Cyclotomic polynomials for primes

By Samdney, Mi 18 Juli 2018, in category Math

cyclotomic polynomials, number theory, primes

Currently, I'm spending my time with cyclotomic polynomials. So, I don't want to miss the possibility also to mention a small note about the connection of this polynomials to the set of prime numbers.

At first, we have a cyclotomic polynomial in the following way

\begin{equation} \phi_{n}\left(x\right) = \prod_{\substack{1 \leq g \leq n \\ ggT\left(g,n\right) = 1}} \left(x - e^{\frac{2\pi i g}{n}}\right) \end{equation}

with \(n, g \in \mathbb{N}\), respectively

\begin{alignat}{3} x^{n} - 1 &= \prod_{1 \leq g \leq n} \left(x - e^{\frac{2\pi i g}{n}}\right) \notag \\ &= \prod_{d | n} \prod_{\substack{1 \leq g \leq n \\ ggT\left(g,n\right) = d}} \left(x - e^{\frac{2\pi i g}{n}}\right) \notag \\ &= \prod_{d | n} \phi_{n/d}\left(x\right) \notag \\ &= \prod_{d | n} \phi_{d}\left(x\right) \end{alignat}

. In the case of pime numbers \(p \in \mathbb{P}\), we also have

\begin{equation} \phi_{p}\left(x\right) = 1 + x + x^{2} + \dots + x^{p-1} = \sum_{k = 0}^{p-1} x^{k} \end{equation}

. From this follows the first rewriting

\begin{alignat}{3} \phi_{p}\left(x\right) &= \sum_{k = 0}^{p-1} x^{k} \notag \\ &= \sum_{k = 0}^{p-1} e^{k\ln\left(x\right)} \notag \\ &= e^{0\ln\left(x\right)} + \sum_{k = 1}^{p-1} e^{k\ln\left(x\right)} \notag \\ &= 1 + \frac{e^{\ln\left(x\right)} \left(e^{\left(p-1\right)\ln\left(x\right)} - 1\right)}{e^{\ln\left(x\right)} - 1} \end{alignat}

, since

$$ \sum_{k = 1}^{n} e^{kA} = \frac{e^{A}\left(e^{An} - 1\right)}{e^{A} - 1} $$

. Now we will solve this equation for \(p\).

\begin{alignat}{3} \phi_{p}\left(x\right) &= 1 + \frac{e^{\ln\left(x\right)} \left(e^{\left(p-1\right)\ln\left(x\right)} - 1\right)}{e^{\ln\left(x\right)} - 1} \notag \\ \left(\phi_{p}\left(x\right) - 1\right)\left(e^{\ln\left(x\right)} - 1\right) &= e^{\ln\left(x\right)}\left(e^{\left(p-1\right)\ln\left(x\right)} - 1\right) \notag \\ \frac{\left(\phi_{p}\left(x\right) - 1\right)\left(e^{e^{\ln\left(x\right)}} - 1\right)}{e^{\ln\left(x\right)}} + 1 &= e^{\left(p-1\right)\ln\left(x\right)} \notag \\ \left(p-1\right)\ln\left(x\right) &= \ln\left(\frac{\left(\phi_{p}\left(x\right) - 1\right)\left(e^{\ln\left(x\right)} - 1\right)}{e^{\ln\left(x\right)}} + 1\right) \notag \\ \left(p-1\right)\ln\left(x\right) &= \ln\left(\frac{\left(\phi_{p}\left(x\right) - 1\right)\left(x - 1\right) + x}{x}\right) \notag \\ \ln\left(x\right)\left(\left(p-1\right) + 1\right) &= \ln\left(\left(\phi_{p}\left(x\right) - 1\right)\left(x - 1\right) + x\right) \notag \\ p &= \ln\left(\frac{\left(\phi_{p}\left(x\right) - 1\right)\left(x - 1\right) + x}{x}\right) \end{alignat}

Additionally, we have our second, very easy to see, connection for prime numbers

\begin{alignat}{3} \phi_{p}\left(x\right) &= x^{p} - 1 \notag \\ &= \sum_{k' = 0}^{p} x^{k'} - \sum_{k'' = 0}^{p-1} x^{k''} - 1 \notag \\ &= \phi_{p+1}\left(x\right) - \phi_{p}\left(x\right) - 1 \end{alignat}

. I have no idea for which this can be useful, but I think it is nice to know :P.