3
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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 $\ge p-1$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

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

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

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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