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

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.

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.

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I have just edited the expression for $x$ to make it (slightly) more general in a quite obvious way, and I think what you suggest should still apply. Only I don't really understand what you presume, particularly what do you mean by "early terms of the expansion"? –  Wolfgang Jan 25 '12 at 18:34
As Charles pointed out, consider $\mathbb{Q}(\zeta_p, i)$, and a prime ideal $\mathcal{P}\subset \mathbb{Q}(\zeta_p,i)$ lying over $(1-\zeta_p)\subset \mathbb{Q}(\zeta_p)$. Then, $\sqrt{p}$, sin, and tan terms are divisible by $\mathcal{P}$ considered as in $\mathcal{P}$-adic integer, while cos term alone is not. Putting $\sqrt{p}$ on other terms will leave cos term alone, so this will not be divisible by $\mathcal{P}$. Anyway, very interesting observation. –  i707107 Mar 8 '12 at 4:37