3

I. Comparison

It doesn't seem to be well-known that the generic cubic (prominent in this MO post) for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

has the nice property that its roots $a,b,c$, if in correct order, obey,

$$(a^2b)^{1/3}+(b^2c)^{1/3}+(c^2a)^{1/3} = 0$$

(I only noticed this after I asked an MO question about the similar-looking Klein quartic $a^3b+b^3c+c^3a=0.$)

Since the generic cubic is intimately connected to the roots of unity for prime $p\equiv 1\,\text{mod}\; 6$, (the case $n=1$ yields $1^{1/7}$), naturally I got curious about its big sister the Emma Lehmer quintic which is for $p\equiv 1\,\text{mod}\; 10$, namely,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1=0$$

It turns out its roots $x_k$ may have a similar property.

II. Question:

Analogous to the generic cubic, is it true that the Emma Lehmer quintic obeys, $$(a^4b^3c^2d)^{1/5} + (b^4c^3d^2e)^{1/5} + (c^4d^3e^2a)^{1/5} + (d^4e^3a^2b)^{1/5} + (e^4a^3b^2c)^{1/5} = 0$$ for the correct ordering of its roots $a,b,c,d,e$?

Update 1: Thanks to Peter Taylor in the comments, and using the fact that $abcde = -1$, we can get rid of the fifth roots and get the equivalent but more elegant form,

$$\frac1{a}-\frac1{ab}+\frac1{abc}-\frac1{abcd}+\frac1{abcde} = 0$$

Or more generally (for $\mu$ an integer),

$$\frac{\mu}{a}-\frac{\mu^{2}}{ab}+\frac{\mu^{3}}{abc}-\frac{\mu^{4}}{abcd}+\frac{\mu^{5}}{abcde} = 0$$

where $\mu^5$ is the constant term of the quintic and the Lehmer quintic the special case $\mu=1$.

Update 2: I've already tested the Hashimoto septic which fortunately has a seventh power $\mu^7$ as its constant term and the analogous relation,

$$\frac{\mu}{a}-\frac{\mu^{2}}{ab}+\frac{\mu^{3}}{abc}-\frac{\mu^{4}}{abcd}+\frac{\mu^{5}}{abcde}-\frac{\mu^{6}}{abcdef}+\frac{\mu^{7}}{abcdefg} = 0$$

and among the $(p-1)! = 720$ permutation of its roots, at least 9 works.

III. Example

Let $n=-1$. Then we get the quintic for $p=11$ and its roots,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$

$$a,b,c,d,e = 2\cos\frac{2\pi k}{11}$$

with $k = 1, 4, 5, 2, 3$ as the correct order. One can then verify it obeys the relation in the question.

P.S. For $p=5$, there are $(p-1)! = 24$ permutations of its roots. It is easy for a computer to find the correct order for any $n$ that I tested. But does it hold true for ALL $n$?

Tito Piezas III
  • 12,044
  • 2
    I haven't worked through the details, but Darmon's Note on a polynomial of Emma Lehmer is suggestive that the correct ordering might be given by the cyclic map between the roots $r \to \frac{n+2 + nr - r^2}{1 + (n+2)r}$ – Peter Taylor Jan 17 '23 at 10:07
  • 1
    @PeterTaylor: Thanks for checking. I'll get rid of the 2nd question to simplify things, as it was just an afterthought anyway. So the conjectured property holds up so far? – Tito Piezas III Jan 17 '23 at 14:03
  • 1
    Why do you write $p=6m+1$ or $p=10m+1$ when there is no visible role for that $m$ at all? At first I thought $m$ was a typo for $n$. Maybe you just mean $p\equiv 1\bmod 6$ and $p\equiv 1\bmod 10$, in which case I think it would be clearer to write “some prime $p\equiv 1\bmod 6$“, say, without mentioning an $m$ that never actually gets used. – KConrad Jan 17 '23 at 14:46
  • @KConrad My apologies. I will fix it. – Tito Piezas III Jan 17 '23 at 16:01
  • Thanks, but what is the connection between $n$ and $p$? You describe the quintic as "for $p \equiv 1 \bmod 10$" but there is no $p$ anywhere in the quintic. Imagine you saw the sentence "For a positive integer $k$, consider $a^2 + 1$." That is how your post reads when you bring up $n$ and $p$. I think there is some kind of information behind these polynomials that is being left unsaid. – KConrad Jan 17 '23 at 16:29
  • @KConrad. For the cubic, I did mention in the post that, ".. the case $n=1$ yields $1^{1/7}"$, If you substitute $n=1$ into the cubic, you get the minimal polynomial for $2\cos(2\pi/7)$ and you have your prime $p=7$. For the quintic, if you substitute $n=-1$, you get the minimal polynomial for $2\cos(2\pi/11)$ and you have your prime $p=11$. – Tito Piezas III Jan 17 '23 at 17:16
  • @CHUAKS: Exactly. However, for the cubic in the post, the discriminant is $(n^2-3n+9)^2$ though obviously it is just a minor negation of the variable. – Tito Piezas III Jan 17 '23 at 17:31
  • @TitoPiezasIII Oh! That helps a lot. Why not actually include this information in your post? Saying "for $n = -1$ we get $p = 7$" is quite opaque by comparison. Please put a direct relation between $n$ and $p$ in your post. – KConrad Jan 17 '23 at 20:13
  • 3
    I have a proof, but it's not very enlightening so I'll hold posting it as an answer for a few days in the hope that someone finds a more insightful one. Factor out $(a^4b^3c^2d)^{1/5}$ to get $1-\frac{1}{a}+\frac{1}{ab}-\frac{1}{abc}+\frac{1}{abcd}$ and compute in $\mathbb{Q}[n]/p(a)$. Sage code can be run online at https://sagecell.sagemath.org but a direct execution link is too long for a comment. – Peter Taylor Jan 17 '23 at 22:20
  • @PeterTaylor Ah, clever! You also used the fact that $abcde = -1$ to get rid of the fifth roots. We can employ that further to make the relation include ALL roots and look more symmetric as, $$\frac1{a}-\frac1{ab}+\frac1{abc}-\frac1{abcd}+\frac1{abcde} = 0$$ I will update the post and credit that alternative form to you. – Tito Piezas III Jan 18 '23 at 02:16
  • @PeterTaylor Perhaps u can write a tentative answer? U can always edit it based on input from other people. I am curious as to what u found. – Tito Piezas III Jan 20 '23 at 00:07

1 Answers1

3

The map $s(r) = \frac{n+2 + nr - r^2}{1 + (n+2)r}$ cyclically permutes the roots. This map is given in [2], and I found it through the reference in [1]. It turns out to give the correct order.

Explicitly, if we pick one root $x_1$ we have

\begin{eqnarray*} D &=& n^3 + 5n^2 + 10n + 7 \\ Dx_2 &=& (n^2 + 4n + 4)x_1^4 + (n^4 + 4n^3 + 4n^2 - n - 2)x_1^3 + \\&& (-2n^5 - 14n^4 - 43n^3 - 76n^2 - 80n - 39)x_1^2 + \\&& (n^6 + 9n^5 + 37n^4 + 89n^3 + 131n^2 + 107n + 36)x_1 + \\&& (n^4 + 8n^3 + 26n^2 + 39n + 22) \\ % Dx_3 &=& (-2n - 3)x_1^4 + (-2n^3 - 4n^2 - 3n - 2)x_1^3 + \\&& (3n^4 + 14n^3 + 31n^2 + 41n + 24)x_1^2 + \\&& (-n^5 - 7n^4 - 21n^3 - 36n^2 - 29n - 6)x_1 + \\&& (-2n^2 - 7n - 6) \\ % Dx_4 &=& (-n^2 - 3n - 2)x_1^4 + (-n^4 - 3n^3 - 2n^2 + n + 1)x_1^3 + \\&& (2n^5 + 12n^4 + 33n^3 + 54n^2 + 53n + 23)x_1^2 + \\&& (-n^6 - 8n^5 - 29n^4 - 62n^3 - 81n^2 - 59n - 18)x_1 + \\&& (-n^5 - 7n^4 - 24n^3 - 47n^2 - 52n - 25) \\ % Dx_5 &=& (n + 1)x_1^4 + (n^3 + 2n^2 + 3n + 3)x_1^3 + \\&& (-n^4 - 4n^3 - 9n^2 - 14n - 8)x_1^2 + \\&& (-n^4 - 7n^3 - 19n^2 - 29n - 19)x_1 + \\&& (n^4 + 6n^3 + 16n^2 + 20n + 9) \end{eqnarray*}

and it is pure computation to show that $$x_1 x_2 x_3 x_4 x_5 - x_2 x_3 x_4 x_5 + x_3 x_4 x_5 - x_4 x_5 + x_5 = 0$$

Online computation with sagecell.sagemath.org

However, this isn't a very satisfying proof because it doesn't address the question of necessary and sufficient properties of the original polynomial. Does a suitable permutation of roots exist for all irreducible polynomials of odd (or maybe odd prime) degree whose Galois group is cyclic? The equivalence between the forms

$$ (x_1^4 x_2^3 x_3^2 x_4)^{1/5} + (x_2^4 x_3^3 x_4^2 x_5)^{1/5} + (x_3^4 x_4^3 x_5^2 x_1)^{1/5} + (x_4^4 x_5^3 x_1^2 x_2)^{1/5} + (x_5^4 x_1^3 x_2^2 x_3)^{1/5} = 0 \\ \frac1{x_1} - \frac1{x_1 x_2}+\frac1{x_1 x_2 x_3} - \frac1{x_1 x_2 x_3 x_4} + \frac1{x_1 x_2 x_3 x_4 x_5} = 0 \\ x_1 x_2 x_3 x_4 x_5 - x_2 x_3 x_4 x_5 + x_3 x_4 x_5 - x_4 x_5 + x_5 = 0 $$ and a couple of other variants relies on the constant coefficient of Emma Lehmer's quintic being $1$. If there is a generalisation, which variant does it require? Or does generalisation require a cyclic Galois group and a unit constant coefficient?

[1] Henri Darmon, Note on a polynomial of Emma Lehmer, Math. Comp. vol. 56, no. 194, April 1991, pp. 795-800.
[2] Rene Schoof and Lawrence C. Washington, Quintic polynomials and real cyclotomic fields with large class numbers, Math. Comp. vol. 50, no. 182, April 1988, pp. 543-556.

Peter Taylor
  • 6,571
  • 2
    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 $1$. – Joachim König Jan 20 '23 at 15:32
  • 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. – Joachim König Jan 20 '23 at 17:55
  • @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. – Tito Piezas III Jan 21 '23 at 10:36
  • @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.) – Tito Piezas III Jan 21 '23 at 10:42
  • @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 $n$. – Joachim König Jan 21 '23 at 15:27
  • @JoachimKönig, I think you should probably write an answer. – Peter Taylor Jan 21 '23 at 15:41
  • @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. – Tito Piezas III Jan 22 '23 at 11:41
  • @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. – Joachim König Jan 26 '23 at 02:16