# Show that a space $A$ is an integral domain.

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 $\mathbb{C}$?

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I have the impression that $d$ is fixed. As you are not working in e.g. a quotient of a space embedded in $\Bbb C$, it is indeed clear that $A$ has no nontrivial zero divisors. –  Lord_Farin Oct 13 '12 at 7:28
Thanks Andre and Farin for catching my mistake. It is clear now. –  Jack Rousseau Oct 13 '12 at 7:53
1) The parameter $d$ is fixed, else we do not have closure under addition.
2) So there is no serious issue about sum, product being of the shape $x+y\sqrt{d}$. However, the $\overline{\mathbb{Z}}$ part needs to be dealt with.