 Unit from norm one or minus one

For a number field $K$ and an algebraic integer $\alpha \in {\mathcal{O}}_K$, the criterion isHere the absolute field norm means the field norm down to $\mathbb{Q}$; its absolute value is one. Necessity follows from $N(\alpha )N({\alpha }^{-1})=1$ with both factors integers. Conversely, multiplication by $\alpha$ on an integral basis is a matrix with integer entries and determinant $\pm 1$. Its inverse also has integer entries, and applying that inverse to $1$ gives ${\alpha }^{-1}\in {\mathcal{O}}_K$. The integrality hypothesis matters: $\alpha =(3+\sqrt{2})/(3-\sqrt{2})=(11+6\sqrt{2})/7$ has field norm one, but is not in $\mathbb{Z}[\sqrt{2}]$.

In $\mathbb{Z}[\sqrt{2}]$, the criterion becomes $a^2-2b^2=\pm 1$. Thus $(1+\sqrt{2}{)}^{-1}=\sqrt{2}-1$ and $(3+2\sqrt{2}{)}^{-1}=3-2\sqrt{2}$ are again algebraic integers. These units produce infinite families of solutions to the Pell equation.