Timeline for Galois groups of truncated $\cosh(x)$ Taylor polynomials and related results?
Current License: CC BY-SA 4.0
24 events
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| Aug 19 at 14:43 | comment | added | Alexei Entin | It solves the case $n=p$ prime, but this is the easiest case (should be similar to Thm 6.3 in mathoverflowUser's thesis but without the annoying $S_r(x)$ factor). | |
| Aug 18 at 19:26 | comment | added | Dror Speiser | This probably doesn't help much, but it seems like $p^{2p-1}||\text{disc}(F_p)$ for odd prime $p$. | |
| Aug 18 at 17:36 | comment | added | Alexei Entin | @SeanEberhard Because Assumption 1 implies Assumption 2 and then the only quadratic subfield of $K$ is $\mathbb Q(\sqrt{\mathrm{disc}(F_n)})$, which by Assumption 1 doesn't contain $\sqrt{(2n)!}$. | |
| Aug 18 at 16:36 | comment | added | Sean Eberhard | Why does it follow from Assumption 1 that $(2n)!$ is not a square in $K$? | |
| Aug 18 at 14:10 | comment | added | mathoverflowUser | Thanks for the clarification. So one still has to prove "Assumption 1" and then one has shown that $Gal(C_n) = S_2 \wr S_n$. | |
| Aug 18 at 12:41 | history | edited | Alexei Entin | CC BY-SA 4.0 |
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| Aug 18 at 8:10 | history | edited | Alexei Entin | CC BY-SA 4.0 |
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| Aug 18 at 6:26 | comment | added | Alexei Entin | @mathoverflowUser Yes. In fact only Assumption 1 is necessary because it implies Assumption 2. | |
| Aug 18 at 2:33 | comment | added | mathoverflowUser | Are you saying with your comment, that combining Thm. 6.2 and your Assumptions 1, 2 it should be possible to prove $Gal(C_n) = S_2 \wr S_n$? | |
| Aug 17 at 20:14 | comment | added | mathoverflowUser | Yes, Thm 6.2 says that $\operatorname{Gal}(F_n)$ contains the alternating group $A_n$. In Thm 6.3 I make a similar observation. | |
| Aug 17 at 19:50 | comment | added | Alexei Entin | @mathoverflowUser My German is still shaky, but do I understand correctly that Thm 6.2 in your thesis shows (using my notation) $\mathrm{Gal}(F_n)\supset A_n$? In this case Assumption 1 implies Assumption 2 and is equivalent to $\mathrm{Gal}(C_n)=S_2\wr S_n$. | |
| Aug 17 at 17:07 | comment | added | Steve D | Thanks for responding. But I guess my real question is how do you know this is a split extension? Thinking about this more it's probably because you can describe the action explicitly in terms of permutations of the $\sqrt{\alpha_i}$. | |
| Aug 17 at 16:48 | comment | added | mathoverflowUser | Thanks for your interest in this question: I think it was a combination of both. If you are intersted and do not mind that it is written in German, you can take a look at part 6 of the thesis here: orges-leka.de/diploma_orges_leka.pdf I am currently trying to translate it to english with ChatGPT, so probably I can post a link in a few days here. Thanks again for your input. | |
| Aug 17 at 16:40 | comment | added | Alexei Entin | @mathoverflowUser Was the main obstacle showing that $\mathrm{Gal}(F_n)\supset A_n$ or obtaining information about $\mathrm{disc}(C_n)$? | |
| Aug 17 at 16:19 | history | edited | Alexei Entin | CC BY-SA 4.0 |
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| Aug 17 at 16:17 | comment | added | Alexei Entin | @SteveD One needs to show that $\mathrm{Gal}(L/K)$ is as large as possible, i.e. $(S_2)^n$, not just elementary abelian. Note that $S_2\wr S_n=(S_2)^n\rtimes S_n$. | |
| Aug 17 at 16:17 | comment | added | mathoverflowUser | Thanks for the update. I also came to a similar conclusion in my diploma thesis but could not finish the proof. | |
| Aug 17 at 16:15 | comment | added | Alexei Entin | @mathoverflowUser You are right, I misread the definition of $C_n$ and answered the wrong question. I updated my answer to address the correct question. | |
| S Aug 17 at 16:14 | review | First answers | |||
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| S Aug 17 at 16:14 | history | edited | Alexei Entin | CC BY-SA 4.0 |
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| Aug 17 at 14:14 | comment | added | Steve D | Sorry if I'm missing something obvious, but why is it enough to show $Gal(L/K)$ is elementary Abelian? | |
| Aug 17 at 13:02 | comment | added | mathoverflowUser | Thanks for your answer. I do not understand why the splitting field of $C_n$ equals $\mathbb{Q}(\sqrt{\alpha_1},\cdots,\sqrt{\alpha_n})$ where $\alpha_1,\cdots,\alpha_n$ are the roots of $E_n$? | |
| S Aug 17 at 11:53 | review | First answers | |||
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| S Aug 17 at 11:53 | history | answered | Alexei Entin | CC BY-SA 4.0 |