Mixing
Two forms of the primitive element theorem.
1
$begingroup$
The primitive element theorem says that a finite separable extension is simple. As I understand it, there is another primitive element theorem that states that an extension is simple iff there are finitely many intermediate fields. How can we deduce one from another?
abstract-algebra
edited Jan 2 at 21:48
user26857
39.3k124183
asked Jan 2 at 20:14
roi_saumonroi_saumon
45628
$endgroup$
add a comment |
1
$begingroup$
The primitive element theorem says that a finite separable extension is simple. As I understand it, there is another primitive element theorem that states that an extension is simple iff there are finitely many intermediate fields. How can we deduce one from another?
abstract-algebra
edited Jan 2 at 21:48
user26857
39.3k124183
asked Jan 2 at 20:14
roi_saumonroi_saumon
45628
$endgroup$
$begingroup$
I think it is a duplicate of this question, together with the above. A finite separable extension has only finitely many intermediate fields and hence is simple, i.e., monogeneous. The converse is clear. Both is mentioned at wikipedia.
$endgroup$
– Dietrich Burde
Jan 2 at 20:34
$begingroup$
The question is indeed similar, but how to prove it using Galois theory?
$endgroup$
– roi_saumon
Jan 2 at 20:34
$begingroup$
In the notes by Milne, chapter $5$ there is a "proof" (well, rather short). Would you mind to delete your edit? It is no longer valid.
$endgroup$
– Dietrich Burde
Jan 2 at 20:38
$begingroup$
Oh, I think I understand : if E/F is finite and separable, take the normal closure N/E/F, then N/F is Galois and so the intermediate fields of E/F which are also intermediate fields of N/F are in 1-1 correspondence with the subgroups of gal(N/F). but since this is a finite group it has only finitely many subgroups. Is that the right way to think?
$endgroup$
– roi_saumon
Jan 3 at 0:19
add a comment |
1
1
1
$begingroup$
The primitive element theorem says that a finite separable extension is simple. As I understand it, there is another primitive element theorem that states that an extension is simple iff there are finitely many intermediate fields. How can we deduce one from another?
abstract-algebra
edited Jan 2 at 21:48
user26857
39.3k124183
asked Jan 2 at 20:14
roi_saumonroi_saumon
45628
$endgroup$
The primitive element theorem says that a finite separable extension is simple. As I understand it, there is another primitive element theorem that states that an extension is simple iff there are finitely many intermediate fields. How can we deduce one from another?
abstract-algebra
abstract-algebra
edited Jan 2 at 21:48
user26857
39.3k124183
asked Jan 2 at 20:14
roi_saumonroi_saumon
45628
edited Jan 2 at 21:48
user26857
39.3k124183
asked Jan 2 at 20:14
roi_saumonroi_saumon
45628
edited Jan 2 at 21:48
user26857
39.3k124183
edited Jan 2 at 21:48
user26857
39.3k124183
edited Jan 2 at 21:48
user26857
39.3k124183
39.3k124183
asked Jan 2 at 20:14
roi_saumonroi_saumon
45628
asked Jan 2 at 20:14
roi_saumonroi_saumon
45628
asked Jan 2 at 20:14
roi_saumonroi_saumon
45628
45628
$begingroup$
I think it is a duplicate of this question, together with the above. A finite separable extension has only finitely many intermediate fields and hence is simple, i.e., monogeneous. The converse is clear. Both is mentioned at wikipedia.
$endgroup$
– Dietrich Burde
Jan 2 at 20:34
$begingroup$
The question is indeed similar, but how to prove it using Galois theory?
$endgroup$
– roi_saumon
Jan 2 at 20:34
$begingroup$
In the notes by Milne, chapter $5$ there is a "proof" (well, rather short). Would you mind to delete your edit? It is no longer valid.
$endgroup$
– Dietrich Burde
Jan 2 at 20:38
$begingroup$
Oh, I think I understand : if E/F is finite and separable, take the normal closure N/E/F, then N/F is Galois and so the intermediate fields of E/F which are also intermediate fields of N/F are in 1-1 correspondence with the subgroups of gal(N/F). but since this is a finite group it has only finitely many subgroups. Is that the right way to think?
$endgroup$
– roi_saumon
Jan 3 at 0:19
add a comment |
$begingroup$
I think it is a duplicate of this question, together with the above. A finite separable extension has only finitely many intermediate fields and hence is simple, i.e., monogeneous. The converse is clear. Both is mentioned at wikipedia.
$endgroup$
– Dietrich Burde
Jan 2 at 20:34
$begingroup$
The question is indeed similar, but how to prove it using Galois theory?
$endgroup$
– roi_saumon
Jan 2 at 20:34
$begingroup$
In the notes by Milne, chapter $5$ there is a "proof" (well, rather short). Would you mind to delete your edit? It is no longer valid.
$endgroup$
– Dietrich Burde
Jan 2 at 20:38
$begingroup$
Oh, I think I understand : if E/F is finite and separable, take the normal closure N/E/F, then N/F is Galois and so the intermediate fields of E/F which are also intermediate fields of N/F are in 1-1 correspondence with the subgroups of gal(N/F). but since this is a finite group it has only finitely many subgroups. Is that the right way to think?
$endgroup$
– roi_saumon
Jan 3 at 0:19
$begingroup$
I think it is a duplicate of this question, together with the above. A finite separable extension has only finitely many intermediate fields and hence is simple, i.e., monogeneous. The converse is clear. Both is mentioned at wikipedia.
$endgroup$
– Dietrich Burde
Jan 2 at 20:34
$begingroup$
I think it is a duplicate of this question, together with the above. A finite separable extension has only finitely many intermediate fields and hence is simple, i.e., monogeneous. The converse is clear. Both is mentioned at wikipedia.
$endgroup$
– Dietrich Burde
Jan 2 at 20:34
$begingroup$
The question is indeed similar, but how to prove it using Galois theory?
$endgroup$
– roi_saumon
Jan 2 at 20:34
$begingroup$
The question is indeed similar, but how to prove it using Galois theory?
$endgroup$
– roi_saumon
Jan 2 at 20:34
$begingroup$
In the notes by Milne, chapter $5$ there is a "proof" (well, rather short). Would you mind to delete your edit? It is no longer valid.
$endgroup$
– Dietrich Burde
Jan 2 at 20:38
$begingroup$
In the notes by Milne, chapter $5$ there is a "proof" (well, rather short). Would you mind to delete your edit? It is no longer valid.
$endgroup$
– Dietrich Burde
Jan 2 at 20:38
$begingroup$
Oh, I think I understand : if E/F is finite and separable, take the normal closure N/E/F, then N/F is Galois and so the intermediate fields of E/F which are also intermediate fields of N/F are in 1-1 correspondence with the subgroups of gal(N/F). but since this is a finite group it has only finitely many subgroups. Is that the right way to think?
$endgroup$
– roi_saumon
Jan 3 at 0:19
$begingroup$
Oh, I think I understand : if E/F is finite and separable, take the normal closure N/E/F, then N/F is Galois and so the intermediate fields of E/F which are also intermediate fields of N/F are in 1-1 correspondence with the subgroups of gal(N/F). but since this is a finite group it has only finitely many subgroups. Is that the right way to think?
$endgroup$
– roi_saumon
Jan 3 at 0:19
add a comment |
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$begingroup$
I think it is a duplicate of this question, together with the above. A finite separable extension has only finitely many intermediate fields and hence is simple, i.e., monogeneous. The converse is clear. Both is mentioned at wikipedia.
$endgroup$
– Dietrich Burde
Jan 2 at 20:34
$begingroup$
The question is indeed similar, but how to prove it using Galois theory?
$endgroup$
– roi_saumon
Jan 2 at 20:34
$begingroup$
In the notes by Milne, chapter $5$ there is a "proof" (well, rather short). Would you mind to delete your edit? It is no longer valid.
$endgroup$
– Dietrich Burde
Jan 2 at 20:38
$begingroup$
Oh, I think I understand : if E/F is finite and separable, take the normal closure N/E/F, then N/F is Galois and so the intermediate fields of E/F which are also intermediate fields of N/F are in 1-1 correspondence with the subgroups of gal(N/F). but since this is a finite group it has only finitely many subgroups. Is that the right way to think?
$endgroup$
– roi_saumon
Jan 3 at 0:19