Let $\zeta_{l}$ be a primitive $l$th root of unity for some integer $l$, $K = \mathbb{Q}(\zeta_l)$ the $l$th cyclotomic field with $\mathcal{O}_{K} = \mathbb{Z}[\zeta_l]$ its ring of integers, and $\Phi_l (x)$ be the $l$th cyclotomic polynomial.
Let $n$ be an integer with several prime divisors, say $n = pqr$ where $p,q,r$ are distinct primes.
We have that
$$ \frac{\mathbb{Z} [\zeta_n]}{(1 - \zeta_p)} = \frac{\mathbb{Z} [\zeta_{qr},\zeta_p]}{(1 - \zeta_p)} \cong \frac{\mathbb{Z} [\zeta_{qr}] [X]}{(1 - X,\Phi_p(X))} = \frac{\mathbb{Z} [\zeta_{qr}] [1]}{(\Phi_p(1))} = \frac{\mathbb{Z} [\zeta_{qr}]}{(p)} $$
where in the last step we have used the fact that $\Phi_l (1) = l$ whenever $l$ is prime.
Similarly, we have that
$$ \frac{\mathbb{Z} [\zeta_{qr}]}{(1 - \zeta_q)} \cong \frac{\mathbb{Z} [\zeta_{r}]}{(q)} $$
Now, I'd be tempted (want) to apply a similar argument to say something like
$$ \frac{\mathbb{Z} [\zeta_{n}]}{(1 - \zeta_p,1 - \zeta_q)} \cong \frac{\mathbb{Z} [\zeta_{qr}]}{(p,1 - \zeta_q)}\cong \frac{\mathbb{Z} [\zeta_{r}]}{(pq)} $$
Is this true? Is there a better way to prove/disprove it?
I'm concerned because I know that $\Phi_q (X)$ is not necessarily the minimal polynomial of $\zeta_q$ over $\mathbb{F}_p$, only when $p$ is a generator of $\mathbb{Z}_q^*$. (And if that is the case, do we know anything about $ m_q (1)$ where $m_q(x)$ is the minimal polynomial of $\zeta_q$ over $\mathbb{F}_p$?)
Any help or pointers to good references is greatly appreciated!