Norm map in Ideles If $L/K$ is a finite extension, then there is a natural norm map from $\mathbf{A}^*_L$ to $\mathbf{A}^*_K$. This is a continuous homomorphism
$$\text{N}^L_K: \mathbf{A}^*_L\rightarrow \mathbf{A}^*_K$$
defined by the prescription that the $v$-component of $\text{N}^L_K$ is
$$\prod_{w\mid v}\text{N}^{L_w}_{K_v}x_w\;.$$
I'm trying to do an example with $L=\mathbb{Q}[i]$ and $K=\mathbb{Q}$. Consider 
$$x=(1,3,1,1,\dots)\in\mathbf{A}^*_L$$
Then 
$$\text{N}^L_K(x)=\left(\text{N}^{\mathbb{Q}[i]_{1+i}}_{\mathbb{Q}_2}1,\, \text{N}^{\mathbb{Q}[i]_{3}}_{\mathbb{Q}_3}3,\, \text{N}^{\mathbb{Q}[i]_{2+i}}_{\mathbb{Q}_5}1,\, \text{N}^{\mathbb{Q}[i]_{2-i}}_{\mathbb{Q}_5}1,\dots\right)=(1,\text{N}^{\mathbb{Q}[i]_{3}}_{\mathbb{Q}_3}3,1,1,\dots)$$
So what is $\text{N}^{\mathbb{Q}[i]_{3}}_{\mathbb{Q}_3}3$? Is it $9$? I thought it maybe $9$ because, to me, the extension $\mathbb{Q}[i]_{3}/\mathbb{Q}_{3}$ is a quadratic extension, so generalising the definition of $N^L_K$ for number fields to be the determinant of the multiplication matrix, we should get $9$.
EDIT (generalisation):
In a more general setting, if we have $\mathfrak{p}$ a prime in $K$ which is inert in $L$, and we consider 
$$x=(\dots,1,1,\pi,1,1,\dots)\in\mathbf{A}^*_L$$
with $\pi$ the uniformiser of $\mathfrak{p}$ in the place induced by $\mathfrak{p}$. So we have
$$\text{N}^L_K(x)=(\dots,1,1,\text{N}^{L_{\mathfrak{p}}}_{K_{\mathfrak{p}}}\pi,1,1,\dots)$$
Now, what is $\text{N}^{L_{\mathfrak{p}}}_{K_{\mathfrak{p}}}\pi$? Is it $\text{N}^L_K\mathfrak{p}$?
Thanks for the help!

By Dedekind theorem on the factorisation of primes, and for $x^2+1\pmod3$ is irreducible, we find that $3\mathbb Z[i]$ is a prime ideal. This tells us that the inertia degree of $3$ is $2$. Hence the local field extension is $ef=2$, where $e$ is the ramification index and $f$ the inertia degree. Consequently, yes, that norm is indeed $9$.
Edit
I cannot understand the reason for taking a uniformiser $\pi$, but, in any case, I think the following suffices.
If by $\pi$ you refer to a uniformiser in $K_\mathfrak p$, then certainly $\text{N}_{K_\mathfrak p}^{L_\mathfrak p}(\pi)=\pi^n$, where $n$ is the degree $[L_\mathfrak p:K_\mathfrak p]$. Since $\mathfrak p$ is inert, we find that $n=[L:K]$.
Hope this helps.  
