Good day all, I am trying to show that, for $d \neq 0,1$ a square-free integer $(d=-1$ is allowed$)$ the space $A$ defined as below, is an integral domain: \begin{equation*} A = \{a + b\sqrt{d} : a,b \in \mathbb{Q} \} \cap \overline{\mathbb{Z}} \end{equation*}
where $\overline{\mathbb{Z}}$ is the set of all roots of monic polynomials with coefficients in $\mathbb{Z}$.
I can prove the majority of properties associated with an integral domain, except for closure under addition and non-existence of zero divisors.
Let $x_1 = a_1 + b_1\sqrt{d_1}$ and $x_2=a_2 + b_2\sqrt{d_2}$ be elements of $A$. Then: \begin{eqnarray*} x_1 + x_2 & = & a_1 + b_1\sqrt{d_1} + a_2 + b_2\sqrt{d_2} \\ & = & a_1 + a_2 + b_1\sqrt{\frac{d_1d_2}{d_2}}+b_2\sqrt{\frac{d_1d_2}{d_1}} \\ & = & a_1 + a_2 + \left(\frac{b_1}{\sqrt{d_2}} + \frac{b_2}{\sqrt{d_2}}\right)\sqrt{d_1d_2} \end{eqnarray*}
But $d_1d_2$ is not necessarily squarefree and $\left(\frac{b_1}{\sqrt{d_2}} + \frac{b_2}{\sqrt{d_2}}\right)$ is definitely not always in $\mathbb{Q}$. How do I remedy this!?
Lastly, would it be alright to state that the non-existence of zero divisors follows because $A$ is a subset of $\mathbb{C}$?