Show if $\mathfrak p$ is a minimal prime ideal of $R$ then $\mathfrak pR_\mathfrak p$ is the only prime ideal of $R_\mathfrak p$

Question

Show if $\mathfrak p$ is a minimal prime ideal of $R$ then $\mathfrak pR_\mathfrak p$ is the only prime ideal of $R_\mathfrak p$.

Here are what I know and don't need to prove: I know $\mathfrak pR_\mathfrak p$ is the maximal ideal of $R_\mathfrak p$. I know $R_\mathfrak p$ is a local ring and so $R_\mathfrak p$ is the unique maximal ideal. And I know the set of maximal ideals is the set of non-units. Also, I know for $\mathfrak p$ is a prime ideal of $R$, $\mathfrak pR_\mathfrak p$ also is a prime ideal of $R_\mathfrak p$. How can I connect these ideas? Thank you.

Answer 1

The prime ideals in a ring of fractions $S^{-1}R$ correspond bijectively to the prime ideals $\mathfrak q$ of $R$ such that $\mathfrak q\cap S=\varnothing$. In the present case, this means $\mathfrak q \subset \mathfrak p$. By the minimality of $\mathfrak p$, this implies $\mathfrak q=\mathfrak p$.