## 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.