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!


1 Answer 1


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 $d$.


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