Timeline for why are subextensions of Galois extensions also Galois?
Current License: CC BY-SA 4.0
19 events
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| S Aug 26, 2023 at 6:31 | history | suggested | Linuxmetel | CC BY-SA 4.0 |
fix a dead link (of the refered thesis)
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| Aug 25, 2023 at 19:33 | review | Suggested edits | |||
| S Aug 26, 2023 at 6:31 | |||||
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Feb 1, 2013 at 15:52 | comment | added | Martin Brandenburg | @ACL: You use that $L/M$ is normal and separable. | |
| Feb 1, 2013 at 14:52 | comment | added | David Benjamin Lim | @ACL Would you like to post your comment above as an answer? | |
| Feb 1, 2013 at 13:43 | history | edited | Mozibur Ullah | CC BY-SA 3.0 |
added 10 characters in body
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| Feb 1, 2013 at 12:02 | comment | added | ACL | @Martin & Mozibur. (Assuming $G=\mathop{\rm Aut}_K(L)$ finite). It is obvious that $M$ is contained in $L^{\mathop{\rm Aut}_M(L)}$, what needs to be shown is the other inclusion. Or: if $x\not\in M$, there exists $g\in\mathop{\rm Aut}_M(L)$ such that $g(x)\neq x$. Now, all conjugates of $x$ over $M$ belong to $L$, and one of them, say $y$, is distinct from $x$. This gives an $M$-linear morphism $M[x]\to L$ such that $x\mapsto y$. Going on, this morphism can be extended to a $M$-linear morphism from $L$ to $L$, which is then an element $g$ of $G$ such that $g(x)=y$. | |
| Feb 1, 2013 at 0:58 | answer | added | Will Sawin | timeline score: 9 | |
| Jan 31, 2013 at 23:40 | comment | added | Zhen Lin | Szamuely uses the same definition of "Galois extension" in Galois groups and fundamental groups, but immediately proves a lemma showing that Galois extensions are exactly the normal separable extensions. It's certainly not obvious to me that $L \mid M$ is automatically Galois if $L \mid K$ is; in my view this is already half of the fundamental theorem of Galois theory! | |
| Jan 31, 2013 at 22:40 | history | edited | Mozibur Ullah | CC BY-SA 3.0 |
added 136 characters in body
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| Jan 31, 2013 at 18:52 | comment | added | Mozibur Ullah | @Brandenburg: ditto. | |
| Jan 31, 2013 at 18:51 | comment | added | Mozibur Ullah | @Quid: Yes, this is what I was getting at. | |
| Jan 31, 2013 at 18:36 | answer | added | Peter Mueller | timeline score: 10 | |
| Jan 31, 2013 at 18:22 | comment | added | Martin Brandenburg | This is an interesting question, which should be read more carefully (especially by the ones who down/close voted it because it seems to be elementary). Here a reformulation: If $L/M/K$ are algebraic field extensions, and $K = L^{\mathrm{Aut}_K(L)}$, how can we prove directly that $M = L^{\mathrm{Aut}_M(L)}$? | |
| Jan 31, 2013 at 17:45 | comment | added | user9072 | The question as I read it seems really not answered on math.SE, possiblty because it was misunderstood. OP does not seem to ask how one can proof this at all but rather: Suppose we define an extension to be Galois if the field fixed under Aut_K(L) is K. Is there then an 'obvious' reason that for an intermideate field M also the extension L over M is Galois. [I am not sure this is an appriate question ATM; an have no time to decide, but in any case I feel the question is partly misunderstood.] | |
| Jan 31, 2013 at 17:23 | comment | added | Mozibur Ullah | @Tveiten: I have, and I still don't think its answered my question. They go via the route of characterising Galois extensions first as algebraic, separable & normal extensions - and then show the property that Delgado uses to characterise a Galois extension follows. Whereas Delgado starts of with this and deduces the characterisation. | |
| Jan 31, 2013 at 17:14 | history | edited | Mozibur Ullah | CC BY-SA 3.0 |
added 167 characters in body
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| Jan 31, 2013 at 14:44 | comment | added | Ketil Tveiten | I can't see how the question wasn't answered in the Stackexchange thread. Try reading it again, and consulting your Galois theory textbook? | |
| Jan 31, 2013 at 13:52 | history | asked | Mozibur Ullah | CC BY-SA 3.0 |