeta_{2n}))$ and $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q})$ ?
$endgroup$
– reuns
Jan 12 at 9:42
$begingroup$
@reuns: Why is $pnmidvarphi(2n)$?
$endgroup$
– sai
Jan 12 at 10:33
$begingroup$
$[mathbb{Q}(2^{1/p}):mathbb{Q}] = ?$ then use the multiplicativity of degree of extension to find $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}]$ and $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})]$
$endgroup$
– reuns
Jan 12 at 14:11
add a comment |
0
0
0
$begingroup$
How to find the Galois group of the extension $mathbb{Q}(2^{1/p},zeta_{2n})$ of $mathbb{Q}$? Here $n$ is a positive integer, and $p$ is an odd prime factor of $n$.
galois-theory
asked Jan 12 at 4:18
saisai
1376
$endgroup$
How to find the Galois group of the extension $mathbb{Q}(2^{1/p},zeta_{2n})$ of $mathbb{Q}$? Here $n$ is a positive integer, and $p$ is an odd prime factor of $n$.
galois-theory
galois-theory
asked Jan 12 at 4:18
saisai
1376
asked Jan 12 at 4:18
saisai
1376
asked Jan 12 at 4:18
saisai
1376
asked Jan 12 at 4:18
saisai
1376
asked Jan 12 at 4:18
saisai
1376
1376
$begingroup$
Did you show that $Gal(mathbb{Q}(zeta_{2n})/mathbb{Q}) = { zeta_{2n}^b mapsto zeta_{2n}^{ab}, a in mathbb{Z}/2nmathbb{Z}^times}$ ? Then assume $pnmid varphi(2n)$, what can you say of $[mathbb{Q}(2^{1/p}):mathbb{Q}] ,[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})],[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}] $ ? So what is $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q}(zeta_{2n}))$ and $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q})$ ?
$endgroup$
– reuns
Jan 12 at 9:42
$begingroup$
@reuns: Why is $pnmidvarphi(2n)$?
$endgroup$
– sai
Jan 12 at 10:33
$begingroup$
$[mathbb{Q}(2^{1/p}):mathbb{Q}] = ?$ then use the multiplicativity of degree of extension to find $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}]$ and $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})]$
$endgroup$
– reuns
Jan 12 at 14:11
add a comment |
$begingroup$
Did you show that $Gal(mathbb{Q}(zeta_{2n})/mathbb{Q}) = { zeta_{2n}^b mapsto zeta_{2n}^{ab}, a in mathbb{Z}/2nmathbb{Z}^times}$ ? Then assume $pnmid varphi(2n)$, what can you say of $[mathbb{Q}(2^{1/p}):mathbb{Q}] ,[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})],[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}] $ ? So what is $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q}(zeta_{2n}))$ and $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q})$ ?
$endgroup$
– reuns
Jan 12 at 9:42
$begingroup$
@reuns: Why is $pnmidvarphi(2n)$?
$endgroup$
– sai
Jan 12 at 10:33
$begingroup$
$[mathbb{Q}(2^{1/p}):mathbb{Q}] = ?$ then use the multiplicativity of degree of extension to find $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}]$ and $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})]$
$endgroup$
– reuns
Jan 12 at 14:11
$begingroup$
Did you show that $Gal(mathbb{Q}(zeta_{2n})/mathbb{Q}) = { zeta_{2n}^b mapsto zeta_{2n}^{ab}, a in mathbb{Z}/2nmathbb{Z}^times}$ ? Then assume $pnmid varphi(2n)$, what can you say of $[mathbb{Q}(2^{1/p}):mathbb{Q}] ,[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})],[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}] $ ? So what is $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q}(zeta_{2n}))$ and $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q})$ ?
$endgroup$
– reuns
Jan 12 at 9:42
$begingroup$
Did you show that $Gal(mathbb{Q}(zeta_{2n})/mathbb{Q}) = { zeta_{2n}^b mapsto zeta_{2n}^{ab}, a in mathbb{Z}/2nmathbb{Z}^times}$ ? Then assume $pnmid varphi(2n)$, what can you say of $[mathbb{Q}(2^{1/p}):mathbb{Q}] ,[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})],[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}] $ ? So what is $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q}(zeta_{2n}))$ and $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q})$ ?
$endgroup$
– reuns
Jan 12 at 9:42
$begingroup$
@reuns: Why is $pnmidvarphi(2n)$?
$endgroup$
– sai
Jan 12 at 10:33
$begingroup$
@reuns: Why is $pnmidvarphi(2n)$?
$endgroup$
– sai
Jan 12 at 10:33
$begingroup$
$[mathbb{Q}(2^{1/p}):mathbb{Q}] = ?$ then use the multiplicativity of degree of extension to find $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}]$ and $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})]$
$endgroup$
– reuns
Jan 12 at 14:11
$begingroup$
$[mathbb{Q}(2^{1/p}):mathbb{Q}] = ?$ then use the multiplicativity of degree of extension to find $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}]$ and $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})]$
$endgroup$
– reuns
Jan 12 at 14:11
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$begingroup$
Did you show that $Gal(mathbb{Q}(zeta_{2n})/mathbb{Q}) = { zeta_{2n}^b mapsto zeta_{2n}^{ab}, a in mathbb{Z}/2nmathbb{Z}^times}$ ? Then assume $pnmid varphi(2n)$, what can you say of $[mathbb{Q}(2^{1/p}):mathbb{Q}] ,[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})],[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}] $ ? So what is $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q}(zeta_{2n}))$ and $Gal(mathbb{Q}(zeta_{2n},2^{1/p})/mathbb{Q})$ ?
$endgroup$
– reuns
Jan 12 at 9:42
$begingroup$
@reuns: Why is $pnmidvarphi(2n)$?
$endgroup$
– sai
Jan 12 at 10:33
$begingroup$
$[mathbb{Q}(2^{1/p}):mathbb{Q}] = ?$ then use the multiplicativity of degree of extension to find $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}]$ and $[mathbb{Q}(zeta_{2n},2^{1/p}):mathbb{Q}(zeta_{2n})]$
$endgroup$
– reuns
Jan 12 at 14:11