What is the localization of $\mathbb{Z}_{(p)}$ at $p\mathbb{Z}_{(p)}$?

Question

Let $p$ be prime so that $\mathbb{Z}_{(p)}$ is the local ring with unique maximal ideal $\mathfrak{m}=p\mathbb{Z}_{(p)}$. What is the localization $(\mathbb{Z}_{(p)})_\mathfrak{m}$? Intuitively I would guess it would be a field, and in particular $\mathbb{F}_p$ but how can I show this?

Answer 1

Suppose $R$ is a local ring with maximal ideal $m$. Then $R_m=R$. Why? Localising at $m$ introduces an inverse for any element $x\not\in m $. But in a local ring, we have $R^{\times} = R \,\backslash\, m$, so any element not in $m$ already has an inverse.

In terms of universal properties, the identity map $id : R \rightarrow R$ maps every element of $S = R \,\backslash\, m$ to a unit in $R$, so if $i : R \rightarrow R_m$ is the canonical inclusion, there exists a unique $j : R_m \rightarrow R$ such that $id = j \circ i$. Hence $i$ is an isomorphism.

However, it is true that $\mathbb{Z}_{(p)}/p\mathbb{Z}_{(p)} = \mathbb{F}_p.$