Question 1.218: Let D be a multiplicative submonoid of R. Prove that there e......

Let \phi denote the map sending a prime ideal I of R to the ideal generated by the image of I in D^{-1} R. It is clear that, if I \cap D \neq \emptyset, then \phi(I)=D^{-1} R. Therefore, we restrict our attention to the prime ideals that do not intersect D.

Let r \in \phi(J) for some other prime ideal J from R. Thus, r is of the form r_{1} s_{1} / d_{1}+ \cdots r_{k} s_{1} / d_{k} for some r_{1}, \ldots, r_{k} \in J, s_{1}, \ldots, s_{k} \in R and d_{1}, \ldots, d_{k} \in D. Since s_{1} r_{1} / d_{1}+ \cdots s_{k} r_{k} / d_{k}=d_{1}^{\prime} s_{1} r_{1} / d+\cdots+d_{k}^{\prime} s_{k} r_{k} / d, where d=\prod d_{i}, we see that r is of the form r^{\prime} / d^{\prime}, where r^{\prime} \in J and d^{\prime} \in D. Thus, we identify the ideal generated by J in D^{-1} R with the set \left\{r^{\prime} / d^{\prime}: r^{\prime} \in J, d^{\prime} \in D\right\}.

Now, if \phi(J)=\phi(I), then r \in \phi(I), hence there exists r^{\prime \prime} \in I, d^{\prime \prime} \in D such that r^{\prime} / d^{\prime}=r^{\prime \prime} / d^{\prime \prime}, which implies that r^{\prime} \in I. In other words, J \subset I. The same argument shows that I \subset J, hence I=J. Therefore, \phi is injective.

Conversely, let U \subset D^{-1} R be a prime ideal and let I=U \cap D^{-1} R. We claim that I is prime, I \cap D=\emptyset, and furthermore, \phi(I)=U. If r, r^{\prime} \in R are two elements from R such that r r^{\prime} \in I, then in particular r r^{\prime} \in D^{-1} R. Thus, either r \in D^{-1} R or r^{\prime} \in D^{-1} R. But r and r^{\prime} are from R, therefore, at least one of them is contained in I. In other words, I is prime. It is clear that I \cap D=\emptyset, otherwise U contains a unit. Finally, r / d \in U, then d \cdot r / d=r \in U. Since r is an element of R, we see that r \in I. It follows that I generates U in D^{-1} R.