Suppose $R$ is a ring and $P$ a prime ideal. If $S$ is a mutliplicative subset, can anyone explain why we have the equality $R_P=(S^{-1}R)_{P(S^{-1}R)}$ when seen as subsets of the quotient field of $R$?
I know $R_P=\{r/s:r\in R, s\in R-P\}$ and $(S^{-1}R)_{P(S^{-1}R)}=\{(r/s)/(q/t):(r/s)\in S^{-1}R,(q/t)\in S^{-1}R-P(S^{-1}R)\}$.
I attempted to do: I think if $r/s\in R_P$, then we could write $(r/s)=(r/s)/(s/s)\in (S^{-1}R)_{P(S^{-1}R)}$ since if $s\notin P$, then $s/s\notin P(S^{-1}R)$.
On the other hand, is $(r/s)/(q/t)\in S^{-1}R_{P(S^{-1}R)}$, by assumption, $s\notin P$ and $q/t\notin P(S^{-1}R)$, so $q\notin P$. This equals $rt/sq$, and since $s\notin P$ and $q\notin P$, $sq\notin P$, so $rt/sq\in R_P$.
Is this the right approach?