Cardinality of prime divisors in cyclotomic fields 1) For $p$ an odd prime, let $K_{n} = \mathbb{Q}[e^{\frac{2\pi i}{p^{n}}}]  \ $ , and let $R_{n}$ be the ring of integers of $K_{n}$. Let $q\mathbb{Z} \ $ be a prime ideal of  $\mathbb{Z} \ $, with $q$ a prime $\neq \ p$. Show that the number of prime divisors $Q_{n}$ of $qR_{n} \ $ is bounded as $ n \rightarrow \infty$.
2) Assume that there exists a sequence of Galois extensions $K_{n}/\mathbb{Q} $ with Gal($K_{n}/\mathbb{Q} $) dihedral of order $2p^{n}$, $n\geq 0$ . Write $R_{n}$ for the ring of integers of $K_{n}$ . Let $S$ be the set of primes $q\mathbb{Z} \ $ which are inert in the quadratic field $K_{0}$ ( i.e. for which the ideal $qR_{0}$ is prime ). For any $q \in S $ , show that the number of prime divisors of $qR_{n}$ has unbounded cardinality as $ n \rightarrow \infty$. Is $S$ infinite ?
 A: I'll do (1), and leave (2) to you.
Since $q$ is unramified in the abelian extension $K_n/\mathbb Q$, there is a well-defined Frobenius element $\sigma_{q, n} \in \text{Gal}(K_n/\mathbb Q) = (\mathbb Z/p^n\mathbb Z)^\times$. This element is nothing but the class of $q$ mod $p^n$. The order of the Frobenius is equal to the degree of the residue field extension of any prime $\mathfrak q$ lying over $q$. Therefore, the number of prime ideals above $q$ in $K_n$ is equal to the index of $\left<q\right>$ in $ (\mathbb Z/p^n\mathbb Z)^\times$, i.e. to the cardinality of the factor group 
$$H_{n} =(\mathbb Z/p^n\mathbb Z)^\times/\left<q\right>$$
Now I claim that $|H_{n}|$ is bounded as $n\to \infty$. In order to show this, let $H$ be the closure of the group generated by $q$ in $\mathbb Z_p^\times$ (the group of units of the $p$-adic integers). I claim that $H$ is open in $\mathbb Z_p^\times$, and therefore has finite index since $\mathbb Z_p^\times$ is compact. Indeed, as topological groups we have $\mathbb Z_p^\times \cong (\mathbb Z/p\mathbb Z)^\times \times \mathbb Z_p$, and it is well-known that every closed subgroup of $\mathbb Z_p$ is an ideal, hence open provided it is not the zero ideal. Therefore $H$ is open in $\mathbb Z_p^\times$. Now for each $n$, reduction mod $p^n$ induces a surjective map
$$\mathbb Z_p^\times /H \to H_{n}.$$
Thus $|H_{n}|$ is bounded by the (finite) index of $H$ in $\mathbb Z_p^\times$.
Remark: I believe that this boundedness is a basic example of a recurring phenomenon in Iwasawa theory, where one considers arithmetic objects over towers of abelian number fields. A general principle says that "bounded ramification locus in an abelian tower implies boundedness". For example, let $S$ be a finite set of primes, and $E/\mathbb Q$ an elliptic curve. Let $K$ be the maximal abelian extension of $\mathbf Q$ unramified outside $S$. Then $E(K)$ is finitely generated.
