# Irreducible implies prime norm in quadratic integer ring

Let's denote $$\mathbb{Z}[\lambda_d]$$ where $$\lambda_d = \begin{cases} \sqrt{d} & \text{ if } d\equiv 2,3 \; (\text{mod }4),\\ \frac{1+\sqrt{d}}{2} & \text{ if } d\equiv 1 \; (\text{mod }4), \end{cases}$$ with $$d\neq 0$$ is a square-free integer.

Let $$\alpha=a+b\lambda_d\in \mathbb{Z}[\lambda_d]$$, with $$a,b\in\mathbb{Z}$$. It is well known that if $$N(\alpha)=\alpha\overline{\alpha}$$ is a prime in $$\mathbb{Z}$$, then $$\alpha$$ is irreducible in $$\mathbb{Z}[\lambda_d]$$. My question is: What restrictions can be given to $$d$$ such that the converse holds? That is for which $$d$$ if $$\alpha$$ is irreducible in $$\mathbb{Z}[\lambda_d]$$, then $$N(\alpha)$$ is a prime in $$\mathbb{Z}$$?

I believe that the answer is no for every $$d$$. For example when $$d=-1$$ the following famous exercise in Marcus' Number Fields:

Let $$\alpha\in\mathbb{Z}[i]$$. Show that if $$N\left(\alpha\right)$$ is a prime in $$\mathbb{Z}$$ then $$\alpha$$ is irreducible in $$\mathbb{Z}[i]$$. Show that the same conclusion holds if $$N\left(\alpha\right)=p^{2}$$, where $$p$$ is a prime in $$\mathbb{Z}$$, $$p\equiv 3\pmod{4}$$.

Any help will be appreciated. Thank you in advance!

Splitting of primes in quadratic extensions is controlled by the quadratic character (the Legendre symbol). More concretely an odd prime $$p\nmid d$$ splits in $$\mathbb{Z}[\lambda_d]$$ if and ony if $$d$$ is a square modulo $$p$$. You can always find a prime $$p$$ such that $$d$$ is not a square modulo $$p$$, so $$p$$ will be inert (so irreducible in $$\mathbb{Z}[\lambda_d]$$) and $$N(p)=p^2$$. In particular for $$d=-1$$ the quadratic character of $$-1$$ tells you that this happens for every prime $$\equiv 3 \pmod{4}$$. The answer to your question is then that the converse does not hold for any $$d$$.