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Let $p$ be an odd prime and $\omega$ be a primitive $p$th root of unity. The question is to prove that:

$$(1-\omega)(1-\omega^2) \cdots (1-\omega^{p-1})=p$$


What I have done so far is:

I can see that this is true for $p=3$

$$(1-\omega)(1-\omega^2)=1-(\omega+\omega^2)+w^3=1-(-1)+1=3=p$$

I am not able to prove this in general....

$$(1-\omega)(1-\omega^2) \cdots (1-\omega^{p-1})=1-(\omega+\omega^2+\cdots+\omega^{p-1})+????+\omega^{\frac{p(p-1)}{2}}$$

I do not have any idea what that $????$ could be but all I can say is:

  • $(\omega+\omega^2+\cdots +\omega^{p-1})=-1$
  • $\omega^{\frac{p(p-1)}{2}}=1$

So, $(1-\omega)(1-\omega^2)\cdots (1-\omega^{p-1})=3+????$

I am not able to do more than this.

There could be some (hard) way of doing it by hand multiplying all those things but I am looking for a more theoretical idea.

Please help me to clear this up.

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Recall the formula for the summation $1+x+x^2+\cdots+x^n = {x^{n+1}-1 \over x-1}$ when $x\neq 1$.

In particular, this shows that $f(x)=x^{p}-1 =(1+x+x^2+\cdots+x^{p-1})(x-1)$.

Since $\omega$ is a root of unity, we have $f(\omega^k) = 0$ for $k=0,...,p-1$.

Since $\omega$ is primitive, $\omega^0,...,\omega^{p-1}$ are distinct and so $f(x) = (x-1)\cdots (x-\omega^{p-1}) $.

Hence $1+x+x^2+\cdots+x^{p-1} = (x-\omega^1)\cdots (x-\omega^{p-1}) $.

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  • $\begingroup$ all i can say is i was stupid.... I was neglecting $(x-1)$ in factorization of $x^p-1$ and that created all that mess... this is natural.... Thank you... $\endgroup$ – user87543 May 23 '14 at 15:36
  • $\begingroup$ Glad to help! ${}{}$ $\endgroup$ – copper.hat May 23 '14 at 15:39
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Hint: what does $x^n-1$ look like factored?

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  • 2
    $\begingroup$ Nice hint that just gives away the right amount of information. Also your hidden argument shows that $p$ could be any integer be it prime or not. $\endgroup$ – Peder May 23 '14 at 13:56
  • $\begingroup$ Yes Yes... I got it.. Thank you :) $\endgroup$ – user87543 May 23 '14 at 15:42
  • $\begingroup$ @Peder : I for got to mention this but this does not hold for any integer but only primes.... $\endgroup$ – user87543 May 24 '14 at 2:28
  • $\begingroup$ @sea : I guess you also mean $n$ to be prime... Factorization and all other holds smoothly but result is totally dependent on $n$ being prime... $\endgroup$ – user87543 May 24 '14 at 2:31
  • $\begingroup$ @Praphulla: Try $n=4$, $\omega=i$: $(1-i)(1-(-1))(1+i)=4$, so it holds for non-primes as well. $\endgroup$ – Peder May 24 '14 at 2:42
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Let $f(x)=x^p-1$ and $g(x)=f(1-x)$. Then $g(1-\omega^k)=f(\omega^k)=0$ for all $k$.

$(1-\omega)(1-\omega^2)\cdots (1-\omega^{p-1})$ is thus the product of all roots of $g$ except $0$. So, expand $g$, factor $x$, and look at the independent term.

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  • $\begingroup$ This is another nice way... Thank you :) $\endgroup$ – user87543 May 23 '14 at 15:39

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