Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have this question on a homework sheet:

Claim:$$\Phi_{p}(x)=1+x+x^2+...+x^{p-1}\space$$ for $p$ prime.

which was followed by the claim that $\Phi_{p^n}(x)=\Phi_p(x^{p^{n-1}})$ which I have done via the Möbius function definition. The unsolved claim is supposed to be easier (that's how our sheets are structured) and presumably related, but I don't know how to go about it. Please help!

share|cite|improve this question
up vote 3 down vote accepted

If $\xi$ is a complex root of $\Phi_p$ then for each $k$, the number $\xi^k$ is a primitive $d$th root of unity for some divisor $d$ of $p$. The only divisor $d\ne p$ is $1$, hence $\xi, \xi^2,\ldots, \xi^{p-1}$ are $p-1$ different roots of $\Phi_p$. We conclude that $\Phi_p$ is a divisor of $\frac{x^p-1}{x-1}$ and is of degree $\ge p-1$.

share|cite|improve this answer
Thank you! This is very clear. – Dexter Nov 6 '12 at 17:23

$\displaystyle\Phi_p(x)=\prod_{\xi^p=1\ , \ \xi\neq 1} (x-\xi)$

$\text{ord}(\xi)\mid p\implies \text{ord}(\xi)\in\{1,p\}$.

Note that $\displaystyle x^p-1=\prod_{\xi^p=1\ } (x-\xi)$

So, $\Phi_p(x)=\frac{x^p-1}{x-1}=x^{p-1}+\cdots+x+1$

share|cite|improve this answer

Putting $\,C_p:=\{\zeta\in\Bbb C\;\;;\;\;\zeta^p=1\,\,\wedge\,\, \zeta^m\neq 1\,\,\,\forall\,m<p\;\;,\;m,p\in\Bbb N\,\,,\,p\,\,\text{a prime}\}\,$ , we have

$$\Phi_p(x):=\prod_{\zeta\in C_p}(x-\zeta)$$

Since $\,\zeta^p=1\Longrightarrow \zeta=e^{\frac{2\pi i}{p}}\,$ , we get that $\,\zeta\in C_p\Longleftrightarrow \zeta^p=1\,\,\wedge\,\,\zeta\neq 1\,$ , so finally


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.