Let $R$ be a commutative ring, and $P,Q \subset R$ prime ideals such that $ P \subseteq Q$. Let $R_{P}$ denote the localization of $R$ at $P$. Prove that $R_{1} = R_{P}$ is isomorphic to $R_{Q}$ localized at $PR_{Q}$, denote this ring as $R_{2}$. I tried to give explicit definitions for $R_{1}$ and $R_{2}$. So $R_{1} = \left\{ \frac{r}{p} : r \in R, p \in P \right\}$, and $R_{2} = \left\{ \frac{\frac{r_{1}}{q_{1}}}{p \frac{r_{2}}{q_{2}}} : r_{1} , r_{2} \in R , q_{1}, q_{2} \in Q , p \in P\right\}$. I am assuming that $\frac{\frac{r_{1}}{q_{1}}}{p \frac{r_{2}}{q_{2}}} \sim \frac{r}{p}$ for some $r \in R$ to get the desired isomorphism, but I am stuck.
In a previous exercise, it was shown that $\phi' : D^{-1}R \rightarrow F^{-1}S$ defined by $\phi'(r/d) = \phi(r) / \phi(d)$ is a ring homomorphism for $\phi : R \rightarrow S$ a ring homomorphism, and $D = \phi^{-1}(F)$ where $F \subset S$ is a multiplicatively closed set.
2026-05-06 09:46:35.1778060795