Let $K/F$ be extensions of $p$-adic fields. We define $e(K)$ to be the index $[v_K(K^\times):v_p(\mathbb{Q}_p^\times)]$, and similarly for $e(K)$. We then define $e=e(K/F)=[v_K(K^\times):v_F(F^\times)]=e(K)/e(F)$ and call it the ramification index of $K/F$. Now let $\mathfrak{p}$ be the unique maximal ideal of $\mathcal{O}_F$, and let $\mathfrak{P}$ be that of $\mathcal{O}_K$. I am trying to understand the proof that $\mathfrak{p}\mathcal{O}_K=\mathfrak{P}^e$.
So we pick the normalized valuation $v$ of $F$ and consider its extension $w$ to $K$. Let $\pi\in F$ be a uniformizer, and let $\Pi\in K$ be such that $w(\Pi)=1/e$. The proof in my book then says that we have $\mathfrak{p}=\pi\mathcal{O}_F$, and $\mathfrak{P}=\Pi\mathcal{O}_K$.
Why does $\mathfrak{P}=\Pi\mathcal{O}_K$? Further, how do we know that we can find such an element $\Pi$ so that $w(\Pi)=1/e$, since we only have $w(K^\times)=(1/e(K))\mathbb{Z}$ and not $(1/e)\mathbb{Z}$?