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For a CM-field $K$ over its maximal real subfield $K^{+}$, the ordinary Hilbert class field $H_{K^{+}}$ is totally real because its real places split. Thus $H_{K^{+}}\cap K=K^{+}$. Its compositum with $K$ is an abelian extension of $K$ of degree $h_{K^{+}}$, and is everywhere unramified by unramified extensions remain unramified under base change. It lies in $H_K$, giving the divisibility. The argument does not require injectivity of extension of ideal classes.

Let $H_k$ be the ordinary Hilbert class field of $k$. Its real places split completely, so $H_k$ is totally real. Since $K/k$ is quadratic and $K$ is totally imaginary, $H_k\cap K$ cannot equal $K$; hence it equals $k$. Therefore the compositum $K{H}_k$ hasand is an abelian extension of $K$, by restriction of the Galois group to $H_k$. At finite primes it is unramified because unramified extensions remain unramified under base change. The field $K$ has no real places, so there is no infinite-place ramification condition left. Thus $K{H}_k\subseteq H_K$, where $H_K$ is the Hilbert class field of $K$. Since $[H_K:K]=h_K$, the tower degree formula proves class number divisibility for a CM extension:This argument uses the ordinary class field with split real places; replacing it silently by the narrow class field would invalidate the claim that $H_k$ is totally real.

Fix a prime $\mathfrak{q}$ of $E$ over $\mathfrak{p}$, and a prime $\mathfrak{r}$ of $LE$ over $\mathfrak{q}$. Let $\mathfrak{P}$ be its restriction to $L$. Inside the completed compositum, putThe assumption means $U/F$ is an unramified extension. Its residue degree is some $f$; the completed extension $(LE{)}_{\mathfrak{r}}/J$ is the corresponding compositum $UJ/J$. It suffices to prove that this is unramified, regardless of the ramification of $J/F$.

Here is a proof of unramified extensions remain unramified under base change using the classification already established. Choose an integral generator $\alpha$ of $U/F$ whose monic minimal polynomial $g\in {\mathcal{O}}_F[X]$ has irreducible separable reduction of degree $f$. Write $k_F={\mathbb{F}}_q$ and $k_J={\mathbb{F}}_{q^r}$. Let $J_f/J$ be the degree-$f$ unramified extension. Its residue field ${\mathbb{F}}_{q^{rf}}$ contains ${\mathbb{F}}_{q^f}$, so $\bar{g}$ has a simple root there. Hensel lemma lifts this root to a root of $g$ in $J_f$. This gives an $F$-embedding $U\hookrightarrow J_f$.

An unramified extension of a local field is Galois, so all $F$-embeddings of $U$ into a fixed algebraic closure have the same image. The original $U$ therefore lies in $J_f$, and $UJ$ is an intermediate field of the unramified extension $J_f/J$. Its ramification index is one. Since $\mathfrak{q}$ and $\mathfrak{r}$ were arbitrary,This proof also explains why no linear-disjointness assumption on $L$ and $E$, and no assumption that $E/K$ is unramified, is needed.