Defining the Artin Map on the Ideles Let $L/K$ be abelian.  There is a natural way to define the Artin reciprocity map on the ideles using the notion of an admissible cycle.  I don't want to go into the details of what that is right now, but essentially what I'm having trouble with is the following.  In order to show that the Artin map is well defined on the ideles $\mathbb I_K$, I must show:

Suppose $x \in K^{\ast}$ is a local norm at each ramified place.  Also suppose that for any real place $v$ of $K$ (and corresponding embedding $ \sigma: K \rightarrow \mathbb{R}$) which has a complex place $w$ lying over it, we have $\sigma(x) > 0$.  Then $$\prod\limits_{v } (\mathfrak p_v, L/K)^{ord_v(x)}$$
  is the identity element of $Gal(L/K)$, where $(\mathfrak p, L/K)$ is the Frobenius element and $v$ runs through all the unramified places.  

One way to do this would be to show that $x$ is a global norm, or at least a local norm at each place $v$ where $ord_v(x) \neq 0$.  But I don't know if this is true.  Any hints?
 A: This is actually pretty simple. The basic properties of the Artin symbol show that
$$(\mathfrak{N}^L_K(\mathfrak{U},L/K)=1$$
for any ideal $\mathfrak{U}$ not divisible by ramified primes since $(\mathfrak{a}, L/K)=(N^K_{k}(\mathfrak{a}), k)$ for $k\subseteq K\subseteq L$.
Now we note that the Artin map is surjective, and that the global cyclic norm index inequality gives, for a conductor $\mathfrak{c}$
$$[I(\mathfrak{c}):P_{\mathfrak{c}}\mathfrak{N}(\mathfrak{c})]\le [L:K]=[\operatorname{Gal}\left(L/K\right):1]$$
since the extension is Galois. But then immediately this implies that the kernel is exactly $P_{\mathfrak{c}}\mathfrak{N}(\mathfrak{c})$, which is exactly what you're trying to show. The inequality is important since showing this is the Artin kernel starts with showing that an admissible cycle exists, but if it does then the rest is easy.

Additional notes: $(1)$The local norm conditions ensure that the product of the $\mathfrak{p}^{v_{\mathfrak{p}}(\mathfrak{p})}$ turn into the norm of the ideal for the primes--multiplicativity means we need only consider $x$ which generate prime ideals, i.e. which localize to a non-unit at one place, since the local valuations are all that affect the local Artin symbols anyways, though the process requires sufficient typing that I don't want to regurgitate it here, since you have a definition on $I$, I assume they taught you the standard procedure, though comment if not.
$(2)$ Without knowing what an admissible cycle is, it's not possible to do this:  the fact that the Artin kernel is exactly things which are congruent to 1 $\mod^* \mathfrak{c}$, this is at the heart of the proof of Artin reciprocity, and involves a lot of auxiliary work. If you want anything with greater detail, I recommend Lang's Algebraic Number Theory which is quite accessible, especially the chapter on the Artin symbol, I learned it there, and it explains the issues in a pretty straightforward way.
