Norms in extended fields let's have some notation to start with:
$K$ is a number field and $L$ is an extension of $K$. Let $\mathfrak{p}$ be a prime ideal in $K$ and let its norm with respect to $K$ be denoted as $N_{\mathbb{Q}}^K(\mathfrak{p})$.

My question is this: If $|L:K|=n$, what is $N_{\mathbb{Q}}^L(\mathfrak{p})$?

I would like to think that $N_{\mathbb{Q}}^L(\mathfrak{p})=\left(N_{\mathbb{Q}}^K(\mathfrak{p})\right)^n$, ie if $L$ is a quadratic extension of $K$, then $N_{\mathbb{Q}}^L(\mathfrak{p})=\left(N_{\mathbb{Q}}^K(\mathfrak{p})\right)^2$. Is this right? I feel that the prove would involve using the definition that $N_K^L(x)$ is the determinant of the multiplication by $x$ matrix (Here, $K$ and $L$ are arbitrary). Thanks!

The queston is simple, if one notices the equality:  

Let $p\in \mathfrak p\cap \mathbb Z$ be a prime integer in $\mathfrak p$. And let $f$ be the inertia degree of $\mathfrak p$. Then $N^K_{\mathbb Q}\mathfrak p=p^f$.  

Now the result follows from the multiplicativity of the inertia degree $f$.
P.S. The above equality could be found in any standard algebraic number theory text.  
