Localization at prime ideals in von Neumann regular rings Let $R$ be a commutative von Neumann regular ring, i.e. for all $r$ in $R$, there exists an $s$ in $R$ such that $r = r^{2}s$.  For every prime ideal $\mathfrak{p}$, $R/\mathfrak{p} \cong R_{\mathfrak{p}}$.  What is the map that induces this isomorphism?
 A: Let $\lambda:R\to R_\mathfrak{p}$ be the localization map.

Claim: $\lambda$ is surjective.
Proof: It suffices to show that $\lambda(t)^{-1}\in\operatorname{im}(\lambda)$ for any $t\in R\setminus\mathfrak{p}$. Indeed, every element of $R_\mathfrak{p}$ is of the form $\lambda(r)\lambda(t)^{-1}$ for some $r\in R$ and $t\in R\setminus\mathfrak{p}$, and if $\lambda(s)=\lambda(t)^{-1}$ then $\lambda(r)\lambda(t)^{-1}=\lambda(rs)\in\operatorname{im}\lambda$.
So fix $t\in R\setminus\mathfrak{p}$. By hypothesis, there exists $s\in R$ with $t=t^2s$, ie with $(1-ts)t=0$. Since $t\notin\mathfrak{p}$, this means $1-ts\in\operatorname{ker}(\lambda)$, so that $1=\lambda(1)=\lambda(ts)=\lambda(t)\lambda(s)$, whence $\lambda(s)=\lambda(t)^{-1}$, as desired. $\blacksquare$

Claim: $\operatorname{ker}\lambda=\mathfrak{p}$.
Proof: For the inclusion $\supseteq$, fix $p\in\mathfrak{p}$. By hypothesis, there exists $s\in R$ with $(1-ps)p=0$. Since $ps\in \mathfrak{p}$, we have $1-ps\notin\mathfrak{p}$, whence $\lambda(p)=0$, as desired.
For the inclusion $\subseteq$, suppose $\lambda(r)=0$ for some $r\in R$; then there exists $t\in R\setminus\mathfrak{p}$ with $tr=0$. Then in particular $tr\in\mathfrak{p}$; since $\mathfrak{p}$ is prime and $t\notin\mathfrak{p}$, this forces $r\in\mathfrak{p}$, as desired.

Now by Noether's first isomorphism theorem we are done.
