Is $\mathbb Z$ the only proper sub-domain ( a subring that is an integral domain ) with unity of the ring $\mathbb Q$? Is $\mathbb Z$ the only proper sub-domain ( a subring that is an integral domain ) with unity of the ring $\mathbb Q$ ? ( I can easily prove that if $D$ is any subring with unity then $\mathbb Z \subseteq D$ but cannot seem to apply integral domain condition to conclude the reverse ) 
 A: Any subring of a field is a domain, so it's not surprising you can't apply the condition.
If $S$ is a multiplicatively closed subset of $\mathbb{Z}$ (with $0\notin S$), then
$$
S^{-1}\mathbb{Z}=\left\{\frac{a}{b}\;\middle|\; a\in\mathbb{Z}, b\in S\right\}
$$
is a subring of $\mathbb{Q}$ containing $\mathbb{Z}$.
A non trivial example of $S$ is $S=\mathbb{Z}\setminus p\mathbb{Z}$, with $p$ a prime, so $S^{-1}\mathbb{Z}$ is (isomorphic to) the localization of $\mathbb{Z}$ at the prime ideal $p\mathbb{Z}$.
Other examples are $S=\{c^n\mid n\ge0\}$, where $c\ne0$.
A: No, it's false: any ring of fractions $S^{-1}\mathbf Z$ of $\mathbf Z$ is strictly between  $\mathbf Z$ and  $\mathbf Q$ as soon as $S$ is different from $\{1-,1\}$ and $S$ is a saturated multiplicative part different from $\mathbf Z\setminus{0}$. Most notable examples are  $\mathbf Z[\frac1n]$ (fractions with denominator a power of $n$) and, for a prime $p$,  $\mathbf Z_{(p)}$, the set of fractions with denominator not divisible by $p$. These are local rings, i.e. they have only one maximal ideal.
