Comparative size of lattice of subgroups and lattice of the intermediate fields
$begingroup$
If an extension of fields $E/F$ is normal and separable, Galois theory tells us that the intermediate fields $E supset M supset F$ are in 1-1 correspondence with the proper of the Galois group $operatorname{Gal}(E/F)$. But if the extension is for instance not normal, then which lattice is in general bigger? The one of the subgroups or the one of the intermediate fields? Are there easy example for the case "the latice of the intermediate fields is bigger"/"the lattice of the subgroups is bigger"?
galois-theory
edited Dec 29 '18 at 13:39
Bernard
123k741117
asked Dec 29 '18 at 13:36
roi_saumonroi_saumon
61838
$endgroup$
add a comment |
$begingroup$
If an extension of fields $E/F$ is normal and separable, Galois theory tells us that the intermediate fields $E supset M supset F$ are in 1-1 correspondence with the proper of the Galois group $operatorname{Gal}(E/F)$. But if the extension is for instance not normal, then which lattice is in general bigger? The one of the subgroups or the one of the intermediate fields? Are there easy example for the case "the latice of the intermediate fields is bigger"/"the lattice of the subgroups is bigger"?
galois-theory
edited Dec 29 '18 at 13:39
Bernard
123k741117
asked Dec 29 '18 at 13:36
roi_saumonroi_saumon
61838
$endgroup$
add a comment |
$begingroup$
If an extension of fields $E/F$ is normal and separable, Galois theory tells us that the intermediate fields $E supset M supset F$ are in 1-1 correspondence with the proper of the Galois group $operatorname{Gal}(E/F)$. But if the extension is for instance not normal, then which lattice is in general bigger? The one of the subgroups or the one of the intermediate fields? Are there easy example for the case "the latice of the intermediate fields is bigger"/"the lattice of the subgroups is bigger"?
galois-theory
edited Dec 29 '18 at 13:39
Bernard
123k741117
asked Dec 29 '18 at 13:36
roi_saumonroi_saumon
61838
$endgroup$
If an extension of fields $E/F$ is normal and separable, Galois theory tells us that the intermediate fields $E supset M supset F$ are in 1-1 correspondence with the proper of the Galois group $operatorname{Gal}(E/F)$. But if the extension is for instance not normal, then which lattice is in general bigger? The one of the subgroups or the one of the intermediate fields? Are there easy example for the case "the latice of the intermediate fields is bigger"/"the lattice of the subgroups is bigger"?
galois-theory
galois-theory
edited Dec 29 '18 at 13:39
Bernard
123k741117
asked Dec 29 '18 at 13:36
roi_saumonroi_saumon
61838
edited Dec 29 '18 at 13:39
Bernard
123k741117
asked Dec 29 '18 at 13:36
roi_saumonroi_saumon
61838
edited Dec 29 '18 at 13:39
Bernard
123k741117
edited Dec 29 '18 at 13:39
Bernard
123k741117
edited Dec 29 '18 at 13:39
Bernard
123k741117
123k741117
asked Dec 29 '18 at 13:36
roi_saumonroi_saumon
61838
asked Dec 29 '18 at 13:36
roi_saumonroi_saumon
61838
asked Dec 29 '18 at 13:36
roi_saumonroi_saumon
61838
61838
add a comment |
add a comment |
1 Answer
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$begingroup$
For a purely inseparable normal extension
(for instance, $k(X,Y)$ over $k(X^p,Y^p)$ for $k$
of characteristic $p$), the Galois group is trivial
but the lattice of intermediate fields isn't.
In general distinct subgroups of the Galois group will determine
subfields as fixed fields, but one need not get all subfields.
answered Dec 29 '18 at 13:40
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
For a purely inseparable normal extension
(for instance, $k(X,Y)$ over $k(X^p,Y^p)$ for $k$
of characteristic $p$), the Galois group is trivial
but the lattice of intermediate fields isn't.
In general distinct subgroups of the Galois group will determine
subfields as fixed fields, but one need not get all subfields.
answered Dec 29 '18 at 13:40
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
For a purely inseparable normal extension
(for instance, $k(X,Y)$ over $k(X^p,Y^p)$ for $k$
of characteristic $p$), the Galois group is trivial
but the lattice of intermediate fields isn't.
In general distinct subgroups of the Galois group will determine
subfields as fixed fields, but one need not get all subfields.
answered Dec 29 '18 at 13:40
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
For a purely inseparable normal extension
(for instance, $k(X,Y)$ over $k(X^p,Y^p)$ for $k$
of characteristic $p$), the Galois group is trivial
but the lattice of intermediate fields isn't.
In general distinct subgroups of the Galois group will determine
subfields as fixed fields, but one need not get all subfields.
answered Dec 29 '18 at 13:40
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
For a purely inseparable normal extension
(for instance, $k(X,Y)$ over $k(X^p,Y^p)$ for $k$
of characteristic $p$), the Galois group is trivial
but the lattice of intermediate fields isn't.
In general distinct subgroups of the Galois group will determine
subfields as fixed fields, but one need not get all subfields.
answered Dec 29 '18 at 13:40
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Dec 29 '18 at 13:40
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Dec 29 '18 at 13:40
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Dec 29 '18 at 13:40
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
add a comment |
add a comment |
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