# About the existence of tamely ramified extensions

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 $$|Z-1|<1$$.