Timeline for why are subextensions of Galois extensions also Galois?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 1, 2013 at 13:41 | comment | added | Mozibur Ullah | @Mueller: I'm beginning to suspect that Delgado is wrong in his claim, particularly the 'immediacy' of the deduction... | |
| Feb 1, 2013 at 10:01 | comment | added | Martin Brandenburg | My comments refer to older versions of the answer. | |
| Feb 1, 2013 at 7:03 | history | edited | Peter Mueller | CC BY-SA 3.0 |
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| Feb 1, 2013 at 0:14 | comment | added | Martin Brandenburg | You show that $a$ is the unique root of $f$, but this only implies that $f(x)=x-a$ if we already knew that $L/M$ is separable. You also use in the proof that $L/M$ is normal. Therefore, again this is just the proof (reducing to the statements for normal and separable) which Mozibur wants to avoid. Of course, I don't claim that this is possible at all, but it would be interesting. | |
| Jan 31, 2013 at 23:37 | history | edited | Peter Mueller | CC BY-SA 3.0 |
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| Jan 31, 2013 at 19:35 | comment | added | Martin Brandenburg | Once again, this doesn't answer the question, and probably Mozibur knows all that (see also the math.SE discussion). Mozibur has asked if there is a proof which avoids the usual characterization of Galois extensions as well as the Main Thm on Galois theory, because Robalo Delgados indicates that this is possible. And even for finite extensions this is an interesting question. | |
| Jan 31, 2013 at 18:36 | history | answered | Peter Mueller | CC BY-SA 3.0 |