Let $p$ be a prime positive integer, $m$ be an integer which does not contain $p$ as a divisor, and $\alpha = e^{2\pi i /m}$. Let $\mathfrak p$ be a prime ideal of $\mathbb{Z}[\alpha]\subset \mathbb{C}$ that contains $p$. Note that the canonical quotient map $\pi:\mathbb{Z}[\alpha]\to \mathbb{Z}[\alpha]/\mathfrak p$ restricts to a homomorphism of unit groups $\pi^\times:\mathbb{Z}[\alpha]^\times\to (\mathbb{Z}[\alpha]/\mathfrak p)^\times$.
Is the order of $\pi^\times(\alpha)$ necessarily $m$ (same as for $\alpha$)?
Some notes:
I've tried a few $m$ and $p$ and nothing has failed yet, so I'm leaning toward yes. Perhaps it is because $\pi^\times$ is injective, but it's not obvious that this is equivalent.
The statement need not hold if $p|m$; for instance if $m=4$ and $p=2$ we can choose $\mathfrak p = \langle 1+i\rangle$ and then $i$ has order $4$ but $\pi(i)$ is the identity.
$\mathbb{Z}[\alpha]/\mathfrak p$ is a finite field of order $p^r$, where I think $r$ is the order of $m$ as an element of $(\mathbb{Z}/p\mathbb{Z})^\times$.
Optimally there would be some proof that doesn't use this, but there is an explicit description of these kinds of primes $\mathfrak{p}$: they are generated by $p$ and some $f_i(\alpha)$, where $\Phi_m(t)=\prod_j f_j(t)$ is an irreducible decomposition of the cyclotomic over $\mathbb{F}_p$ .
Note that $K=\mathbb{Z}[\alpha]/\mathfrak{p}$ is a finite field extension of $\mathbb{F}_p$.
We know $\pi(\alpha) \in K$ is a root of the reduction $\psi_m$ of the cyclotomic polynomial $\Phi_m \in \mathbb{Z}[X]$. In particular $\pi(\alpha)^m=1$. Assume now that $\pi(\alpha)^d=1$ for some $d < m$ dividing $m$.
Then $\pi(\alpha)$ is a root in $K$ of $\psi_d$ and $\psi_m$, thus $\alpha$ is a multiple root of (the reduction mod $p$ of) $X^m-1$ in $K$. However, note that since $p$ and $m$ are coprime, $X^m-1$ is separable mod $p$, a contradiction.