A question on a proof in the Ralph Greenberg's paper "On a Certain l-Adic Representation"

Question

I'm currently reading the paper "On a Certain l-Adic Reprersentation" written by Ralph Greenberg.(Inventiones 1973) And I'm stuck with a proof of the Proposition 2.

Here $k$ is a totally imaginary abelian extension of $\mathbb{Q}$, $K/k$ is the cyclotomic $\mathbb{Z}_{l}$-extension. $A_{n}$ is the $l$-primary subgroup of the ideal class group of $k_{n}$.

First of all, I don't know why there exists a "root of unity" $w$ in $k_{n}$ such that $N_{n,0}(\beta)=N_{n,0}(w)$.

Instead, Using the condition $c \in A_{n}^{-}$, and as the complex conjugation $J$ on $k_{n}$ acts on $\sigma$ (a generator of the galois group $G(k_{n}/k)$) by $J \sigma J = {\sigma}^{-1}$. We can see that there exists a "unit" $u$ and an element $\gamma$ such that $\beta = u \gamma^{\sigma -1}$.

Secondly, I don't understand the line "If $b=a^{1-J}$, then $b^{\sigma - 1}=(\gamma)^{\sigma -1}$."

Here is my computation (that might be wrong.)

$b=a \overline{a}^{-1}$

As $c \in A_{n}^{-}$, there exists $\delta \in k_{n}$ such that $a \overline{a}=(\delta)$.

$\sigma(a) \sigma(\overline{a}) = (\sigma(\delta))$.

$b^{\sigma -1} = \sigma(a) \overline{a} \sigma(\overline{a})^{-1} a^{-1} = (\sigma(a) a^{-1})^{2} (\delta^{1-\sigma})=(\alpha)^{2}(\delta^{\sigma-1})$.

For the proof of proposition 2 to be true, we need $\alpha^{2}$ must be a unit times an element of the form $\phi^{\sigma-1}$. But I don't see why.

Answer 1

First of all, I realized that I had made a big mistake. The Galois extension $k_{n}/k^{+}$ is abelian (since it is a composite of $k/k^{+}$ and $\mathbb{B}_{m}/\mathbb{Q}$ for some m.) hence the complex conjugation commutes with any element of $G(k_{n}/k)$.

And the first question on existence of roots of unity $w$ $\in k_{n}$ such that $N_{n,0}(w) = N_{n,0}(\beta)$ was right as the author claimed and was actually important.

In general, for a number field $k$ and the nth-layer field $k_{n}$ of the cyclotomic $\mathbb{Z}_{l}$-extension, for any root of unity $\alpha \in k$, there exists a root of unity $w \in k_{n}$ such that $N_{n,0}(w)=\alpha$.

If the multipative order of $\alpha$ is prime to $l$, then as $\alpha$ is invariant under Galois action, the existence of $w$ is trivial. Hence we can assume that the multiplicative order of $w$ is a power of $l$.

If the order is strictly larger than 1, then $k$ contains $\mathbb{Q}(\mu _{l})$. By the theory of $\mathbb{Z}_{l}$-extension, the intersection of $k$ and "the cylclotomic $\mathbb{Z}_{l}$ extension of $\mathbb{Q}$" is $\mathbb{Q}(\mu _{l^p})$ for some $p$, which means $k_{n}$ is the extension field of $k$ generated by adjoining $l^{p+n}$th roots of unity.

Then we're done.