# Does there always exist a non-rational algebraic integer in a number field whose discriminant divides its norm?

Let $$K$$ be a number field of degree $$n$$ over the rationals. Under what conditions does there exist a non-rational algebraic integer $$\alpha$$ in $$K$$ such that the discriminant of $$\alpha$$ divides the norm of $$\alpha$$?

This question was first asked on Math StackExchange, Question 2923849, two weeks ago.

Consider the cyclotomic field $$K=\mathbb{Q}(\zeta_5)$$.
Proposition. There is no $$\alpha \in \mathcal{O}_K$$ of degree 4 such that $$D(\alpha)$$ divides $$N(\alpha)$$.
Proof. Assume such an $$\alpha$$ exists. Since the discriminant of $$K$$ is $$\Delta_K=5^3$$, we must have $$5^3|D(\alpha)$$ and thus $$5^3|N(\alpha)$$. Let $$\pi$$ be the unique prime ideal above $$5$$ in $$\mathcal{O}_K$$, so that $$5\mathcal{O}_K=\pi^4$$. Then $$\alpha \in \pi^3$$. But we have surjective maps $$\begin{equation*} \frac{\mathcal{O}_K}{\mathbb{Z}[\alpha]} \to \frac{\mathcal{O}_K}{\mathbb{Z}+\alpha\mathcal{O}_K} \to \frac{\mathcal{O}_K}{\mathbb{Z}+\pi^3}. \end{equation*}$$ The last group has cardinality $$5^2$$ since the image of $$\mathbb{Z}$$ in $$\mathcal{O}_K/\pi^3$$ is isomorphic to $$\mathbb{Z}/5\mathbb{Z}$$. We deduce that the index of $$\mathbb{Z}[\alpha]$$ in $$\mathcal{O}_K$$ is divisible by $$5^2$$. It follows that $$5^4 \Delta_K | D(\alpha)$$ and thus $$5^7 | N(\alpha)$$. Repeating the process, we get $$N(\alpha)=0$$, a contradiction.
I guess that if you take a quartic field $$K$$ with no intermediate subfield and in which some prime $$p$$ is sufficiently ramified, then the same reasoning will give you an example of number field for which the answer to your question is negative.