Let $E(n)$ denote an nth root of unity. (For convenience, we may take $E(n) = \exp(\frac{2πi}{n})$.)

Prove that for any prime $p$ and any natural number $r$, we have $$ \prod_{\substack{j\\ \gcd(p^r, j) = 1}} \left(1 – E(p^r)^j\right) = p,$$ without using the formula to get the cyclotomic polynomial of $p^r$.

  • 3
    $\begingroup$ Without using what formula? And why without using it? $\endgroup$ – Phira Nov 11 '12 at 10:49
  • $\begingroup$ @Phira, because with using the formula it would be really direct. I wanted to see a proof without using the formula. The formula is Phi(x) = sum i= 1 to p-1 ( x^(i*p^(k-1))) $\endgroup$ – John Chang Nov 11 '12 at 13:20
  • $\begingroup$ It should be $\displaystyle{\prod_{\left(p^r,j\right)=1} \left(1 – E\left(p^r\right)^j\right) = p^r}$. $\endgroup$ – P.. Nov 12 '12 at 11:57
  • $\begingroup$ @Pambos, the cyclotomic polynomial of p^r, Phi(x) = sum i= 1 to p-1 ( x^(i*p^(k-1))) gives us that it is indeed p. I think. But I wanted to see a more detailed proof without using this formula. $\endgroup$ – John Chang Nov 12 '12 at 12:00

Let $n$ be a natural number and let $E_n=e^{2\pi i/n}$. Then the polynomial $x^n - 1$ splits in $\mathbb{Q}(E_n)$ as

$x^n - 1 = (x - 1)(x - E_n)\cdots(x-E_n^{n-1})$, which yields:

$1+x+\cdots+x^{n-1} = \frac{x^n-1}{x-1} = (x - E_n)(x - E_n^2)\cdots(x-E_n^{n-1})$

and so we get the following result.

$\prod_{j=1}^{n-1}(1-E_n^j) = n$

Now let $n = p^m$, for some natural number $m$. Thus,

$p^m = \prod_{j=1}^{p^m-1}(1-E_{p^m}^j) = \prod_{(j, p^m)=1}(1-E_{p^m}^j)\prod_{(j, p^{m-1})=1}(1-E_{p^{m-1}}^j)\cdots\prod_{(j, p)=1}(1-E_p^j)$

Now look at the right side of the last equation. Each product $\prod_{(j, p^r)=1}(1-E_{p}^j)$ is greater than 1, and there are a total of $m$ factors of this form. Since the product of these $m$ factors is $p^m$, then each factor must be equal to $p$.

Therefore, $\prod_{(j, p^r)=1}(1-E_{p^r}^j) = p$.


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