Timeline for Cubic polynomials with "nice" roots, which can be expressed by trig functions of rational angles
Current License: CC BY-SA 4.0
12 events
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| Apr 10, 2023 at 9:50 | comment | added | The Amplitwist |
The link to Girstmair's article at springerlink.com is broken, but it can be found at doi:10.1007/BF01223707 (Zbl 0442.12004).
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| Jul 16, 2022 at 16:53 | history | edited | Glorfindel | CC BY-SA 4.0 |
broken link fixed, cf. https://math.meta.stackexchange.com/a/34713/228959
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 24, 2012 at 8:59 | comment | added | François Brunault | @spanferkel: The thing about the divisibility is not completely clear to me. This should be related to the ramification properties of $K/\mathbf{Q}$ at $p$, and to the fact that $4a^3-27b^2$ is square and divisible by $p^2$ (but this last condition alone is not sufficient to ensure that $a$ and $b$ are divisible by $p$). Maybe this could be turned into an interesting question, but one would need to compute more examples for this. | |
| Jan 23, 2012 at 14:49 | comment | added | Wolfgang | @François: OK, sorry, I see now. Thanks for clarifying. (I think I'll have to read more about Galois theory...) A last question: if there are $p$th roots of unity involved, is it straightforward to prove that $p$ divides $a$ and $b$? | |
| Jan 23, 2012 at 12:14 | comment | added | François Brunault | @spanferkel. I don't understand your last comment. Could you clarify what you mean by "feasible" ? Anyway, to clarify, every irreducible degree 3 polynomial over the rationals with square discriminant is "nice" in your terminology. This was the content of Kevin's first comment. | |
| Jan 23, 2012 at 11:21 | comment | added | Wolfgang | @François: I am not really after that, that was just a little remark and BTW now the equivalence is very obvious to me. The real question is the one which $(a,b)$ are feasible. Is each such polynomial with a square determinant "nice"? I don't think so... | |
| Jan 23, 2012 at 10:53 | comment | added | François Brunault | In fact every cyclic extension of odd degree has to be contained in $\mathbf{Q}(e^{2i\pi/N}) \cap \mathbf{R} = \mathbf{Q}(\cos 2\pi/N)$ so that the desired result is indeed true. Explicit expressions can be obtained by Gauss's method. See for example galois-group.net/theory/Galois_Kraus.pdf (p.24, "Annexe") | |
| Jan 23, 2012 at 3:16 | comment | added | Benjamin Steinberg | I agree with Qiaochu and voted to close. | |
| Jan 22, 2012 at 21:36 | comment | added | Qiaochu Yuan | As Kevin's answer shows, this follows by some fairly standard facts in algebraic number theory. I think the question is more appropriate for math.SE (if Kevin's comment hasn't already cleared it up for you). | |
| Jan 22, 2012 at 21:19 | comment | added | Kevin Buzzard | What is going on is that the discriminant of the polynomial (which is $4a^3−27b^2$ in your notation) is a square in the "nice" examples -- then by Galois theory the roots generate an $A_3$, and hence abelian, extension of the rationals, and one can get explicit formulae for the roots just using roots of unity, by the Kronecker-Weber theorem. Finally, the roots of unity can be converted into sines and cosines via the usual methods. | |
| Jan 22, 2012 at 21:04 | history | asked | Wolfgang | CC BY-SA 3.0 |