Timeline for A similar relationship between the generic cubic and the Lehmer quintic?
Current License: CC BY-SA 4.0
11 events
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| Jan 26, 2023 at 2:16 | comment | added | Joachim König | @PeterTaylor I refrained from writing an answer because even though the above observation can easily generate polynomials where the construction works, it doesn't explain why this particular polynomial works. For that I don't have a better argument than "the computer says so". For completeness, the root $a$ of the Lehmer quintic is $a=-\sigma(x)/x$ with $<\sigma> = Gal(Q(n)(a)/Q(n))$ and $x$ a root of $p(n,X):=X^5-zX^3-(n+2)zX^2-nzX+z$, where $z:=n^4+5n^3+15n^2+25n+25$, but I don't have a real “explanation" for the crucially necessary fact that the coefficient of $p$ at $X^4$ is vanishing. | |
| Jan 25, 2023 at 16:44 | history | bounty ended | Tito Piezas III | ||
| Jan 25, 2023 at 16:43 | vote | accept | Tito Piezas III | ||
| Jan 22, 2023 at 11:41 | comment | added | Tito Piezas III | @JoachimKönig I finished the quintic version. The relation discussed here plays a role in the new post. In contrast to the cubic method, the quintic method generates FOUR solutions. | |
| Jan 21, 2023 at 15:41 | comment | added | Peter Taylor | @JoachimKönig, I think you should probably write an answer. | |
| Jan 21, 2023 at 15:27 | comment | added | Joachim König | @TitoPiezasIII It's not specific to these few families, you can generate every cyclic field of odd prime degree by $a=\sigma(x)/x$ with $x$ of trace $0$ (i.e. vanishing second-highest term of minimal polynomial) , and then the minimal polynomial of $a$ will do what you want, by easy computation. I can also write down $x$ such that the resulting $a$ becomes the root of the Lehmer quintic, and to me the polynomial for $x$ looks even nicer. Why people nevertheless prefer to give the minimal polynomial for the norm-$1$ element, I don't know for sure, maybe it minimizes the degree in $n$. | |
| Jan 21, 2023 at 10:42 | comment | added | Tito Piezas III | @Peter Thanks for the answer. Ah, so that essentially is a 4th deg Tschirnhausen transformation between the selected root and the rest. I've also tested the $7$th deg Hashimoto septic and its analogous relation also works. But, unlike the Lehmer quintic where only 1 of the $(p-1)! = 24$ root permutations works, for the Hashimoto at least 9 of the $(p-1)! =720$ permutations will do. So one has to account for the "extra" solutions. (Kindly see the updated post re the septic.) | |
| Jan 21, 2023 at 10:36 | comment | added | Tito Piezas III | @JoachimKönig Thanks for testing. I've also noticed that thing about the constant term being a $5$th power and I've updated the "translated" form. I noticed that when testing the Hashimoto septic (with constant term fortunately a $7$th power) and its analogous relation works. (See link in post.) But it works _too well_ since among the $(p-1)=720$ permutations, at least 9 will do. | |
| Jan 20, 2023 at 17:55 | comment | added | Joachim König | Actually, the conjecture in my previous comment is too strong. Take an element $a$ of norm 1 (e. g. the negative of a root of a polynomial as above) in a $C_5$ field, and use Hilbert 90 to set $a=\sigma(x)/x$. This simplifies the expression dramatically, but it then turns out that one needs $x$ to be of trace $0$ for the whole thing to vanish. | |
| Jan 20, 2023 at 15:32 | comment | added | Joachim König | Out of interest, I ran the (roughly 1000) polynomials with group $C_5$ stored in LMFDB database. The ones showing the property discussed here (in original form, without using $abcde=-1$ in the translation) were exactly those whose constant coefficient was a fifth power. I didn't give it any further thought, but this seems suggestive; in particular the "translated" form should probably hold for all monic $C_5$ quintics with constant coefficient $1$. | |
| Jan 20, 2023 at 13:13 | history | answered | Peter Taylor | CC BY-SA 4.0 |