Normal extension and Galois group

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There is an important theorem states:
$E/F/K$ are a chain of extensions. And $E/K$ and $F/K$ are normal. Then we have $textGal(E/F)trianglelefteq textGal(E/K)$ and $textGal(F/K)congtextGal(E/K)/textGal(E/F)$.



For the normality of subgroups we need $F/K$ is normal. For the isomorphism. Is is true seems that we just need $E/F$ to be normal? Of course, $E/K$ normal implies that. Because we just need to extend $F$ to $E$ (we can skip $K$). Just need a polynomial in $F[X]$ splits in $E$.



Of course, $E/K$ normal implies $E/F$ normal and the former is stronger the latter in general. Is there any reason to state a weaker theorem?




abstract-algebra galois-theory




share|cite|improve this question





















  • If the extensions aren't normal then what is $mathrmGal(E/F)$, etc.? Is this just alternative notation for $mathrmAut(E/F)$?
    – Alex Mathers
    Jul 29 at 22:35







  • 1




    What is B in the question?
    – Stefan4024
    Jul 29 at 22:35










  • Is $B$ really $F$?
    – Steve D
    Jul 29 at 22:36










  • OH yes, it was a typo..
    – Xavier Yang
    Jul 29 at 22:58










  • Can you explain what happens when $K=mathbbQ$, $F=mathbbQ(sqrt2)$, and $E=mathbbQ(sqrt[4]2)$?
    – Steve D
    Jul 30 at 0:27














up vote
2
down vote

favorite
2












There is an important theorem states:
$E/F/K$ are a chain of extensions. And $E/K$ and $F/K$ are normal. Then we have $textGal(E/F)trianglelefteq textGal(E/K)$ and $textGal(F/K)congtextGal(E/K)/textGal(E/F)$.



For the normality of subgroups we need $F/K$ is normal. For the isomorphism. Is is true seems that we just need $E/F$ to be normal? Of course, $E/K$ normal implies that. Because we just need to extend $F$ to $E$ (we can skip $K$). Just need a polynomial in $F[X]$ splits in $E$.



Of course, $E/K$ normal implies $E/F$ normal and the former is stronger the latter in general. Is there any reason to state a weaker theorem?




abstract-algebra galois-theory




share|cite|improve this question





















  • If the extensions aren't normal then what is $mathrmGal(E/F)$, etc.? Is this just alternative notation for $mathrmAut(E/F)$?
    – Alex Mathers
    Jul 29 at 22:35







  • 1




    What is B in the question?
    – Stefan4024
    Jul 29 at 22:35










  • Is $B$ really $F$?
    – Steve D
    Jul 29 at 22:36










  • OH yes, it was a typo..
    – Xavier Yang
    Jul 29 at 22:58










  • Can you explain what happens when $K=mathbbQ$, $F=mathbbQ(sqrt2)$, and $E=mathbbQ(sqrt[4]2)$?
    – Steve D
    Jul 30 at 0:27












up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





There is an important theorem states:
$E/F/K$ are a chain of extensions. And $E/K$ and $F/K$ are normal. Then we have $textGal(E/F)trianglelefteq textGal(E/K)$ and $textGal(F/K)congtextGal(E/K)/textGal(E/F)$.



For the normality of subgroups we need $F/K$ is normal. For the isomorphism. Is is true seems that we just need $E/F$ to be normal? Of course, $E/K$ normal implies that. Because we just need to extend $F$ to $E$ (we can skip $K$). Just need a polynomial in $F[X]$ splits in $E$.



Of course, $E/K$ normal implies $E/F$ normal and the former is stronger the latter in general. Is there any reason to state a weaker theorem?




abstract-algebra galois-theory




share|cite|improve this question













There is an important theorem states:
$E/F/K$ are a chain of extensions. And $E/K$ and $F/K$ are normal. Then we have $textGal(E/F)trianglelefteq textGal(E/K)$ and $textGal(F/K)congtextGal(E/K)/textGal(E/F)$.



For the normality of subgroups we need $F/K$ is normal. For the isomorphism. Is is true seems that we just need $E/F$ to be normal? Of course, $E/K$ normal implies that. Because we just need to extend $F$ to $E$ (we can skip $K$). Just need a polynomial in $F[X]$ splits in $E$.



Of course, $E/K$ normal implies $E/F$ normal and the former is stronger the latter in general. Is there any reason to state a weaker theorem?





abstract-algebra galois-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 29 at 22:55
























asked Jul 29 at 22:29









Xavier Yang

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  • If the extensions aren't normal then what is $mathrmGal(E/F)$, etc.? Is this just alternative notation for $mathrmAut(E/F)$?
    – Alex Mathers
    Jul 29 at 22:35







  • 1




    What is B in the question?
    – Stefan4024
    Jul 29 at 22:35










  • Is $B$ really $F$?
    – Steve D
    Jul 29 at 22:36










  • OH yes, it was a typo..
    – Xavier Yang
    Jul 29 at 22:58










  • Can you explain what happens when $K=mathbbQ$, $F=mathbbQ(sqrt2)$, and $E=mathbbQ(sqrt[4]2)$?
    – Steve D
    Jul 30 at 0:27
















  • If the extensions aren't normal then what is $mathrmGal(E/F)$, etc.? Is this just alternative notation for $mathrmAut(E/F)$?
    – Alex Mathers
    Jul 29 at 22:35







  • 1




    What is B in the question?
    – Stefan4024
    Jul 29 at 22:35










  • Is $B$ really $F$?
    – Steve D
    Jul 29 at 22:36










  • OH yes, it was a typo..
    – Xavier Yang
    Jul 29 at 22:58










  • Can you explain what happens when $K=mathbbQ$, $F=mathbbQ(sqrt2)$, and $E=mathbbQ(sqrt[4]2)$?
    – Steve D
    Jul 30 at 0:27















If the extensions aren't normal then what is $mathrmGal(E/F)$, etc.? Is this just alternative notation for $mathrmAut(E/F)$?
– Alex Mathers
Jul 29 at 22:35





If the extensions aren't normal then what is $mathrmGal(E/F)$, etc.? Is this just alternative notation for $mathrmAut(E/F)$?
– Alex Mathers
Jul 29 at 22:35





1




1




What is B in the question?
– Stefan4024
Jul 29 at 22:35




What is B in the question?
– Stefan4024
Jul 29 at 22:35












Is $B$ really $F$?
– Steve D
Jul 29 at 22:36




Is $B$ really $F$?
– Steve D
Jul 29 at 22:36












OH yes, it was a typo..
– Xavier Yang
Jul 29 at 22:58




OH yes, it was a typo..
– Xavier Yang
Jul 29 at 22:58












Can you explain what happens when $K=mathbbQ$, $F=mathbbQ(sqrt2)$, and $E=mathbbQ(sqrt[4]2)$?
– Steve D
Jul 30 at 0:27




Can you explain what happens when $K=mathbbQ$, $F=mathbbQ(sqrt2)$, and $E=mathbbQ(sqrt[4]2)$?
– Steve D
Jul 30 at 0:27















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