I'm trying to understand the proof of the existence of tamely ramified extensions. For this, the theorem from my book says:
Let $K$ be a complete field with respect to a discrete valuation, and let $\Omega/K$ be a totally ramified extension of degree $n=e(\Omega/K)$. Let the characteristic of the residue field $k$ be $p>0$ and suppose that $n=n_{0}p^{l}$ with $(n_{0},p)=1$. Then there exists a unique extension $V$ of $K$ with $K\subset V\subset \Omega$ such that $[V:K]=n_{0}$. Moreover, $V=K(\sqrt[n_{0}]\pi)$ where $\pi$ is an element of $K$ such that $\mathfrak{p}_{K}=\pi\mathcal{O}_{K}$
I understood the proof except by one fact which might be obvious, but I can't see why it happens. I'll write how the proof begins:
Since $\Omega/K$ is totally ramified, $\omega=k$ (the residue field o $\Omega$, ehich is $\omega$ coincides with the residue field of $K$.), and if $G_{K}=\langle |\pi|\rangle, G_{\Omega}=\langle |\Pi|\rangle$ then $\Omega=K(\Pi)$ and $$\mathcal{O}_{\Omega}=\mathcal{O}_{K}+\mathcal{O}_{K}\Pi+\cdots+\mathcal{O}_{K}\Pi^{e-1},$$ with $e=n=n_{0}p^{l}$. (All this facts were viewed in previous theorems)
It follows that $\Pi^{n_{0}p^{l}}=\pi U$ with $U\in\mathcal{O}_{\Omega}$ satisfying $|U|=1$ (This fact were proved in a previous theorem)
Now, here it comes the part which I don't understand:
Since $\omega=k$ we may write $U=uZ$ where $u\in K$ satisfies $|u|=1$ and $Z\in\mathcal{O}_{\Omega}$ satisfies $|Z-1|<1$.
In some try to understand this I have the following:
think in $\overline{U}$ in $k$, then $\overline{U}=\overline{u}$ with $u\in\mathcal{O}_{K}$, also $\overline{u}=\overline{u}\overline{1}$, now we view this equality in $\omega$ and if we lift both sides we have the desired result.
Are correct my last argument? Any hint for obtain the result? I think that is the easier fact on the proof, but is the only one which I couldn't understand
As a remark, $G_{K}$ is the value group of $|-|$.
Thanks
Yes, you're right. Since $K$ and $\Omega$ has the same residual field, hence thinking $\bar{U}$ in $k$, we have $\bar{U} = \bar{u}$ for some $u\in \mathcal{O}_K$. Let $Z = U/u \in \mathcal{O}_\Omega$, then $\bar{Z} = \bar{1}$, hence $|Z-1|<1$.