Question: Let $R$ be a commutative ring with identity that has exact one prime ideal $P$. Show that $R\cong R_P$, the ring of quotients of $R$ with respect to the multiplicative set $R-P$
This is an old question, and the first part of the question was to show that $R/P$ is a field. So, we consider maximal ideal $M$ of $R$ such that $P\subseteq M\subsetneq R$. Thus, $M$ is prime and so $M=P$ by uniqueness of $P$. Thus $P$ is maximal and so $R/P$ is a field. Now, for the question, I'm not sure if it is relevant, but whenever I see "multiplicative set", I think of the theorem:
Let $R$ be a nonzero commutative ring with identity and $S$ a multiplicative subset of $R$ not containing $0$. If $P$ is maximal in the set of ideals of $R$ not intersecting $S$, then $P$ is prime.
But, I don't know if this will help. To show the isomorphism, I imagine I need to come up with a map, but I can't quite see one that would work. Any help is greatly appreciated! Thank you.
Best Answer
Since maximal ideals are prime, there is also at most one maximal ideal. And since there exists a maximal ideal (unless $R=0$, in which case there would be no prime ideal), there is exactly one maximal ideal, which is at the same time the unique prime ideal, making $R$ a local ring.
And it is known that the localisation of a local ring at its maximal ideal is just the local ring.