Class numbers of cyclotomic fields and their maximal totally real subfields

Question

Let $\zeta_p$ be a $p$-th root of unity for a prime $p$, let $L:=\mathbb{Q}(\zeta_p)$ and $K$ the maximal totally real subfield of $L$, i.e. $K:=\mathbb{Q}(\zeta_p+\zeta_p^{-1})$. I am trying to prove that the narrow class number of $K$ divides the class number of $L$ i.e. $h_{K}^+\mid h_L$.

I was trying to show that if $F$ is an extension of $K$ which is unramified at all finite primes, then $F(\zeta_p)$ will be an extension of $L$ which is unramified at all primes. However, I am not sure how to prove this. Any help would be much appreciated!

Answer 1

To prove that $F(\zeta_p)$ is an extension of $L$ which is unramified at all primes, it is enough to show that $F(\zeta_p)$ is an extension of $L$ which is unramified at all finite primes, because all finite primes are the only primes that can ramify.

To show that $F(\zeta_p)$ is an extension of $L$ which is unramified at all finite primes, we can use the following criterion: If $F$ is a Galois extension of $K$ which is unramified at all finite primes, then $F(\zeta_p)$ is an extension of $L$ which is unramified at all finite primes. This criterion can be proved as follows:

Let $\mathfrak{p}$ be a prime of $L$ which lies above a prime $\mathfrak{P}$ of $F$. We want to show that $\mathfrak{p}$ is unramified in $F(\zeta_p)$.

Since $F$ is a Galois extension of $K$, the decomposition group of $\mathfrak{P}$ in $F$ is equal to the Galois group of $F/K$. Since $F$ is unramified at all finite primes, the decomposition group of $\mathfrak{P}$ in $F$ is a subgroup of the inertia group of $\mathfrak{P}$ in $F$.

On the other hand, the Galois group of $F(\zeta_p)/L$ is equal to the Galois group of $F/K$ by the Galois correspondence, so the decomposition group of $\mathfrak{p}$ in $F(\zeta_p)$ is equal to the Galois group of $F/K$.

Since the decomposition group of $\mathfrak{p}$ in $F(\zeta_p)$ is equal to the decomposition group of $\mathfrak{P}$ in $F$, which is a subgroup of the inertia group of $\mathfrak{P}$ in $F$, it follows that the decomposition group of $\mathfrak{p}$ in $F(\zeta_p)$ is a subgroup of the inertia group of $\mathfrak{p}$ in $F(\zeta_p)$. This means that $\mathfrak{p}$ is unramified in $F(\zeta_p)$.