# Show that $R\cong R_P$, the ring of quotients of $R$ with respect to the multiplicative set $R-P$ if $R$ has exactly one prime ideal $P$.

abstract-algebrafield-theorymaximal-and-prime-idealsring-theory

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.

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.