Show that the subset $\overline{I} = \{\overline{x}:x \in I\}$ is an ideal. Assume that $I$ is an ideal of the ring $\mathcal{O}_d = \left\{
        \begin{array}{ll}
            \mathbb{Z} [\sqrt{d}] & \text{ if } d \text{ is even } \\
            \mathbb{Z} [ \frac{1 + \sqrt{d}}{2}] & \text{ if } d \text{ is odd. }
        \end{array} 
    \right.$
Show that subset $\overline{I} = \{\overline{x}:x \in I\}$ is also an ideal of this ring.
The additive subgroup properties are obvious (I think), so we just need to check that multiplication by elements in the field by elements in $\overline{I}$ remain in $\overline{I}.$ 
Note that for $\overline{a} \in \mathcal{O}_d$ and $x \in I$ one has $ \overline {a} x \in I.$ Thus $\overline{\overline{a} x} = a \overline{x} \in \overline{I}, $ demonstrating that $I$ is closed under multiplication by elements from $\mathcal{O}_d.$
 A: Fields don't have proper nontrivial ideals. (Sometimes one abuses the term and by primes of a number field one is referring to the prime ideals of the ring of integers.)
Consider the ring of integers ${\cal O}_K$ of a quadratic field $K=\Bbb Q(\sqrt{d})$. Suppose $I$ is an ideal. We can form the set $\bar{I}=\{\bar{x}:x\in I\}$ as the image of $I$ under the conjugation map $x\mapsto\bar{x}$ (i.e. the unique nontrivial symmetry coming from the Galois group). Additivity is obvious, as $\overline{x}+\overline{y}=\overline{x+y}\in I$ for any two elements $\bar{x},\bar{y}\in \bar{I}$. Now check that $\bar{I}$ is closed under ambient multiplication. Suppose we have some number $a\in{\cal O}_K$ and some element $\bar{x}\in\bar{I}$. We must show $a\bar{x}\in I$, i.e. that $a\bar{x}$ is the conjugate of some element of $I$: notice $a\bar{x}=\overline{\bar{a}x}$ and $\bar{a}x\in I$ (since $I$ is an ideal).
In general any surjective ring homomorphism sends ideals to ideals. Suppose $\phi:R\to S$ is an onto ring homomorphism and $I$ is an ideal of $R$. As $\phi(x)+\phi(y)=\phi(x+y)\in\phi(I)$ for any choice of elements $\phi(x),\phi(y)\in\phi(I)$, additivity holds. Let's check closure under ambient multiplication: suppose $\phi(x)\in\phi(I)$ and $s\in S$ is arbitrary. We must show $s\,\phi(x)\in I$ as well, i.e. that $s\,\phi(x)$ is the image under $x$ of some element of $I$. Pick any $r\in R$ for which $r\mapsto s$, so $s\,\phi(x)=\phi(rx)$; since we know $rx\in I$ (as $x\in I,r\in R$ and $I$ is an ideal of $R$) we know $\phi(rx)\in\phi(I)$.
You should check that $x\mapsto\bar{x}$ (i.e. $a+b\sqrt{d}\mapsto a-b\sqrt{d}$) is a ring automorphism if you haven't already. (We used this fact - $\overline{xy}=\bar{x}\bar{y}$, and also $\overline{\bar{x}}=x$ - in our proof, after all.)
