Let $p\in\mathbb{Z}$ be a prime number, $\mathfrak{p}\subset \mathbb{Z}$ be the prime ideal it generates and let $\mathbb{Z}_{\mathfrak{p}}$ be the localization of $\mathbb{Z}$ at $\mathfrak{p}$, i.e. the fractions whose denominators don't lie in $\mathfrak{p}$.
Then the localization $\mathfrak{p}_{\mathfrak{p}}\subset \mathbb{Z}_{\mathfrak{p}}$ of the ideal $\mathfrak{p}$ is the unique maximal ideal of $\mathbb{Z}_{\mathfrak{p}}$.
Which field is $\mathbb{Z}_{\mathfrak{p}}/\mathfrak{p}_{\mathfrak{p}}$? Is it $\mathbb{F}_p$? Is there an easy way to see this?
Thank you!
Best Answer
In general, if $R$ is a ring (by which I mean commutative with identity), $\mathfrak{p}$ a prime of $R$, and $k(\mathfrak{p})=R_\mathfrak{p}/\mathfrak{p}R_\mathfrak{p}$ the residue field of $\mathfrak{p}$, then the natural ring map $R/\mathfrak{p}\rightarrow k(\mathfrak{p})$ identifies the target with the field of fractions of the domain $R/\mathfrak{p}$. So, in particular, if $\mathfrak{p}$ is maximal, i.e. $R/\mathfrak{p}$ is already a field, this is an isomorphism.