What are conditions to satisfied by rational prime p so that every prime lying above p is a prime of order 1 and generates class group?

Question

I was reading a paper on Euclidean ideals by H Graves and M. Ram Murthy. I have a problem in understanding one of the claims.

setup

Let $K$ be a number field and $H(K)$ is its Hilbert class field. Suppose $H(K)/\Bbb Q$ is abelian and $f(K)$ is the conductor of K which is defined to be the smallest even number $n$ such that $K \subset \Bbb Q(\zeta_n)$. Suppose that the class group $\mathrm{Cl}_K$ of $K$ is cyclic and let $C\in\mathrm{Cl}_K$ be an ideal class generating the ideal class group.

On page 2 they claim that there exists some integer $a$, $0<a<f(K),(a, f(K)) = 1,$ such that if $p \equiv a\pmod{f(K)},$ then $\mathfrak{p}$ is of degree $1$ and $[\mathfrak{p}]=[C]$. Here $\mathfrak{p}$ is a prime in $K$ lying above $p$.

I tried to establish a relation between the class $p \pmod{f(K)}$ and the ideal class of the primes above $p$ by tracking down the map given there, but I was not successful. Any help in this regard is very appreciable.

Answer 1

First of all, as they observe, the assumption that $H(K)/\mathbb{Q}$ is abelian ensures that $H(K)\subseteq\mathbb{Q}(\zeta_{f(K)})$ and not only $K\subseteq\mathbb{Q}(\zeta_{f(K)})$. Also, they work under the assumption that the class group of $K$ is cyclic, and I have added this to your question.

Now, Artin reciprocity gives an isomorphism between the ideal class group $\mathrm{Cl}_K$ of $K$ and the Galois group $\operatorname{Gal}(H(K)/K)$: this is one of the main results of Global Class Field Theory. In particular, there is an isomorphism (let me call it $\varphi$) between $\mathrm{Cl}_K$ and a subgroup of $\operatorname{Gal}(\mathbb{Q}(\zeta_{f(K)})/\mathbb{Q})$, as a consequence of the inclusion $H(K)\subseteq\mathbb{Q}(\zeta_{f(K)})$.

On the other hand, the theory of cyclotomic fields yields an isomorphism $$ \tau\colon (\mathbb{Z}/f(K))^\times\cong \operatorname{Gal}(\mathbb{Q}(\zeta_{f(K)})/\mathbb{Q}). $$ But more is true: the isomorphism $\tau$ associates to every prime $p\nmid f(K)$ (hence, whose class $[p]$ lies in $(\mathbb{Z}/f(K))^\times$) the element $\mathrm{Frob}(p,\mathbb{Q}(\zeta_{f(K)}/\mathbb{Q})$.

Now we have this class $C$: through $\varphi$ it corresponds to a unique element $\varphi(C)=\gamma\in\operatorname{Gal}(H(K)/K)$. The Chebotarev Density theorem says that there exists a positive density of primes $\mathfrak{q}$ in $K$ such that their Frobenius element $\operatorname{Frob}(\mathfrak{q},H(K)/K)$ equals $\gamma$. Since primes of degree $>1$ have density $0$, the above set of primes contains infinitely many primes of degree $1$, lest its density would be $0$. So we can chose one prime $\mathfrak{p}$ in that set, and it will

1. Have degree $1$ in $K/\mathbb{Q}$;
2. Have Frobenius element $\mathrm{Frob}(\mathfrak{p},H(K)/K)$ equal to $\gamma=\varphi(C)$;
3. Setting $p=\mathrm{Norm}^K_\mathbb{Q}(\mathfrak{p})$, this number will be prime (by 1.) and its Frobenius element $\tau(p)=\mathrm{Frob}(p,H(K)/\mathbb{Q})$ will restrict to $\mathrm{Frob}(\mathfrak{p},H(K)/K)=\varphi(C)$ and will depend only on $p\pmod{f(K)}$: letting $a$ be the integer satisfying $a\equiv p\pmod{f(K)}$ and $0<a<f(K)$ you get your element.