How to prove $N_{L/K }(\mathfrak{P})=\mathfrak{p}^{f_{\mathfrak{P}/\mathfrak{p}}}$? Let $L/K$ be a finite extension of number fields and the prime $\mathfrak{P}$ lies above the prime$(\mathfrak{p})$ of $K.$ To prove $N_{L/K }(\mathfrak{P})=\mathfrak{p}^{f_{\mathfrak{P}/\mathfrak{p}}}$, I tried using ideal norm arguments, but I was not successful.
Any help is much appreciated.
 A: Let $L/K/\Bbb{Q}$ be a tower of finite extensions and $\mathfrak{P}\subset O_L$ a maximal ideal. Let $N_{L/K}(\mathfrak{P})$ be the $O_K$ ideal generated by the norms of all elements $\in \mathfrak{P}$.
$N_{L/K}(\mathfrak{P})$ is a power of $\mathfrak{p}=\mathfrak{P}\cap O_K$: this is because for each $a\in O_K-\mathfrak{p}$ there is $b\in \mathfrak{p}$ such that  $(a,b)=O_K$ thus $(a,N_{L/K}(\mathfrak{P}))\supset (a,b^{[L:K]})=O_K$ and hence $N_{L/K}(\mathfrak{P})$ is comaximal with $(a)$.
From our knowledge of Dedekind domains this implies that $N_{L/K}(\mathfrak{P})=\mathfrak{p}^r$ for some $r$.
$N_{L/\Bbb{Q}}$ is defined the same way, but we also know that
$$N_{L/\Bbb{Q}}(\mathfrak{P})=|O_L/\mathfrak{P}|\Bbb{Z}=p^{f(\mathfrak{P}/p)}\Bbb{Z},\qquad N_{L/\Bbb{Q}}(\mathfrak{P})=N_{K/\Bbb{Q}}(\mathfrak{p}^r)=p^{r\ f(\mathfrak{p}/p)}\Bbb{Z}$$
Whence $$r = \frac{f(\mathfrak{P}/p)}{f(\mathfrak{p}/p)}=f(\mathfrak{P/p})$$
where $f(\mathfrak{P/p})$ means $[O_L/\mathfrak{P}:O_K/\mathfrak{p}]$.

I think it is a bit non obvious that $N_{L/\Bbb{Q}}(\mathfrak{P})=N_{K/\Bbb{Q}}(\mathfrak{p}^r)$:
by definition $N_{L/\Bbb{Q}}(\mathfrak{P})\subset N_{K/\Bbb{Q}}(\mathfrak{p}^r)$, for the reverse inclusion I would take $c,d\in \mathfrak{P}$ such that $(c)=\mathfrak{P} I,(d)=\mathfrak{P} J$, $(N_{L/\Bbb{Q}}(I),N_{L/\Bbb{Q}}(J))=(1)$, therefore $\mathfrak{p}^r=(N_{L/K}(c),N_{L/K}(d)), N_{K/\Bbb{Q}}(\mathfrak{p}^r)=(N_{L/\Bbb{Q}}(c),N_{L/\Bbb{Q}}(d))$.
$N_{L/\Bbb{Q}}(\mathfrak{P})=|O_L/\mathfrak{P}|\Bbb{Z}$ follows from the same idea.
