Let $K\subseteq L$ be number fields over the field of rationals $\Bbb Q$.
with rings of integers $\mathcal{O}\_K\subseteq \mathcal{O}\_L$.
Let $P$ be a prime ideal of $\mathcal{O}\_L$, let $p$ be a prime ideal of $\mathcal{O}\_K$, such that $P$ is over $p$.


The residue class degree $f$ is defined to be $f=[\mathcal{O}\_L/P:\mathcal{O}\_K/p]$.
The norm of $P$ is the ideal $N(P)=p^f$.
This is the usual definition of norm of an ideal. (See Serre's *Local fields* and Serge Lang's *Algebraic tumber theory*.)


Swinnerton-Dyer's *A brief guide to algebraic number theory* has a different definition of norm of an ideal (page 25).
Namely if $A$ is an ideal of $\mathcal{O}\_L$, it is defined as $N(A)$ = ideal in $\mathcal{O}\_K$ generated by elements $N(a)$ where $a\in A$.


I don't know why these two definitions are the same. Swinnerton-Dyer claims 
so in his book. Can anyone here give a hint, an explanation or anything 
else?



A problem in algebraic number theory, norm of ideals
**[Edit: this answer is incomplete/incorrect, see rather [this one](https://mathoverflow.net/a/97720/14094)]**


Ok, here's the argument:
First recall that the usual norm for non-zero elements of a field is transitive in towers; thus the same is true for your second definition of the norm of an ideal. In particular, $N\_{K|Q}\circ N\_{L|K} = N\_{L|Q}$. The fact that the norm $N\_{L|Q}(\mathfrak{P}) = [\mathcal{O}\_L:\mathfrak{P}] \cdot \mathbb{Z}$ is easy to see for a prime $\mathfrak{P}$ in $\mathcal{O}\_L$; edit: and thus the same is true for any integral ideal $\mathfrak{a}$. Now let $\mathfrak{p} = \mathcal{O}\_K \cap \mathfrak{P}$ and $(p) = \mathbb{Z} \cap \mathfrak{P}$.


We have $N\_{L|Q}(\mathfrak{P}) = p^{f(\mathfrak{P}|p)} = N\_{K|Q}N\_{L|K}\mathfrak{P}$. In particular, we deduce that $N\_{L|K}\mathfrak{P} = \mathfrak{p}^d$ for some $d$. Moreover, we know that


$N\_{K|Q}\mathfrak{p}^d = p^{d \cdot f(\mathfrak{p}|p)} = p^{f(\mathfrak{P}|p)}.$


But then $d = f(\mathfrak{P}|p) / f(\mathfrak{p}|p) = f(\mathfrak{P}|\mathfrak{p})$ as required. 


