Fraction field of $R/P$

Question

It may be a simple question seeming too easy, but I seek a help:

If $P$ is a prime ideal of a commutative ring $R$, could one say that $R_P/PR_P$ is the field of fractions of $R/P$?

Thanks a lot!

Answer 1

Let $S = R \setminus P$. We know that if $M$ is an R-module, then $$S^{-1}M \cong S^{-1}R \otimes_{R}M$$ So consider the exact sequence $$0 \rightarrow P \rightarrow R \rightarrow R/P \rightarrow 0$$ $S^{-1}R$ is a flat $R$-module, so applying the exact functor $S^{-1}R \otimes -$ we obtain the exact sequence $$0 \rightarrow PR_{P} \rightarrow R_{P} \rightarrow (R/P)_{P} \rightarrow 0$$ Then $$R_{P} / PR_{P} \cong (R/P)_{P}$$ and so $R_{P} / PR_{P}$ is the quotient field of $(R/P)_{P}$