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Separability over intersection of intermediate fields

“Intersection” of separable subfieldsProve this field extension is separableExtension over the intersection of intermediate fields is also separableGalois extension of intersection of fieldsSplitting fields of polynomials over finite fieldsIntermediate fields of splitting fieldNumber of intermediate field of a finite separable extensionExistence of an intermediate field $Ksubseteq M subseteq L$ such that $[L:M]=p$Intermediate field of a Galois extensionA finite extension is simple iff the purely inseparable closure is simple?Classification of fields with only separable finite degree fields extensionsLet $L:K$ be a Galois extention, show that $L:M$ is a normal.Comparative size of lattice of subgroups and lattice of the intermediate fieldsExtension over the intersection of intermediate fields is also separable

2

$begingroup$

Let $Emathop/F$ be a finite-degree normal extension of fields. Let $K$ and $L$ be intermediate fields (that is, $Fsubseteq Ksubseteq E$ and $Fsubseteq Lsubseteq E$) such that $E$ is separable over both $K$ and $L$. Show that $E$ is separable over $Kcap L$.

So I was thinking that if $f=min_Kcap L(alpha)$. Then $min_K(alpha)$ divides $f$. But if $f$ has repeated roots then I don't see what goes wrong.

Thanks!

galois-theory

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edited Mar 22 at 2:45

Saad

20.3k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

$endgroup$

add a comment | 

2

$begingroup$

Let $Emathop/F$ be a finite-degree normal extension of fields. Let $K$ and $L$ be intermediate fields (that is, $Fsubseteq Ksubseteq E$ and $Fsubseteq Lsubseteq E$) such that $E$ is separable over both $K$ and $L$. Show that $E$ is separable over $Kcap L$.

So I was thinking that if $f=min_Kcap L(alpha)$. Then $min_K(alpha)$ divides $f$. But if $f$ has repeated roots then I don't see what goes wrong.

Thanks!

galois-theory

share|cite|improve this question

edited Mar 22 at 2:45

Saad

20.3k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

$endgroup$

add a comment | 

$begingroup$

Let $Emathop/F$ be a finite-degree normal extension of fields. Let $K$ and $L$ be intermediate fields (that is, $Fsubseteq Ksubseteq E$ and $Fsubseteq Lsubseteq E$) such that $E$ is separable over both $K$ and $L$. Show that $E$ is separable over $Kcap L$.

So I was thinking that if $f=min_Kcap L(alpha)$. Then $min_K(alpha)$ divides $f$. But if $f$ has repeated roots then I don't see what goes wrong.

Thanks!

galois-theory

share|cite|improve this question

edited Mar 22 at 2:45

Saad

20.3k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

$endgroup$

share|cite|improve this question

edited Mar 22 at 2:45

Saad

20.3k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

share|cite|improve this question

edited Mar 22 at 2:45

Saad

20.3k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

edited Mar 22 at 2:45

Saad

20.3k92352

edited Mar 22 at 2:45

Saad

20.3k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

asked Jan 29 '15 at 8:58

ZeusZeus

936

1 Answer
1

active

oldest

votes

1

$begingroup$

$E/F$ is normal implies $E/K$ and $E/L$ are both normal. Since they are also separable, we get that $E/K$ and $E/L$ are Galois. Set $G_1=Gal(E/K)$ and $G_2= Gal(E/L)$. Note that $E^G_1=K$ and $E^G_2=L$.

Now consider $G=Aut(E/Kcap L)$ and $E^G$. Obviously, $Kcap Lleq E^G$. Since obviously $G_1leq G$ and $G_2leq G$, we get that $E^Gleq E^G_1=K$ and $E^Gleq E^G_2=L$, hence $E^Gleq Kcap L$.

So, $E^G=Kcap L$, i.e. $E^Aut(E/Kcap L)=Kcap L$, hence $E/Kcap L$ is Galois, and therefore separable.

share|cite|improve this answer

edited Jan 30 '15 at 22:03

answered Jan 30 '15 at 21:55

SMMSMM

2,978511

$endgroup$

add a comment | 

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1

$begingroup$

$E/F$ is normal implies $E/K$ and $E/L$ are both normal. Since they are also separable, we get that $E/K$ and $E/L$ are Galois. Set $G_1=Gal(E/K)$ and $G_2= Gal(E/L)$. Note that $E^G_1=K$ and $E^G_2=L$.

Now consider $G=Aut(E/Kcap L)$ and $E^G$. Obviously, $Kcap Lleq E^G$. Since obviously $G_1leq G$ and $G_2leq G$, we get that $E^Gleq E^G_1=K$ and $E^Gleq E^G_2=L$, hence $E^Gleq Kcap L$.

So, $E^G=Kcap L$, i.e. $E^Aut(E/Kcap L)=Kcap L$, hence $E/Kcap L$ is Galois, and therefore separable.

share|cite|improve this answer

edited Jan 30 '15 at 22:03

answered Jan 30 '15 at 21:55

SMMSMM

2,978511

$endgroup$

add a comment | 

1

$begingroup$

$E/F$ is normal implies $E/K$ and $E/L$ are both normal. Since they are also separable, we get that $E/K$ and $E/L$ are Galois. Set $G_1=Gal(E/K)$ and $G_2= Gal(E/L)$. Note that $E^G_1=K$ and $E^G_2=L$.

Now consider $G=Aut(E/Kcap L)$ and $E^G$. Obviously, $Kcap Lleq E^G$. Since obviously $G_1leq G$ and $G_2leq G$, we get that $E^Gleq E^G_1=K$ and $E^Gleq E^G_2=L$, hence $E^Gleq Kcap L$.

So, $E^G=Kcap L$, i.e. $E^Aut(E/Kcap L)=Kcap L$, hence $E/Kcap L$ is Galois, and therefore separable.

share|cite|improve this answer

edited Jan 30 '15 at 22:03

answered Jan 30 '15 at 21:55

SMMSMM

2,978511

$endgroup$

add a comment | 

$begingroup$

$E/F$ is normal implies $E/K$ and $E/L$ are both normal. Since they are also separable, we get that $E/K$ and $E/L$ are Galois. Set $G_1=Gal(E/K)$ and $G_2= Gal(E/L)$. Note that $E^G_1=K$ and $E^G_2=L$.

Now consider $G=Aut(E/Kcap L)$ and $E^G$. Obviously, $Kcap Lleq E^G$. Since obviously $G_1leq G$ and $G_2leq G$, we get that $E^Gleq E^G_1=K$ and $E^Gleq E^G_2=L$, hence $E^Gleq Kcap L$.

So, $E^G=Kcap L$, i.e. $E^Aut(E/Kcap L)=Kcap L$, hence $E/Kcap L$ is Galois, and therefore separable.

share|cite|improve this answer

edited Jan 30 '15 at 22:03

answered Jan 30 '15 at 21:55

SMMSMM

2,978511

$endgroup$

share|cite|improve this answer

edited Jan 30 '15 at 22:03

answered Jan 30 '15 at 21:55

SMMSMM

2,978511

answered Jan 30 '15 at 21:55

SMMSMM

2,978511

answered Jan 30 '15 at 21:55

SMMSMM

2,978511