Let $L:K$ be a Galois extention, show that $L:M$ is a normal.
Assume the field extension $L:K$ is Galois (i.e. finite, normal and separable), with $M$ an intermediate field.
- Show that $L:M$ and $M:K$ are finite separable field extensions.
Attempt: Both $M:K$ and $L:M$ are algebraic ($M:K$ since $M subset L$, so every $alpha in M$ is algebraic over $K$ and $L:M$ since any $alpha in L$ is algebraic over $K$, so also over $M$).
Then $M:K$ is separable again since $M subset L$ (every element of $L$ so in particular every element of $M$ is separable over $K$).
For $L:M$, take any $alpha in L$, then min$_M(alpha)$ divides min$_K(alpha)$ in $M[X]$. Since min$_K(alpha)$ has no multiple zeros in a splitting field, neither does min$_M(alpha)$, i.e. $alpha$ is separable over $M$.
I'm not quite about the proof for finite, would it involve the Tower Law?
- Show that $L:M$ is a normal extension.
I'm a bit stuck here, any help for this one?
galois-theory extension-field normal-subgroups separable-extension
edited Jun 2 at 17:40
Jyrki Lahtonen
108k12166367
asked Jun 2 at 14:37
user12002
775
add a comment |
Assume the field extension $L:K$ is Galois (i.e. finite, normal and separable), with $M$ an intermediate field.
- Show that $L:M$ and $M:K$ are finite separable field extensions.
Attempt: Both $M:K$ and $L:M$ are algebraic ($M:K$ since $M subset L$, so every $alpha in M$ is algebraic over $K$ and $L:M$ since any $alpha in L$ is algebraic over $K$, so also over $M$).
Then $M:K$ is separable again since $M subset L$ (every element of $L$ so in particular every element of $M$ is separable over $K$).
For $L:M$, take any $alpha in L$, then min$_M(alpha)$ divides min$_K(alpha)$ in $M[X]$. Since min$_K(alpha)$ has no multiple zeros in a splitting field, neither does min$_M(alpha)$, i.e. $alpha$ is separable over $M$.
I'm not quite about the proof for finite, would it involve the Tower Law?
- Show that $L:M$ is a normal extension.
I'm a bit stuck here, any help for this one?
galois-theory extension-field normal-subgroups separable-extension
edited Jun 2 at 17:40
Jyrki Lahtonen
108k12166367
asked Jun 2 at 14:37
user12002
775
1
Your proof will depend on the definition of normality that you’re using. What’s yours?
– Lubin
Jun 2 at 19:37
@Lubin "A field extension $L:K$ is normal if every irreducible polynomial over $K$ which has at least one zero in $L$ splits in $L$."
– user12002
Jun 2 at 21:28
Take an $M$-irreducible polynomial $f$ with a root $alpha$ in $L$. Look at the minimal (irreducible) $K$-polynomial for $alpha$, call it $g$. What do you know about $f$ versus $g$? (You might look at some examples.)
– Lubin
Jun 3 at 1:56
add a comment |
Assume the field extension $L:K$ is Galois (i.e. finite, normal and separable), with $M$ an intermediate field.
- Show that $L:M$ and $M:K$ are finite separable field extensions.
Attempt: Both $M:K$ and $L:M$ are algebraic ($M:K$ since $M subset L$, so every $alpha in M$ is algebraic over $K$ and $L:M$ since any $alpha in L$ is algebraic over $K$, so also over $M$).
Then $M:K$ is separable again since $M subset L$ (every element of $L$ so in particular every element of $M$ is separable over $K$).
For $L:M$, take any $alpha in L$, then min$_M(alpha)$ divides min$_K(alpha)$ in $M[X]$. Since min$_K(alpha)$ has no multiple zeros in a splitting field, neither does min$_M(alpha)$, i.e. $alpha$ is separable over $M$.
I'm not quite about the proof for finite, would it involve the Tower Law?
- Show that $L:M$ is a normal extension.
I'm a bit stuck here, any help for this one?
galois-theory extension-field normal-subgroups separable-extension
edited Jun 2 at 17:40
Jyrki Lahtonen
108k12166367
asked Jun 2 at 14:37
user12002
775
Assume the field extension $L:K$ is Galois (i.e. finite, normal and separable), with $M$ an intermediate field.
- Show that $L:M$ and $M:K$ are finite separable field extensions.
Attempt: Both $M:K$ and $L:M$ are algebraic ($M:K$ since $M subset L$, so every $alpha in M$ is algebraic over $K$ and $L:M$ since any $alpha in L$ is algebraic over $K$, so also over $M$).
Then $M:K$ is separable again since $M subset L$ (every element of $L$ so in particular every element of $M$ is separable over $K$).
For $L:M$, take any $alpha in L$, then min$_M(alpha)$ divides min$_K(alpha)$ in $M[X]$. Since min$_K(alpha)$ has no multiple zeros in a splitting field, neither does min$_M(alpha)$, i.e. $alpha$ is separable over $M$.
I'm not quite about the proof for finite, would it involve the Tower Law?
- Show that $L:M$ is a normal extension.
I'm a bit stuck here, any help for this one?
galois-theory extension-field normal-subgroups separable-extension
galois-theory extension-field normal-subgroups separable-extension
edited Jun 2 at 17:40
Jyrki Lahtonen
108k12166367
asked Jun 2 at 14:37
user12002
775
edited Jun 2 at 17:40
Jyrki Lahtonen
108k12166367
asked Jun 2 at 14:37
user12002
775
edited Jun 2 at 17:40
Jyrki Lahtonen
108k12166367
edited Jun 2 at 17:40
Jyrki Lahtonen
108k12166367
edited Jun 2 at 17:40
Jyrki Lahtonen
108k12166367
108k12166367
asked Jun 2 at 14:37
user12002
775
asked Jun 2 at 14:37
user12002
775
asked Jun 2 at 14:37
user12002
775
775
1
Your proof will depend on the definition of normality that you’re using. What’s yours?
– Lubin
Jun 2 at 19:37
@Lubin "A field extension $L:K$ is normal if every irreducible polynomial over $K$ which has at least one zero in $L$ splits in $L$."
– user12002
Jun 2 at 21:28
Take an $M$-irreducible polynomial $f$ with a root $alpha$ in $L$. Look at the minimal (irreducible) $K$-polynomial for $alpha$, call it $g$. What do you know about $f$ versus $g$? (You might look at some examples.)
– Lubin
Jun 3 at 1:56
add a comment |
1
Your proof will depend on the definition of normality that you’re using. What’s yours?
– Lubin
Jun 2 at 19:37
@Lubin "A field extension $L:K$ is normal if every irreducible polynomial over $K$ which has at least one zero in $L$ splits in $L$."
– user12002
Jun 2 at 21:28
Take an $M$-irreducible polynomial $f$ with a root $alpha$ in $L$. Look at the minimal (irreducible) $K$-polynomial for $alpha$, call it $g$. What do you know about $f$ versus $g$? (You might look at some examples.)
– Lubin
Jun 3 at 1:56
1
1
Your proof will depend on the definition of normality that you’re using. What’s yours?
– Lubin
Jun 2 at 19:37
Your proof will depend on the definition of normality that you’re using. What’s yours?
– Lubin
Jun 2 at 19:37
@Lubin "A field extension $L:K$ is normal if every irreducible polynomial over $K$ which has at least one zero in $L$ splits in $L$."
– user12002
Jun 2 at 21:28
@Lubin "A field extension $L:K$ is normal if every irreducible polynomial over $K$ which has at least one zero in $L$ splits in $L$."
– user12002
Jun 2 at 21:28
Take an $M$-irreducible polynomial $f$ with a root $alpha$ in $L$. Look at the minimal (irreducible) $K$-polynomial for $alpha$, call it $g$. What do you know about $f$ versus $g$? (You might look at some examples.)
– Lubin
Jun 3 at 1:56
Take an $M$-irreducible polynomial $f$ with a root $alpha$ in $L$. Look at the minimal (irreducible) $K$-polynomial for $alpha$, call it $g$. What do you know about $f$ versus $g$? (You might look at some examples.)
– Lubin
Jun 3 at 1:56
add a comment |
1 Answer
1
active
oldest
votes
Yes, to show that $L/M$ and $M/K$ are finite extensions use the Tower Law,
$$
[L:K] = [L:M][M:K].
$$
To show that $L/M$ is a normal extension, let $f in M[x]$ be an irreducible polynomial which has a root $alpha$ in $L$. We want to show that $f$ splits over $L$. Let $g in K[x]$ be the minimal polynomial of $alpha$ over $K$. Then, $f$ divides $g$. Since $L/K$ is normal, $g$ splits over $L$ and therefore so does $f$.
answered Nov 24 at 11:53
Brahadeesh
6,11742360
1
Happy crusading :-)
– Jyrki Lahtonen
Nov 29 at 4:50
@JyrkiLahtonen Thank you for the encouragement :)
– Brahadeesh
Nov 29 at 7:10
add a comment |
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Yes, to show that $L/M$ and $M/K$ are finite extensions use the Tower Law,
$$
[L:K] = [L:M][M:K].
$$
To show that $L/M$ is a normal extension, let $f in M[x]$ be an irreducible polynomial which has a root $alpha$ in $L$. We want to show that $f$ splits over $L$. Let $g in K[x]$ be the minimal polynomial of $alpha$ over $K$. Then, $f$ divides $g$. Since $L/K$ is normal, $g$ splits over $L$ and therefore so does $f$.
answered Nov 24 at 11:53
Brahadeesh
6,11742360
1
Happy crusading :-)
– Jyrki Lahtonen
Nov 29 at 4:50
@JyrkiLahtonen Thank you for the encouragement :)
– Brahadeesh
Nov 29 at 7:10
add a comment |
Yes, to show that $L/M$ and $M/K$ are finite extensions use the Tower Law,
$$
[L:K] = [L:M][M:K].
$$
To show that $L/M$ is a normal extension, let $f in M[x]$ be an irreducible polynomial which has a root $alpha$ in $L$. We want to show that $f$ splits over $L$. Let $g in K[x]$ be the minimal polynomial of $alpha$ over $K$. Then, $f$ divides $g$. Since $L/K$ is normal, $g$ splits over $L$ and therefore so does $f$.
answered Nov 24 at 11:53
Brahadeesh
6,11742360
1
Happy crusading :-)
– Jyrki Lahtonen
Nov 29 at 4:50
@JyrkiLahtonen Thank you for the encouragement :)
– Brahadeesh
Nov 29 at 7:10
add a comment |
Yes, to show that $L/M$ and $M/K$ are finite extensions use the Tower Law,
$$
[L:K] = [L:M][M:K].
$$
To show that $L/M$ is a normal extension, let $f in M[x]$ be an irreducible polynomial which has a root $alpha$ in $L$. We want to show that $f$ splits over $L$. Let $g in K[x]$ be the minimal polynomial of $alpha$ over $K$. Then, $f$ divides $g$. Since $L/K$ is normal, $g$ splits over $L$ and therefore so does $f$.
answered Nov 24 at 11:53
Brahadeesh
6,11742360
Yes, to show that $L/M$ and $M/K$ are finite extensions use the Tower Law,
$$
[L:K] = [L:M][M:K].
$$
To show that $L/M$ is a normal extension, let $f in M[x]$ be an irreducible polynomial which has a root $alpha$ in $L$. We want to show that $f$ splits over $L$. Let $g in K[x]$ be the minimal polynomial of $alpha$ over $K$. Then, $f$ divides $g$. Since $L/K$ is normal, $g$ splits over $L$ and therefore so does $f$.
answered Nov 24 at 11:53
Brahadeesh
6,11742360
edited Nov 28 at 17:42
answered Nov 24 at 11:53
Brahadeesh
6,11742360
answered Nov 24 at 11:53
Brahadeesh
6,11742360
answered Nov 24 at 11:53
Brahadeesh
6,11742360
6,11742360
1
Happy crusading :-)
– Jyrki Lahtonen
Nov 29 at 4:50
@JyrkiLahtonen Thank you for the encouragement :)
– Brahadeesh
Nov 29 at 7:10
add a comment |
1
Happy crusading :-)
– Jyrki Lahtonen
Nov 29 at 4:50
@JyrkiLahtonen Thank you for the encouragement :)
– Brahadeesh
Nov 29 at 7:10
1
1
Happy crusading :-)
– Jyrki Lahtonen
Nov 29 at 4:50
Happy crusading :-)
– Jyrki Lahtonen
Nov 29 at 4:50
@JyrkiLahtonen Thank you for the encouragement :)
– Brahadeesh
Nov 29 at 7:10
@JyrkiLahtonen Thank you for the encouragement :)
– Brahadeesh
Nov 29 at 7:10
add a comment |
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Your proof will depend on the definition of normality that you’re using. What’s yours?
– Lubin
Jun 2 at 19:37
@Lubin "A field extension $L:K$ is normal if every irreducible polynomial over $K$ which has at least one zero in $L$ splits in $L$."
– user12002
Jun 2 at 21:28
Take an $M$-irreducible polynomial $f$ with a root $alpha$ in $L$. Look at the minimal (irreducible) $K$-polynomial for $alpha$, call it $g$. What do you know about $f$ versus $g$? (You might look at some examples.)
– Lubin
Jun 3 at 1:56