Norm compatibility of ideals and ideles Let $L|K$ be a finite separable extension of global fields, $\mathbb{I}$ the idele group and $\mathcal{I}$ the group of fractional ideals. We have a surjective homomorphism from $\mathbb{I}\to\mathcal{I}$ given by 
$a\mapsto\prod\mathfrak{p}^{v_{\mathfrak{p}}(a)}$. The idele norm between 
$\mathbb{I}_{L}$ and $\mathbb{I}_{K}$ is given by $N_{L|K}(b_{w})=(a_v)$ where each $a_{v}$ is given by $a_{v}:=\prod_{w|v}N_{{L_{w}|K_{v}}}(b_{w})$. The ideal norm is given by $\mathfrak{P}\mapsto \mathfrak{p}^{f(\mathfrak{P}|\mathfrak{p})}$ and is extended linearly.
Why is the diagram
$\require{AMScd}$
\begin{CD}
\mathbb{I}_{L} @>>> \mathcal{I}_{L}\\
@V{N_{L|K}}VV @V{N_{L|K}}VV \\
\mathbb{I}_{K} @>>> \mathcal{I}_{L};
\end{CD}
commutative? 
 A: I will show the commutativity for an idele $x = (..., 1, \pi_P, 1, ...) \in \mathbb I_L$ where $\pi_P$ is a uniformizer in $L_P$, i.e. $v_P(\pi_P) = 1 / e(P/p)$ (it has minimal postitive valuation — see step 0 just below).
Here $p$ denotes a prime of $K$, and $P$ is a prime of $L$ above $p$.
 It is not difficult to conclude then the commutativity for an arbitrary idele of $L$.
Step 0.
Recall first that we define
$$v_P(b_P) := \dfrac{1}{[L_P : K_p]} v_p(N_{L_P/K_p}(b_P))$$
for any $b_P \in L_P^{\times}$, and recall that 
$$[L_P : K_p] = e(P/p) f(P/p).$$
Step 1.
We first apply the idele norm to $x$. The only non-trivial component of the idele $N_{L/K}(x) \in \Bbb I_K$ is the component of the prime $p$ :
$$ (N_{L/K}(x))_p = \prod_{Q \mid p} N_{L_Q / K_p}(x_Q) = N_{L_P / K_p}(x_P)$$
because $x_Q = 1$ whenever $Q \neq P$.
We then apply the bottom arrow to $N_{L/K}(x)$, and we get the follwoing ideal of $K$ :
$$ p^{v_p(N_{L_P / K_p}(x_P))} = p^{f(P/p)} $$
Here I used the identites from Step 0.
Step 2.
We now follow the other path in the diagram.
First apply the top arrow to our idele $x$. This gives directly $P$.
[Here there is a subtelty : the top arrow is given as follows : for an idele $(b_P)$, we can write $b_P = u_P \pi_P^{n_P}$ for a unique $u_P \in O_{L_P}^{\times}$ and a unique $n_P \in \Bbb Z$, and then this idele is sent to the ideal $\prod_P P^{n_P}$. We have to be careful that $n_P$ is not the same as $v_P(b_P)$ in general!]
Apply the ideal norm to this, and you get
$$p^{f(P/p)},$$
which  is exactly what we've found before. That's it!
$\hspace{7cm}\blacksquare$
