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

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.

• You can read about primes in rings after localisation in any textbook about commutative algebra. The assertion is tautological with that basic knowledge. – MooS Feb 16 '15 at 8:59

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 $$\mathfrak q=\mathfrak p$$.
• thank you for answering, but isn't $q$ is a sub-ring of $R$, how come $q∩R=∅$? – user138017 Feb 16 '15 at 9:42
• No, a proper ideal isn't a subring: a subring (in commutative algebra) must contain $1$. – Bernard Feb 16 '15 at 9:45
• Sorry for the late comment, but the OP is right to question $\mathfrak{q} \cap R = \varnothing$ (which is nonsensical since $\mathfrak{q} \subseteq R$): it should be $\mathfrak{q} \cap S = \varnothing$. – Viktor Vaughn Dec 29 '16 at 5:10