# Showing $\mathbb{Z}+\mathbb{Z}\left[\frac{1+\sqrt m}{2}\right]$ is a Euclidean domain

Does anyone know an elementary proof for the following proposition? It is stated without proof in my textbook:

Let $m$ be a negative squarefree integer with $m \equiv 1 \pmod 4$. Then the integral domain $\mathbb{Z} + \mathbb{Z}[\frac{1 + \sqrt m}{2}]$ is Euclidean with respect to $\phi_m$ if and only if $m = -3, -7, -11$. Here, $\phi_m:\mathbb{Q}(\sqrt m) \to \mathbb{Q}$ is defined by $\phi_m(r + s\sqrt m) = |r^2 - ms^2|$ for all $r, s \in \mathbb{Q}$.

• Isn't $\mathbb{Z} + \mathbb{Z}\left[\frac{1 + \sqrt{m}}{2}\right] = \mathbb{Z}\left[\frac{1 + \sqrt{m}}{2}\right]$? Oct 31, 2014 at 7:13

A short proof is given here, see see Theorem $1$ and Theorem $2$. Here the first Theorem proves that these rings of integers $\mathcal{O}_K$ for $K=\mathbb{Q}(\sqrt{-m})$, $m=1,2,3,7,11$ are Euclidean, and the second Theorem shows that these are the only ones, for imaginary quadratic number fields, i.e. for $K=\mathbb{Q}(\sqrt{-m})$, $m>0$. The two cases $m=-1,-2$ do not satisfy the congruence $m\equiv 1 (4)$ of course. For $m\equiv 1(4)$ we have $\mathcal{O}_K=\mathbb{Z}+\mathbb{Z}\bigl(\frac{1+\sqrt m}{2}\bigr)$.