Class numbers of cyclotomic fields and their maximal totally real subfields

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!

• Please use a high-level tag like "nt.number-theory". I added this tag now. Commented Dec 19, 2022 at 1:06
• Since $L/K$ is ramified at $p$, the Hilbert class field in the strict sense is disjoint from $L/K$. This proves the claim since any ramification at infinite primes is killed by $L/K$. Commented Dec 19, 2022 at 11:13

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 $$F(\zeta_p)$$.

• Thank you very much for your answer. I am wondering, what if $F$ is not a Galois extension, is it possible to extend the argument, or it is not needed when considering the maximal unramified extension (for the narrow class number)?
– did
Commented Dec 19, 2022 at 8:59