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Short version of the question: Guess the next terms in the sequence : $D_{17},D_{19},D_{23}$ etc where

$$ \begin{array}{lcl} D_3 &=& (a\pm 1) \\ D_5 &=& (a\pm 1) (a^2-1 \pm 11a) \\ D_7 &=& (a\pm 1) (a^3-57a \pm 289a^2) (a^3-289a \pm (57a^2-1)) \\ D_{11} &=& (a\pm 1) \\ & & (a^5-7682623a^3+1013a \pm (1006734a^4-378283a^2+1)) \\ & & (a^5-378283a^3+1006734a \pm (1013a^4-7682623a^2+1)) \\ & & (a^5-14823682a^3+52529a \pm (397443a^4-40074443a^2-1)) \\ & & (a^5-40074443a^3+397443a \pm (52529a^4-14823682a^2-1)) \\ D_{13} &=& (a\pm 1) \\ & & (a^6-9355414620a^4+8689296a^2-1 \pm (107661336a^5-1738683444a^3+4083a)) \\ & & (a^6-8689296a^4+9355414620a^2-1 \pm (4083a^5-1738683444a^3+107661336a)) \\ & & (a^6-83308449621a^4+109554241470a^2+1 \pm (9398391a^5-176361556196a^3+586968a)) \\ & & (a^6+109554241470a^4-833084496210a^2+1 \pm (586968a^5-176361556196a^3+9398391a)) \\ & & (a^6+4464038148a^4-4464038148a^2+1 \pm (49433934a^5-57173803449a^3+49433934a)) \\ \end{array} $$

Detailed version : Let $p>2$ be a prime, and let $a\in {\mathbb Q}$ such that $v=\sqrt[p]{a}$ is irrational. Let $\omega$ be a primitive $p$-th root of unity. It is well-known then that ${\mathbb Q}(v,\omega)={\mathbb Q}(v+\omega)$ has degree $p(p-1)$ over $\mathbb Q$ (see for example, this MSE question). So there are uniquely defined rational fractions $f_0(a),f_1(a),f_2(a),\ldots,f_{p(p-1)-1}(a)$ such that the identity

$$ v=\sum_{k=0}^{p(p-1)-1} f_k(a)(v+\omega)^k $$

holds for all but finitely many $a$. Denote the (monic) lowest common denominator of all the $f_k$ by $D_p$. Then, calculations in PARI-GP for $p=3,5,7,11,13$ yield the values given above (where we use the condensed notation $e_1\pm e_2$ for $(e_1+e_2)(e_1-e_2)$).

Those results naturally suggest several questions (about which I am equally clueless) :

(1) What is the degree of $D_p$ in general ?

(2) Is it true that $a-1,a+1$ are always factors of $D_p$ ?

(3) Is it true that all the irreducible factors of $D_p$ (excepting $a\pm 1$) come in pairs (odd polynomial)$\pm$(even polynomial), that all those factors have degree $\frac{p-1}{2}$, cyclic Galois groups and constant term equal to $-1$ or $1$ ?

(4) Is it true that if $F$ is an irreducible factor of $D_p$ (other than $a\pm 1$) then so is the reciprocal polynomial $a^{\frac{p-1}{2}}F(\frac{1}{a})$ ?

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Ewan Delanoy
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1 Answers1

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The expression of $v$ in terms of the $v+\omega$ is the interpolation polynomial that gives the output $v^\sigma$ for the input $(v+\omega)^\sigma$ for each $\sigma \in G = Gal(\Bbb Q(a,\omega,v) : \Bbb Q(a))$.

When specializing $a$, this polynomial may not exist only when the $(v+\omega)^\sigma$ are not distinct, so the denominators of the coefficients should vanish only when there are $\sigma_1 \neq \sigma_2$ such that $(v+\omega)^{\sigma_1} = (v+\omega)^{\sigma_2}$. Or equivalently, they are factors of the discriminant of the minimal polynomial of $v+\omega$.

If $\sigma_1(v) = \sigma_2(v)$ then $(v+\omega)^{\sigma_1} = (v+\omega)^{\sigma_2}$ reduces to $\omega^{\sigma_1} = \omega^{\sigma_2}$. If $\sigma_1 \neq \sigma_2$ those two are different so this doesn't require anything from $v$.

If $\sigma_1(\omega) = \sigma_2(\omega)$ this reduces to $v(\omega^{a_1}-\omega^{a_2}) = 0$, so $v=0$ : when $v=0$, we only have $p-1$ different inputs instead of $p(p-1)$ for the interpolation polynomial. However, the outputs required are all $0$, so things work out in this case and every coefficient is a multiple of $z$. And in fact, since the polynomial is defined on $\Bbb Q(a)$, it should be a multiple of $a$.

When $v$ does one small loop around $0$,the points $((v+\omega)^\sigma,v^\sigma)$ defining the polynomial also each make an individual loop. However since the polynomial is obtained after forgetting the order between the points of the graph and putting them in a set, we see that the polynomial does loops of index $p$ : rotating $v$ by $2\pi/p$ doesn't change the graph itself, so the polynomial is a multiple of $v^p = a$ but not of $v^{p+1}$ (nor $v^{2p} = a^2$)

If $\sigma_1,\sigma_2$ act differently on both $v$ and $\omega$, we can solve for $v$ in $(v+\omega)^{\sigma_1} = (v + \omega)^{\sigma_2}$ and we get forbidden values for $v$ of the form $v = \omega^i \frac {\omega^j - 1}{\omega^k - 1}$ with $j,k \in (\Bbb Z/p\Bbb Z)^*$ and $i \in \Bbb Z/p\Bbb Z$ (each expression is obtained $p-2$ times).
If $v$ is one of those values then the interpolating polynomial wants to give two different outputs $v^{\sigma_i}$ for the same input $(v+\omega)^{\sigma_i}$, so the denominator must vanish somewhere.

Now the set of forbidden $v$ values contains $1$ (pick $i=0,j=k$) and is stable by inversion (swap $j$ and $k$) and by multiplication by a $p$th root of unity (just increase $i$).

Since $\frac {\omega - 1}{\omega^{-1}-1} = - \omega$, we see that the set of forbidden values is also stable by negation (replace $(i,k)$ with $(i-k,-k)$)

Moreover, if $\iota$ is the automorphism of $\Bbb Q(\omega)$ induced by $\omega \mapsto \omega^{-1}$ (i.e. complex conjugation), then $ (\omega^i \frac {\omega^j - 1}{\omega^k - 1}) / \iota (\omega^i \frac {\omega^j - 1}{\omega^k - 1}) = \omega^{2i} \frac {- \omega^j}{- \omega^k} = \omega^{2i+j-k}$, so when taking the $p$th powers for looking at forbidden values of $a$, we get that they are invariant by $\iota$ :

All the roots of your polynomials in $a$ are in $\Bbb Q(\cos\frac{2\pi}p)$ so their minimal polynomial has degree dividing $(p-1)/2$, and they are $p$th powers of elements of $\Bbb Q(\omega)$

If there is no additional relationship between the conjugates of the forbidden values and if the minimal polynomials are all of degree $(p-1)/2$ then we get $(p-3)$ polynomials in $a$ besides the two factors $(a\pm 1)$

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