A finite field extension $\mathbb{F}/\mathbb{Q}$ is called Norm-Euclidean if the absolute norm function $\alpha \mapsto |\prod \sigma_i(\alpha)|$ induce a Euclidean norm on the integer ring $\mathcal{O}_\mathbb{F}$. The simplest example is of course $\mathbb{Z}$, but there are also $\mathbb{Z}[i],\mathbb{Z}[\sqrt{2}i]$ and $\mathbb{Z}[\omega]$ where $\omega$ is a primitive third root of unity.
I know that in the quadratic case there is a full finite list of the quadratic field extensions which are Norm-Euclidean. Is it true that for any degree there are only finitely many Norm-Euclidean fields (even considering only totally real extensions)?
It is well known that for any fixed $M>0$, there are only finitely many fields (of a fixed degree) that have discriminant $<M$, and it seems that if the discriminant is big enough, then the field should not be Norm-Euclidean. In a sense the other direction is true - if we believe Minkowski's conjecture for the Euclidean minima, then whenever the discriminant of an extension of degree $d$ is at most $2^{d/2}$, then that extension is Norm-Euclidean.
