Q:

$\ell$ splits in $K/\mathbb{Q}$ $\iff$ $\ell$ splits in $K(\rho)/\mathbb{Q}(\rho)$

Let $p \equiv 1(3)$ be prime and $\rho$ a primitive $3$rd root of unity. Let $K\subset \mathbb{Q}(\zeta_p)$ have $[K:\mathbb{Q}]=3$, and $\ell \equiv 2(3)$ prime (so that $\ell$ is inert in $\mathbb{Q}(\rho)$). 
Why do we have that $\ell$ splits in $K/\mathbb{Q}$ $\iff$ $\ell$ splits in $K(\rho)/\mathbb{Q}(\rho)$?
Is this true in general for extensions of Dedekind domains? That is, if $R/S$ is an extension of Dedekind domains, for what sorts of elements $e,a\in R$ is it true that $(a)$ splits in $R/S\Leftrightarrow (a)$ splits in $R(e)/S(e)$?

A:

Here is a proof using class field theory.
Recall that a prime is split iff the Frobenius is trivial.
Consider the diagram (note that every extension is Galois abelian):
$\begin{matrix} 
 & K(\rho) & \\
/ & & \backslash \\
K & &\mathbb{Q}(\rho) \\
\backslash & & / \\
& \mathbb{Q} & \\
\end{matrix}$
The consistency property of the Artin map shows that
$$\mathrm{res}_K(\ell,K(\rho)/\mathbb{Q}(\rho)) = (\ell,K/\mathbb{Q})^{ f(\ell,\mathbb{Q}(\rho)/\mathbb{Q}) }.$$
(1) Since $f(\ell,\mathbb{Q}(\rho)/\mathbb{Q}) = 2$ and $\mathrm{Gal}(K/\mathbb{Q})$ is of order $3$, the element $(\ell,K/\mathbb{Q})^{ f(\ell,\mathbb{Q}(\rho)/\mathbb{Q})}$ is trivial iff $(\ell,K/\mathbb{Q})$ is trivial.
(2) Since $K(\rho)$ is the compositum of $K$ and $\mathbb{Q}(\rho)$, the element $\mathrm{res}_K(\ell,K(\rho)/\mathbb{Q}(\rho))$ is trivial iff $(\ell,K(\rho)/\mathbb{Q}(\rho))$ is trivial.
Hence $(\ell,K/\mathbb{Q})$ is trivial iff $(\ell,K(\rho)/\mathbb{Q}(\rho))$ is trivial.

