If the norm of an element over a subgroup of the Galois group of a Galois extension is $1$, then so is the...
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Let $mathbb E/mathbb F$ be a finite Galois extension with Galois group $G$. If $H$ is a subgroup of $G$, then let $N_H(x):=prod_{sigmain H}sigma(x), forall xin mathbb E$. Notice that if $mathbb K$ is the fixed field of $H$, then $H=Gal(mathbb E/mathbb K)$, and then $N_H$ is just the usual norm $N_{mathbb E/mathbb K}$.
Then how to prove that :
If there exists a subgroup $H$ of $G$ and $ain mathbb E$ such that $N_H(a)=1$, then $N_G(a)=1$ ?
field-theory galois-theory extension-field galois-extensions
asked Nov 12 at 23:48
user521337
486113
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Let $mathbb E/mathbb F$ be a finite Galois extension with Galois group $G$. If $H$ is a subgroup of $G$, then let $N_H(x):=prod_{sigmain H}sigma(x), forall xin mathbb E$. Notice that if $mathbb K$ is the fixed field of $H$, then $H=Gal(mathbb E/mathbb K)$, and then $N_H$ is just the usual norm $N_{mathbb E/mathbb K}$.
Then how to prove that :
If there exists a subgroup $H$ of $G$ and $ain mathbb E$ such that $N_H(a)=1$, then $N_G(a)=1$ ?
field-theory galois-theory extension-field galois-extensions
asked Nov 12 at 23:48
user521337
486113
1
$N_G(x) = prod_{sigma in G/H}sigma( N_H(x))$ where $G = bigcup_{sigma in G/H} sigma H$ (disjoint union). Also $N_G = N_{G/H} circ N_H$ where $N_H = N_{E/ E^H}$ and $N_{G/H} = N_{E^H/ E^G}$
– reuns
Nov 13 at 0:18
add a comment |
up vote
0
down vote
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up vote
0
down vote
favorite
Let $mathbb E/mathbb F$ be a finite Galois extension with Galois group $G$. If $H$ is a subgroup of $G$, then let $N_H(x):=prod_{sigmain H}sigma(x), forall xin mathbb E$. Notice that if $mathbb K$ is the fixed field of $H$, then $H=Gal(mathbb E/mathbb K)$, and then $N_H$ is just the usual norm $N_{mathbb E/mathbb K}$.
Then how to prove that :
If there exists a subgroup $H$ of $G$ and $ain mathbb E$ such that $N_H(a)=1$, then $N_G(a)=1$ ?
field-theory galois-theory extension-field galois-extensions
asked Nov 12 at 23:48
user521337
486113
Let $mathbb E/mathbb F$ be a finite Galois extension with Galois group $G$. If $H$ is a subgroup of $G$, then let $N_H(x):=prod_{sigmain H}sigma(x), forall xin mathbb E$. Notice that if $mathbb K$ is the fixed field of $H$, then $H=Gal(mathbb E/mathbb K)$, and then $N_H$ is just the usual norm $N_{mathbb E/mathbb K}$.
Then how to prove that :
If there exists a subgroup $H$ of $G$ and $ain mathbb E$ such that $N_H(a)=1$, then $N_G(a)=1$ ?
field-theory galois-theory extension-field galois-extensions
field-theory galois-theory extension-field galois-extensions
asked Nov 12 at 23:48
user521337
486113
asked Nov 12 at 23:48
user521337
486113
asked Nov 12 at 23:48
user521337
486113
asked Nov 12 at 23:48
user521337
486113
asked Nov 12 at 23:48
user521337
486113
486113
1
$N_G(x) = prod_{sigma in G/H}sigma( N_H(x))$ where $G = bigcup_{sigma in G/H} sigma H$ (disjoint union). Also $N_G = N_{G/H} circ N_H$ where $N_H = N_{E/ E^H}$ and $N_{G/H} = N_{E^H/ E^G}$
– reuns
Nov 13 at 0:18
add a comment |
1
$N_G(x) = prod_{sigma in G/H}sigma( N_H(x))$ where $G = bigcup_{sigma in G/H} sigma H$ (disjoint union). Also $N_G = N_{G/H} circ N_H$ where $N_H = N_{E/ E^H}$ and $N_{G/H} = N_{E^H/ E^G}$
– reuns
Nov 13 at 0:18
1
1
$N_G(x) = prod_{sigma in G/H}sigma( N_H(x))$ where $G = bigcup_{sigma in G/H} sigma H$ (disjoint union). Also $N_G = N_{G/H} circ N_H$ where $N_H = N_{E/ E^H}$ and $N_{G/H} = N_{E^H/ E^G}$
– reuns
Nov 13 at 0:18
$N_G(x) = prod_{sigma in G/H}sigma( N_H(x))$ where $G = bigcup_{sigma in G/H} sigma H$ (disjoint union). Also $N_G = N_{G/H} circ N_H$ where $N_H = N_{E/ E^H}$ and $N_{G/H} = N_{E^H/ E^G}$
– reuns
Nov 13 at 0:18
add a comment |
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1
$N_G(x) = prod_{sigma in G/H}sigma( N_H(x))$ where $G = bigcup_{sigma in G/H} sigma H$ (disjoint union). Also $N_G = N_{G/H} circ N_H$ where $N_H = N_{E/ E^H}$ and $N_{G/H} = N_{E^H/ E^G}$
– reuns
Nov 13 at 0:18