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](https://math.stackexchange.com/questions/2923849/), two weeks ago. 



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


