A subfield of $\mathbb{Q}(\zeta_{pq})$ with some ramification conditions

Question

Let $p, q\in \mathbb{Z}$ be distinct primes such that

• $p$ splits into $r$ distinct primes in $\mathbb{Q}(\zeta_q)$
• $\mathbb{Q}(\zeta_q)$ contains a subfield $F$ of degree $rf$ over $\mathbb{Q}$
• $e\mid p-1$

I have to show that $\mathbb{Q}(\zeta_{pq})$ contains a subfield in which $p$ splits into $r$ primes with ramification index $e$ and inertia degree $f$.

Let $L$ be the subfield of $\mathbb{Q}(\zeta_{pq})$ with degree $ref$ over $\mathbb{Q}$. Suppose that $p$ splits in $L$ into $r'$ primes with ramification index $e'$ and inertia degree $f'$. I want to prove that $r'=r$, $e'=e$ and $f'=f$. It is easy to show that $r'e'f'=ref$, $r'\ge r$, $e\mid e'$ and $f\mid f'e'$. Unfortunately this is not sufficient. How can I proceed?

Answer 1

The key fact to show what you ask is the following; this is proposition III.1.6. from Janusz's book:

Suppose $K$ is a number field, and $L$ is a finite Galois extension of $K$. Let $\mathfrak p$ be a prime ideal in $\mathcal O_K$. Suppose $\mathfrak p$ factors into $r$ distinct primes in $\mathcal O_L$, let $\mathfrak P$ be such a prime. If $G(\mathfrak P)$ is a normal subgroup of Gal$(L/K)$, then in $L^{G(\mathfrak P)}$, $\mathfrak p$ has factorization $$\mathfrak p_1\cdots\mathfrak p_r, \text{ with } e(\mathfrak p_i/\mathfrak p)=1=f(\mathfrak p_i/\mathfrak p) \text{ for each }i,(1)$$ $G(\mathfrak P)$ being the decomposition group of $\mathfrak P$ in Gal$(L/K)$.

So let $K:=\Bbb Q(\zeta_q)^{G(\mathfrak P)}$, for some prime $\mathfrak P$ of $\mathbb Z[\zeta_q]$ extending $p$; $\mathcal O_{\Bbb Q(\zeta_q)}=\mathbb Z[\zeta_q] $. Then as $\Bbb Q(\zeta_q)/\Bbb Q$ is an abelian extension, we get that in $\mathcal O_K$, $p$ factors as in $(1)$, so that $[K:\mathbb Q]=r$ by the fundamental equality.

As Gal$(\Bbb Q(\zeta_q)/\Bbb Q)$ is a cyclic group of order $q-1$, for each positive integer $n$ such that $n\mid (q-1)$, there is a unique subgroup of order $n.(2)$

Now, by your assumption, $\Bbb Q(\zeta_q)$ has a subfield of degree $rf$, $L$, so $rf\mid (q-1)$. Let $H$ and $G$ be the subgroups of Gal$(\Bbb Q(\zeta_q)/\Bbb Q)$ of orders $(q-1)/rf$ and $(q-1)/r$, resp. Hence by $(2)$ and the fundamental theorem of Galois theory it follows that $L=\mathbb Q(\zeta_q)^H$ and $K=\mathbb Q(\zeta_q)^G$, however by $(2)$ $H\leq G$ and thus $K\subseteq L.$

As $p$ doesn't ramify in $\Bbb Z[\zeta_q]$, neither does in $\mathcal O_L$, hence as $K\subseteq L\subseteq\mathbb Q(\zeta_q)$, we obtain that $p$ factors in $\mathcal O_L$ as a product of distinct primes $\mathfrak q_1\cdots \mathfrak q_r$. However as $[L:\Bbb Q]=rf$, we get from the fundamental equality that $f(\mathfrak q_i/p)=f$ for each $i$. (3)

For each $n$ such that $n\mid p-1$, there is a unique subgroup $H_n$ of Gal$(\Bbb Q(\zeta_p)/\Bbb Q)$ of order $(p-1)/n$, and thus $L_n:=\Bbb Q(\zeta_p)^{H_n}$ is a subfield of $\Bbb Q(\zeta_p)$ of degree $n$ over $\Bbb Q$. Now, as $p$ totally ramifies in $\Bbb Z[\zeta_p]$, so it does in $\mathcal O_{L_n}$.Therefore for the only prime ideal $\mathfrak p$ of $\mathcal O_{L_n}$ containing $p$ we have $e(\mathfrak p/p)=n.(4)$

Now, let $e\mid (p-1)$, and consider the field $F:=LL_e\subseteq \mathbb Q(\zeta_{pq})$. As $\mathbb Q(\zeta_p)$ and $\mathbb Q(\zeta_q)$ are linearly disjoint over $\mathbb Q$, so are $L$ and $L_e$, thus $[F:\Bbb Q]=efr$. Let $\mathfrak P$ be a prime ideal of $\mathcal O_F$ containing $p$. Since $L\subseteq F$ we get $f(\mathfrak P/p)\geq f$ by (3), and since $L_e\subseteq F$, $e(\mathfrak P/p)\geq e$ by $(4)$.

But as $\mathfrak P$ was arbitrary, there are at least $r$ distinct primes in $F$ containing $p$; as $K\subseteq F$, and $[F:\Bbb Q]=efr$, we obtain

$$p\mathcal O_F=(\mathfrak P_1\cdots\mathfrak P_r)^e\text{, with }f(\mathfrak P_i/p)=f\text{ for each }i,$$

by the fundamental equality.