On the factorization of prime ideals of $\mathbb{Z}$ in $\mathbb{Z}[\zeta]$ Let $\mathbb{Q(\zeta)}/\mathbb{Q}$ be a galois extension of degree $p-1$ where $\zeta=e^{\frac{2 \pi i}{p}} $ and let $G=\operatorname{Gal}(\mathbb{Q(\zeta)}/\mathbb{Q})$ be its Galois group. Suppose we have $m\mid p-1 $ and that $q$ is a prime in $\mathbb{Z}$ different from $p$. Then by Dedekind lemma, we can factorize the ideal generated by $q$ as
$$\langle q\rangle
=\mathfrak{b}_{1}^{e}......\mathfrak{b}_{r}^{e},$$
where $\mathfrak{b}_{i} $are  ideals in $\mathbb{Z}[\zeta]$. By ramification theory we have $rfe=p-1$ where $ f $ the is inertial degree and $e$ is the ramification index.
Can we find a prime ideal $\langle q\rangle$ in $\mathbb{Z}$ such that $ r=m$ or $e=m $? If yes how we can find it?
 A: Let $q\in\Bbb{Z}$ be a prime number distinct from $p$, and let $\mathfrak{q}\subset\Bbb{Z}[\zeta_p]$ be a prime ideal lying over $q$, and
$$\mathfrak{q}=\mathfrak{b}_1^{e_1}\cdots\mathfrak{b}_r^{e_r},$$
its factorization into prime ideals, and $f_i:=\dim_{\Bbb{F}_q}\Bbb{Z}[\zeta_p]/\mathfrak{b}_i$. As you already note we have $e_1=\ldots=e_r$, and $f_1=\ldots=f_r$. Then for $e=f_1$ and $f=f_1$ we have $rfe=p-1$.
Next note that the decomposition of $q$ in $\Bbb{Z}[\zeta_p]/\mathfrak{b}_i$ is determined by the factorization of $\Phi_p$, the $p$-th cyclotomic polynomial, over $\Bbb{F}_q$, by the Kummer-Dedekind theorem. Of course $\Phi_p$ is separable over $\Bbb{F}_q$ because $X^p-1$ is, so $q$ does not ramify and hence $e=1$. This shows that if $m\neq1$ then we cannot find a prime $q$ such that $e=m$. If $m=1$ then every prime $q\neq p$ has $e=m$.
It follows that $rf=p-1$ and so we can find a prime $q$ with $r=m$ for every divisor $m$ of $p-1$ if and only if we can find a prime $q'$ with $f=m$ for every divisor $m$ of $p-1$. Of course $f$ is precisely the multiplicative order of $q'$ mod $p$, so such a prime number exists for every $m$ by Dirichlet's theorem on prime numbers in arithmetic progressions.
