Rings of fractions over a PID are PIDs

Prove that if R is a principal ideal domain and D \subseteq R a set of denominators, then the set D^{-1}R of fractions with denominators in D is a principal ideal domain.


D^{-1}R is a subring of a field (the field of fractions of R), and thus is an integral domain.

Let I \subseteq D^{-1}R be an ideal. Let S = \{a \in R \ |\ \frac{a}{b} \in I \}. We claim that S is an ideal of R. To see this, note that \frac{0}{d} \in I, so that 0 \in S. Now let \frac{a}{d}, \frac{b}{e} \in I; then \frac{a}{d} - \frac{e}{d}\frac{b}{e} = \frac{a-b}{d}, so that a-b \in S. Finally, let r \in R and a \in S, with d \in D. Then \frac{r}{d} \frac{a}{d} = \frac{ra}{d^2}, so that ra \in S. Thus S is an ideal of R, and thus we have S = (\alpha). Let d \in D such that \frac{\alpha}{d} \in I; then (\frac{\alpha}{d}) \subseteq I. Now let \frac{a}{b} \in I with a = c\alpha; we have \frac{a}{b} = \frac{cd}{b} \frac{\alpha}{d}, so that I = (\frac{\alpha}{d}) is principal.

Thus D^{-1}R is a principal ideal domain.

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