In Cassels and Frohlich, the chapter Global Class Field Theory by Tate, section 12 Proof of the Existence Theorem, on page number 201, it says
We have already shown that if $L/K$ is abelian,then $N_{L/K}C_L$ is an open subgroup of $C_K$, of finite index.
Although I could not find this proof in the book. Can anyone tell me where the proof of the above is located in the book (or perhaps somewhere else)?
Finite index follows as part of second inequality. But I need proof of openness.
Here's a simple enough argument.
It's enough to show that it contains an open neighborhood of the unit element. So, if we consider the "full" idele group $I_K$ (without quotienting by the field elements -- sorry, I didn't check the most standard notations), we want to show that $\nu=N_{L/K}(I_L)$ contains a neighborhood of $1 \in I_K$.
We'll show that $\nu$ contains a product $\prod_v{T_v}$ where $T_v=\mathbb{R}^{+*}$ for complex or real places, $T_v=\mathcal{O}_v^{\times}$ for all but finitely many $v$, and for all finite $v$, $T_v$ is an open subgroup of $\mathcal{O}^{\times}_v$.
We consider instead the subgroup $N_{L/K}(I'_K) \subset \nu$ (where $I'_K \subset I_L$ is the subgroup of elements where all the coordinates are in the appropriate completion of $K$ inside the given completion of $L$) : it contains indeed the product of the $\mathbb{R}^{+*}$ (at the real/complex places), and the $\mathcal{O}_v^{\times,[L:K]}$ at the finite places.
Let $v$ be a finite place of $K$ for which every prime dividing $[L:K]$ is invertible. Then it follows from Hensel's lemma that $\mathcal{O}_v^{\times,[L:K]}=\mathcal{O}_v^{\times}$.
For the other finite places (there are finitely many of them), we show that $\mathcal{O}_v^{\times,[L:K]}$ contains a $1+\mathfrak{m}_v^k\mathcal{O}_v$ for some $k \geq 1$ and is thus an open subgroup of $\mathcal{O}_v^{\times}$. Let $p$ be the prime above which $v$ is, with $p|[L:K]$ with valuation $\alpha$ and $e$ be the ramification index of $v$ above $\mathbb{Q}$. Let $x=1+y$ with $v(y) \geq e+1+e\alpha$.
Let $z=\sum_{k=1}^{\infty}{\frac{(-1)^{k-1}y^k}{k}}$, then $v(z) \geq e+1+e\alpha$ by ultrametric inequality, and we can define $z'=\frac{z}{[L:K]} \in \mathfrak{m}_v^{e+1}\mathcal{O}_v$ and $x'=\sum_{k=0}^{\infty}{\frac{z'^k}{k!}}$, and it's easy to check that $x'^{[L:K]}=x$, and this concludes.