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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!

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3 Answers

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 $\ge p-1$.

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Thank you! This is very clear. –  Dexter Nov 6 '12 at 17:23
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$\displaystyle\Phi_p(x)=\prod_{\xi^p=1\ , \ \xi\neq 1} (x-\xi)$

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

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

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

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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

$$\Phi_p(x)=\frac{x^p-1}{x-1}=1+x+x^2+...+x^{p-1}$$

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