I am struggling with the proof of the Lemma 2.3 in the appendix of Rubin in the Lang's book "Cyclotomic fields I and II". The setting is the following: $F=\mathbb{Q}(\mu_m)^{+}$, where $m$ is a positive integer, $O_F$ its ring of integers. For a prime $l$, $\mathscr{I_l}$ denotes the $l$ part of the group of fractional ideals of $F$, that is, the free abelian group generated by the ideals of $F$ above $l$. $M$ is a positive odd integer. The lemma is the following:
If $l$ splits completely in $F$ and $l\equiv 1\pmod M$, then there is a unique $G=\text{Gal}(F/\mathbb{Q})$-equivariant surjection $$\varphi_l\colon(O_F/lO_F)^{\times}\to \mathscr{I_l}/M \mathscr{I_l}$$ such that, for every $x\in F(\mu_l)^{\times}$, $$\varphi_l((1-\sigma_l))x=[N_l(x)]_l,$$ where $\sigma_l$ is the generator of the cyclic group $G_l=\text{Gal}(\mathbb{Q}(\mu_l)/\mathbb{Q})=\text{Gal}(F(\mu_l)/F)$, $(1-\sigma_l)x=x/\sigma_l(x)$ (additive notation), $N_l$ is the norm from $F(\mu_l)$ to $F$ and $[y]_l$ denotes the projection of the ideal generated by $y$ to $\mathscr{I_l}/M \mathscr{I_l}$.
The proof is the following: $[F(\mu_l):F]=l-1$, and since $l$ splits completely in $F$, every prime $\lambda$ of $F$ above $l$ is totally ramified and tamely ramified in $F(\mu_l)$. Rubin says that due to this, the maps $1-\sigma_l$ and $[N_l(\cdot)]_l$ are surjective, but I don't know how to see this fact (in this context, does tamely ramified means that the inertia of a prime above $\lambda$ is not divisible by $l$? This is clear, since the inertia is the whole Galois group, so it has order $l-1$). Then he says that the kernel of $1-\sigma_l$ are the elements of $F(\mu_l)^{\times}$ which order at prime above $l$ is divisible by $l-1$, and this is contained in the kernel of $[N_l(\cdot)]_l$, so we can conclude, but I am stuck here too...
Edit: I think that the map $x\to x/\sigma_l(x)$ has to be seen in this way: since a prime over $l$ is totally ramified in $F(\mu_l)$, the decomposition group is all, every isomorphism fixes the valuation, so $x/\sigma_l(x)$ has valuation $0$ in the completion of $F(\mu_l)$ with respect to a prime $\lambda'$ above $l$. Hence $x/\sigma_l(x)$ is a unit in the valuation ring $O_{F(\mu_l),\lambda'}$, we can quotient with the maximal ideal $P$, and find an element in $$(O_{F(\mu_l),\lambda'}/P)^\times\cong (O_{F(\mu_l)}/\lambda')^{\times}\cong (O_{F}/\lambda)^\times,$$ for every prime $\lambda$ of $F$ above $l$...