In an integral domain, the ring of fractions with respect to any set of denominators is a subring of the field of fractions

Let R be an integral domain and let D \subseteq R be a set of denominators. Prove that D^{-1}R is isomorphic to a subring of the field of fractions of R. (In particular, D^{-1}R is an integral domain.)


We begin with a lemma.

Lemma: Let R be a commutative ring. Let D_1, D_2 be sets of denominators in R with D_1 \subseteq D_2. Then the map \varphi : D_1^{-1}R \rightarrow D_2^{-1}R given by \varphi(\frac{r}{d}) = \frac{r}{d} is an injective ring homomorphism. Proof: (Well-defined) Suppose \frac{r_1}{d_1} = \frac{r_2}{d_2} in D_1^{-1}R. Then r_1d_2 = r_2d_1 in R, and thus \frac{r_1}{d_1} = \frac{r_2}{d_2} in D_2^{-1}R. Thus \varphi is well-defined. (Preserves sums) Note that \varphi(\frac{r_1}{d_1} + \frac{r_2}{d_2}) = \frac{r_1}{d_1} + \frac{r_2}{d_2} = \varphi(\frac{r_1}{d_1}) + \varphi(\frac{r_2}{d_2}). (Preserves products) Similar to the sum-preservation proof. (Injective) Suppose \varphi(\frac{r_1}{d_1}) = \varphi(\frac{r_2}{d_2}). Then r_1d_2 = r_2d_1, so that \frac{r_1}{d_1} = \frac{r_2}{d_2} in D_1^{-1}R. \square

The present exercise follows because every set of denominators is contained in R \setminus 0. Because the field of fractions of R is a field, any subring D^{-1}R is an integral domain.

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