Artin symbol property In Serge Lang's Algebraic Number Theory, Chapter 10, it defines Artin symbol as follows:

Let $k$ a number field, $K/k$  be an abelian Galois extension, and $p$ be a prime of $k$ which is unramified in $K$, let $B$ be a prime in $K$ lying above $p$, then there exists a unique element $\sigma$ of the Galois group $G$, lying in the decomposition group $G_B$ having the effect $\sigma a \equiv a^{N_p} 
 \left(\mod B \right), a\in o_K.$ ($N_p = |o_k/p|$ )
This element $\sigma$ depends only on $p$, and is denoted by $(p,K/k)$ , and will be called Artin symbol of $p$ in $G$.

Then it states a property:

Let $K/k$ be abelian and $E/k$ be finite, both Galois. Let $p$ be a prime in $k$ unramified in $K$ and let $q$ be a prime of $E$ lying above $p$. Then $\mbox{res}_K (q, KE/E) = (p, K/k)^f$ where $f$ is the residue class degree $[o_E/q:o_k/p].$

But in order for $(q,KE/E)$ to make sense, $q$ must be unratified in $KE$, why is it the case here? 
 A: Let $\mathfrak P$ be a prime ideal in $\mathcal O_{KE}$ lying above $p$. Also let $\mathfrak p = \mathfrak P \cap (K \cap E)$, $\mathfrak q = \mathfrak P \cap E$. Additionally it's well-known that $KE$ is Galois over $k$, so it makes sense to talk about inertia fields and so on.
Now let $L_I$ be the inertia field of $\mathfrak P/\mathfrak p$ and $L_E$ the inertia field of $\mathfrak q/\mathfrak p$. As $L_I$ contains every intermediate field in which ramification doesn't occur we get that $L_E \subseteq L_I$. In the same manner we have that $L_I \cap E \subseteq L_E$, as no ramification occurs in $K \cap E \subseteq E \cap L_E$. Thus $L_E = E \cap L_I$. Moreover as $\text{Gal}\left(\frac{KE}{K} \right)\cong \text{Gal}\left(\frac{E}{K \cap E}\right)$ we get that $L_I \leftrightarrow L_E$ under the natural isomorphism.
Finally we get that $e(\mathfrak P|\mathfrak p) = [KE:L_I] = [E:L_E] = e(\mathfrak q|\mathfrak p)$. However by multiplicativity $e(\mathfrak P|\mathfrak p) = e(\mathfrak P|\mathfrak q)e(\mathfrak q|\mathfrak p)$ and thus $e(\mathfrak P|\mathfrak q)=1$. In other words we get that $\mathfrak q$ doesn't ramify in $\mathcal O_{KE}$.
Hence the proof.
REMARK: Note as $k \subseteq E$ is finite extension all the ramification indices and extension degrees we worked with are finite integers, so it makes sense to perform all those operations.
