Primes inert in quadratic field of class number one Let $K = \mathbb{Q}(\sqrt{-d})$ be an imaginary quadratic field of class number one (i.e. every ideal in $\mathcal{O}_K$ is principal, i.e. $\mathcal{O}_K$ is a principal ideal domain). Let $d_K$ be the discriminant of $K$. > How does one prove that all primes less than $\frac{1 + |d_K|}{4}$ are inert in $K$?

Let $p$ a prime. If ${\frak P}$ is a prime ideal of $O_K$ above $p$, then the assumption on the class number implies that ${\frak P}=\langle \alpha\rangle$ for some $\alpha\in O_K$. Let $\alpha=a+b\theta$, where $\theta$ is either $\sqrt{-d}$ or $(1+\sqrt{-d})/2$ depending. Show that if $\alpha$ is not a rational integer, then $N(\alpha)\ge(1+|d_K|)/4$.

If $p$ splits, then we must have $N(\alpha)=p$. This is clearly impossible for a rational integer $\alpha$. So the above result shows that $p\ge(1+|d_K|)/4$.