ring of integers of cyclotomic extension of rationals: residue class degree Let $\zeta$ be a $p$-th root of unity, $p$ odd prime. Let $q\neq p$ be prime.
Consider $L=\mathbb{Z}[\zeta]$. Let $I$ be a prime ideal of $L$ divisible by $(q)=q\mathbb{Z}$ in $L$. The field of fractions (fof) of $L$ is Galois extension of $\mathbb{Q}$, so the residue class number $$f=f(I/(q))$$ is the same regardless of the particular prime factor $I$. I also know that $q$ is unramified in $L$. The the number of distinct prime ideal factors of $qL$ is $p-1\over f$. How can I prove that $f$ is the smallest positive integer so that $p|q^f-1$? Any help is great appreciated. Thanks.
 A: By Kummer-Dedekind factorization theorem cited in this question, let $p$ be the minimal polynomial of $\zeta$, then if 
$$ p(x)\equiv\prod_ip_i(x)^{e_i}\pmod q,$$
where $p_i$ are irreducible polynomials with coefficients in $\mathbb F_q=\mathbb Z/q\mathbb Z$,
then 
$$(q)=\prod_i(q,p_i(\zeta))^{e_i}$$
is the prime ideal factorization of $(q)$ in $\mathbb Z[\zeta]$, and the inertia degree of $(q,p_i(\zeta))$ is the degree of $p_i$. Here we know each $e_i=1$, as an aside.
Since $p(x):=x^{p-1}+\cdots+x+1=\frac{x^p-1}{x-1}$ is irreducible (by Eisenstein criterion applied to $p(x+1)$), it is the minimal polynomial for $\zeta$.
Now we try to determine the degrees of each $p_i$. By Galois theory, this is independent of $i$. So we consider only one irreducible factor $p_i$.
Suppose $\deg(p_1)=k$, and let $\theta$ be a root of $p_1$. Then $\mathbb F_q[\theta]/\mathbb F_q$ is a field extension of degree $k$. By the theory of finite fields we know that $\mathbb F_q[\theta]\cong \mathbb F_{q^k}$ and the Galois group $\operatorname{Gal}(\mathbb F_q[\theta]/\mathbb F_q)$ is generated by the Frobenius automorphism $\varphi_q:x\mapsto x^q$. That is, it sends $\theta$ to $\theta^q$. Since the order of the Galois group is just $k$, we conclude that $k$ is the smallest integer such that $\theta^{q^k}=\theta$.
But $\theta$ is a root of $p(x)$, so it has order $p$; that is, if $\theta^n=\theta$, then $p\mid n-1$. As a consequence, we see that $k$ is the smallest integer $k$ such that $p\mid q^k-1$.

Hope this helps.
A: Actually the K-D. factorisation theorem is not needed, you have only to apply standard properties of ramification theory, which I recall briefly. If $K/\mathbf Q$ is galois of degree $n$, the $r$ prime ideals $Q_i$ of the ring of integers $O_K$ above a given prime ideal $q\mathbf Z$ are permuted by the $G=Gal(K/\mathbf Q)$, and the prime decomposition of  $qO_K$ reads $qO_K= (\prod Q_i )^e$, with $n=ref$, where $e$ (resp. $f$) is the common ramification (resp. inertia) indices of the $Q_i$'s above $q$. The Galois interpretation of $r$ is that it is the common order of any of the subgroups of $G$ fixing a $Q_i$ (decomposition groups). Also, $f$ is the degree of any of the residue fields $k_i:=O_K/Q_i$ over $\mathbf F_q$, i.e. the common order of the decomposition subgroups.  
In the special cyclotomic case here, $G$ is abelian, hence the decomposition (resp. inertia) subroups are mutually equal. Call them resp. $D_q$ and $I_q$. If moreover  $q$ is unramified, $I_q$ is trivial and $D_q \cong Gal(k_i/\mathbf F_q)$ (independently from $i$) is cyclic of order $f$, generated by an automorphism $s$ which reduces to the Frobenius automorphism $\phi:x \to x^q$ of the residual extension $k_i/\mathbf F_q$. But the image of $\zeta_p$ generates $k_i/\mathbf F_q$ because $q\neq p$. Comparing the actions of $s$ and $\phi$, we get what is wanted. 
