Proposition 2

If $\mathbf{A}$ is an integral domain, the total ring of fractions $\mathbf{K}$ of $\mathbf{A}$ is a commutative ring.

The ring $\mathbf{K}$ is commutative. It is non-zero since $\mathbf{A} \neq \{0\}$. As $\mathbf{A}$ is an integral domain, every non-zero element of $\mathbf{A}$ is cancellable and $\mathbf{K}$ consists of the fractions $a/b$ with $b \neq 0$. Now $a/b \neq 0$ implies $a \neq 0$ and the fraction $b/a$ is then inverse of $a/b$.

The total ring of fractions of an integral domain is called its field of fractions. Such a ring is identified with its image in its field of fractions.