Timeline for A simple proof of the fundamental theorem of Galois theory
Current License: CC BY-SA 4.0
22 events
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| Nov 15, 2022 at 10:25 | comment | added | Martin Brandenburg | @JochenWengenroth Thank you for your kind words! Actually I've been out of academia since 2014. And the math game is an on/off since then. | |
| Nov 15, 2022 at 10:24 | comment | added | Martin Brandenburg | @YCor Thanks for the reference. Actually all three lemmas in this section appear in the paper by Bialynicki-Birula, Browkin, Schinzel which I cite. I just repeated the proofs for the convenience of the reader. | |
| Nov 15, 2022 at 9:07 | comment | added | Jochen Wengenroth | That your days as an active mathematician are over probably means that you continue your career outside academia. Good luck, Martin! I hope you are staying active in mathematics and, in particular, on MO. | |
| Nov 15, 2022 at 7:28 | comment | added | YCor | About Lemma 3.2: it's contained in a result of B.H. Neumann (Groups covered by finitely many cosets. Publ. Math. Debrecen 3 (1954), 227-242) which says that if a group is covered by finitely many cosets, it's covered by just those of finite index. In particular if the cover is not redundant, then all have finite index. So, if a group is covered in a non-redundant way by finitely many subgroups $G_i$, then the intersection $\bigcap G_i$ has finite index. | |
| Nov 15, 2022 at 2:03 | vote | accept | Martin Brandenburg | ||
| Nov 15, 2022 at 2:00 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Nov 13, 2022 at 22:33 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Nov 8, 2022 at 19:42 | answer | added | Peter Mueller | timeline score: 10 | |
| Nov 8, 2022 at 19:24 | comment | added | coudy | The "neat combinatorial argument" is in Bourbaki, algebre, ch. V.40 (the section on the primitive element). | |
| Nov 8, 2022 at 17:47 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Nov 6, 2022 at 15:27 | comment | added | Benjamin Steinberg | I first saw it used for the primitive element in Algebre Theories Galoisiennes by Douady | |
| Nov 6, 2022 at 15:25 | comment | added | Benjamin Steinberg | In Prop 4.2 you have L is separable over L^H since each element satisfies the separable polynomial where you take the product over all distinct images of the element over H | |
| Nov 6, 2022 at 14:57 | comment | added | Martin Brandenburg | @Pierre-YvesGaillard Thanks! I also found your very short proof of the fundamental theorem. vixra.org/abs/1207.0051. But it uses a very specific setup, so probably doesn't apply here(?). | |
| Nov 6, 2022 at 14:55 | comment | added | Martin Brandenburg | @R.vanDobbendeBruyn Thank you. Yes a lot of Galois theory is missing there, but the aim of the note is to prove the Galois correspondence, not (much) more. But maybe I will add this equivalent def' in the future. | |
| Nov 6, 2022 at 14:51 | comment | added | Martin Brandenburg | @BenjaminSteinberg This is interesting, can you reference a proof along these lines? Also, what is the connection to the proof of the fundamental theorem? I guess that this offers an alternative way to finish the proof of Prop. 4.2 in the special case of finite separable extensions? | |
| Nov 6, 2022 at 14:48 | comment | added | R. van Dobben de Bruyn | One small thing that I'm missing is equivalence with the other definition of Galois extension: a finite extension $K \to L$ is Galois if there exists a finite subgroup $G \subseteq \operatorname{Aut}(L)$ such that $K = L^G$. (This is not the right definition in the infinite case, for instance $K(T)$ has an automorphism $\sigma \colon T \mapsto T+1$ with $K(T)^{\sigma} = K$ if $K$ is infinite, but of course $K \to K(T)$ is not Galois. Can you fix this by asking for a profinite subgroup $G \subseteq \operatorname{Aut}(L)$ somehow?) | |
| Nov 6, 2022 at 14:37 | comment | added | R. van Dobben de Bruyn | I agree with @BenjaminSteinberg. That said, although I cannot comment on novelty, your proof of Prop. 4.2 is very neat, and substantially easier than the one I know. It reminds me of how Galois theory for symmetric polynomials is proven (i.e. showing that $K(\sigma_1,\ldots,\sigma_n) \to K(x_1,\ldots,x_n)$ is Galois with group $S_n$), maybe in that it takes the extension field $L$ as the starting point instead of the base field $K = L^G$. | |
| Nov 6, 2022 at 10:54 | comment | added | Carl-Fredrik Nyberg Brodda | For the history part, it’s worth noting that the fundamental theorem of Galois theory wasn’t proved (or stated) until a century after Galois died, by Artin. Galois himself did not find a correspondence between subgroups of automorphism groups and intermediate fields. | |
| Nov 6, 2022 at 0:09 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Nov 4, 2022 at 21:23 | comment | added | Benjamin Steinberg | I 've definitely seem that a vector space cannot be written as a finite union of proper subspaces to prove prove the primitive element theorem | |
| Nov 4, 2022 at 19:09 | comment | added | Pierre-Yves Gaillard | +1. Let me mention the link math.stackexchange.com/a/89576/660 even if it doesn't answer your question. | |
| Nov 4, 2022 at 18:43 | history | asked | Martin Brandenburg | CC BY-SA 4.0 |