I'm currently working on the following problem, part (a) specifically. I have the forward way, if $a$ is prime in $R$ then $A$ is a prime number, by contradiction. I ran into problems going the opposite way. Contradiction doesn't seem like the right idea, and I'm getting stuck going straight forward. I was also questioning if there is a way to prove this with if and only if arrows.
For part (b), I'm pretty sure I want $\mathbb{Z}[\sqrt{5}]$.
Let $c,d \in \mathbb{Z}$, with $d$ not a square, and $R = \mathbb{Z}[\sqrt{d}].$ Let $a = c + \sqrt{d}$ and let $A$ be the absolute value of $c^{2} - d$.
(a) Show that $a$ is prime in $R$ if and only if $A$ is a prime number.
$(\Rightarrow)$ Assume $a$ is prime, then $(a)$ is a prime ideal. Assume for a contradiction that $A$ is not prime. Let $A=qp \text{ for some } q,p \in \mathbb{Z}$. Since $A = a \bar{a} \text{ we have } A \in (a)$ we note that $\bar{a} = c - \sqrt{d} \in \mathbb{Z}[\sqrt{d}]$ Since $a$ is a prime ideal and $A \in (a)$ this implies that $pq \in (a) \text{ so } p \text{ or } q \in (a)$. Without loss of generality we assume that $ p \in (a)$. Then we have $p = x (c + \sqrt{d}) \in \mathbb{Z}$ then $x=y(c - \sqrt{d})$ where $ y \in \mathbb{Z}$. Therefore $A = qp = q y (c - \sqrt{d})( c + \sqrt{d}) = q y A$. Thus $q$ and $y$ must be units, so $A \in \mathbb{Z}$ is irreducible and $\mathbb{Z}$ is a Unique Factorization Domain, so irreducible implies prime. This is a contradiction to our assumption that A was not prime. Hence A must be prime.
$(\Leftarrow)$ $?$
(b) Give an example (with proof) where $a$ is irreducible but not prime.
(c) Prove that for any such example, there must exist an ideal $I$ in $R$ that is not principal and that contains $a$.