Let $K \supset \mathbb{Q}_p$ be the $p$-adic field. Let $O$ be its ring of integers. Let us denote by $O_{(p)}$ be the localisation of $O$ at the prime/maximal ideal $(p)$. Consider the another ring $O\left[\frac{1}{p} \right]$ obtained by inverting the prime $p$.
My question:
What is the difference between the rings $O_{(p)}$ and $O\left[\frac{1}{p} \right]$ ?
Edit: If $O=\mathbb{Z}_p$, then $O\left[\frac{1}{p} \right]=\mathbb{Q}_p$. But $O_{(p)}=(O \setminus \{(p)\})^{-1} O \subset \mathbb{Q}_p$.
For $K/\Bbb{Q}_p$ a finite extension and $O$ its ring of integers then $(p)$ doesn't have to be a prime ideal, for example with $K=\Bbb{Q}_p(p^{1/2}),O=\Bbb{Z}_p[p^{-1/2}]$, the unique maximal ideal is $(p^{1/2})$. This is what we mean with "an uniformizer $\pi_K$": an element generating the maximal ideal.
We always have $(O-(\pi_K))^{-1}O = O$ and $O[p^{-1}]=K$.