[Math] minimal polynomials of trig functions of ($k \pi/p$) and divisibility of coefficients by p

galois-theorynt.number-theorypolynomialsspecial functions

Take an odd prime $p$ and put $x_0:=\sum\limits_{j=0}^{p-1}\left(a_{j}\sqrt{p}\cos\dfrac{j\pi}p+b_{j}\sin\dfrac{j\pi}p +c_{j}\tan\dfrac{j\pi}p\right)$, where the $a_{ij}$ are integers. If $f$ denotes the minimal polynomial of $x_0$, can we prove that $p$ divides all coefficients of $f$ except the leading one?

I have quite a bit of numerical evidence for this. Note that it obviously doesn't hold without the $\sqrt p$ factor, but more interestingly, it is also false if $\sqrt p$ goes with the other terms instead of the $\cos$ term. Moreover, it seems that in those cases, none of the non-zero coefficients is divisible by $p$.

(More generally, I think those coefficients are divisible by $p$ if we replace $\dfrac{j\pi}p$ by $\dfrac{j\pi}{p^r},\ r\in\mathbb N$ and do the sum over $j=0,…,p^r-1$.)

If all but one of the $a_j,b_j,c_j$ are $0$, the claim is quite easy to prove (and not new). For instance, for $x_0=\sin\dfrac{j\pi}p$ with any fixed $j$, we have explicitly $$f(x)=\sum\limits_{i=0}^k(-1)^k\dbinom p{2i+1}(1-x^2)^{k-i}x^{2i},$$ where $p=2k+1$. So the claim is obvious here.

Added: It should be clear from Galois theory that in general, the conjugates of $x$ are the sums obtained by replacing all the $j$'s by $kj$ for a fixed $k=2,…,p-1$.

Literature:

Beslin, S., de Angelis, V., 2004. The minimal polynomials of sin(2π/p) and cos(2π/p). Mathematical Magazine 77, 146–149.

Heierman, William E., Minimal polynomials for trig functions of angles rationally commensurate with π

Lang, Wolfdieter, Minimal Polynomials of sin (2π/n)

Surowski, David, and McCombs, Paul, Homogenous polynomials and the minimal polynomial of cos(2π/n)

W. Watkins and J. Zeitlin, The Minimal Polynomial of cos(2Pi/n), Am. Math. Monthly 100,5 (1993) 471-4.

Best Answer

I can't quite do this all in my head, but clearly enough you should should write 1 - ζ = π and do π-adic analysis in the cyclotomic field of p-th roots of unity, where ζ is a non-trivial p-th root of 1. Putting the factor in front of the cos term means it doesn't affect the early terms of the expansion. (To be safe you have to put the square root of -1 in the field also.) The rest I presume can be read off the Newton polygon somehow, perhaps with a bit more work.

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