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Given a prime $p$, Let $n = p^d -1$ and let $f$ be an irreducible polynomial dividing the $n$-th cyclotomic polynomial in $\mathbb{F}_p[t]$. Let $\alpha = t + (f)$ in $\mathbb{F}_p[t]/(f)$ where $(f)$ is the ideal generated by $f$.

We have that $\alpha^n = 1$. I would like to show that the multiplicative order of $\alpha$ is $n$ by showing that if $\alpha^k = 1$ for some $k$ dividing $n$ with $k<n$ then $\alpha$ is a repeated root of $t^n - 1$ which cannot happen.

I know that $(t^k - 1)|(t^n - 1)$. (say by considering $t^m - 1 = \prod_{d|m} \Phi_d(t)$, where $\Phi_d$ is the $d$-th cyclotomic polynomial) Following this I tried to show that $\alpha$ must be a root of $g(t) := t^{n-k} + t^{n-2k} + ... + 1$, which would give the desired result since $(t^n - 1) = (t^k - 1)g(t)$ however I didn't really get anywhere with this.

I also tried considering $h(t) = t^{n-1} + \alpha t^{n-2} + ... + \alpha^{n-1}$ since $(t-\alpha)h(t) = (t^n - 1)$ but $h(\alpha) = n \alpha^{n-1} \neq 0$.

It feels like this should be fairly straightforward to solve and I'm just missing something obvious. Is anything I've done so far on the right path or am I just missing something entirely?

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I now have an answer to this question which I'll post here.

Take $k$ to be minimal with the property that $\alpha^k = 1$ where $k$ divides $n$ and $k<n$. Then $\alpha$ is a primitive $k$-th root of unity so $\alpha$ is a root of $\Phi_k$, by definition of $\Phi_k$. We also have that $\alpha$ is a root of $f$ and hence a root of $\Phi_n$. Then note $t^n - 1 = \prod_{d|n} \Phi_d(t)$ gives the desired result as $\Phi_k$ and $\Phi_n$ are included in this product.

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