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.

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    $\begingroup$ You can read about primes in rings after localisation in any textbook about commutative algebra. The assertion is tautological with that basic knowledge. $\endgroup$ – 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 $\mathfrak q=\mathfrak p$.

  • $\begingroup$ thank you for answering, but isn't $q$ is a sub-ring of $R$, how come $q∩R=∅$? $\endgroup$ – user138017 Feb 16 '15 at 9:42
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    $\begingroup$ No, a proper ideal isn't a subring: a subring (in commutative algebra) must contain $1$. $\endgroup$ – Bernard Feb 16 '15 at 9:45
  • $\begingroup$ 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 $\mathfrak{q} \cap S = \varnothing$. $\endgroup$ – Viktor Vaughn Dec 29 '16 at 5:10

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