Let $K$ be a number field, $n$ be a positive integer and $\zeta_n$ a primitve $n^{th}$ root of unity.
How does one show that if a prime ideal $\mathfrak{p}$ of $K$ is ramified in $K(\zeta_n)$ then $n \in \mathfrak{p}$ ?
(This is not homework).
In fact I realize that there doesn't seem to be much information out there about cyclotomic extension of general number fields, does anyone know a good reference ?
This follows from the general fact that the inertia group is well-behaved in extensions. Suppose $L/\mathbf Q$ is a finite Galois extension with Galois group $G$, and $K/\mathbf Q$ is any finite extension. Then $KL/K$ is Galois with Galois group isomorphic to $\text{Gal}(L/K\cap L) \subseteq G$. If $\mathfrak p \subseteq K$ ramifies in $KL$, pick some prime $\mathfrak q \subseteq KL$ above it, and consider the inertia group $I(\mathfrak q|\mathfrak p)$. Then
$$I(\mathfrak q \cap L|\mathfrak p \cap \mathbf Q) = I(\mathfrak q|\mathfrak p) \cap G = I(\mathfrak q|\mathfrak p).$$
So, if $\mathfrak q|\mathfrak p$ is ramified, so is $\mathfrak q \cap L|\mathfrak p \cap \mathbf Q$. This implies your statement when $L=\mathbf Q(\zeta_n)$ since a prime of $\mathbf Q(\zeta_n)$ is ramified over $\mathbf Q$ precisely when it contains $n$.
As for a reference for cyclotomic extensions of a general number field, do a google search for "$\mathbf Z_p$-extensions" (in plaintext, "Z_p extensions"). You'll find a lot of material about cyclotomic towers over an arbitrary number field (though I'm not sure exactly what you're looking for).
Let $L$ be the splitting field over $K$ of the polynomial $X^n-1$, and fix a choice of primitive $n$-th root of unity $\zeta$ in $L$ (so $L=K(\zeta)$).
Let $K_\mathfrak{p}$ be the completion of $K$ at $\mathfrak{p}$. Fix a prime $\mathfrak{P}$ of $L$ above $\mathfrak{p}$ and let $L_\mathfrak{P}$ be the corresponding completion. We get a canonical embedding $K_\mathfrak{p}\hookrightarrow L_\mathfrak{P}$ over the inclusion $K\hookrightarrow L$, and $L_\mathfrak{P}=K_\mathfrak{p}(\zeta)$. Since $\mathfrak{p}$ is ramified in $L$ and $L/K$ is Galois, $L_\mathfrak{P}/K_\mathfrak{p}$ is ramified (we're not really using that $L/K$ is Galois in a serious way because the assumption that $\mathfrak{p}$ is ramified in $L$ means that some $\mathfrak{P}$ over $\mathfrak{p}$ is ramified, so we could have just chosen that one, but in this case, since $L/K$ is Galois, all primes above $\mathfrak{p}$ are ramified). It follows that $p\mid n$, where $p$ is the rational prime lying below $\mathfrak{p}$ (otherwise the local extension would be unramified). So $n\in p\mathbf{Z}$, and thus $n\in p\mathscr{O}_K\subseteq\mathfrak{p}$.
