 Integral basis for coprime biquadratic radicands congruent to two and one

Let $m,n$ be coprime square-free integers, $m\equiv 2(\mathrm{mod}4)$ and $n\equiv 1(\mathrm{mod}4)$, with $n\ne 1$. Put ${\alpha }^2=m$, ${\beta }^2=n$, and $\omega =(1+\beta )/2$. Then $1,\alpha ,\omega ,\alpha \omega$ is an integral basis of $\mathbb{Q}(\alpha ,\beta )$. All four entries are algebraic integers. For spanning, field traces to the three quadratic subfields show that an arbitrary algebraic integer has form $(a+b\alpha +c\beta +d\alpha \beta )/2$ with $a,b,c,d\in \mathbb{Z}$. Its field norm to $\mathbb{Q}(\beta )$ is $A/4+(ac-mbd)\beta /2$, where $A=a^2+n{c}^2-m{b}^2-mn{d}^2$. The ring of integers of a quadratic field formula forces $A$ even and $A\equiv 2ac-2mbd(\mathrm{mod}4)$. These give $a\equiv c(\mathrm{mod}2)$ and then $b\equiv d(\mathrm{mod}2)$. The element is consequently $(a-c)/2+((b-d)/2)\alpha +c\omega +d\alpha \omega$. The coefficient determinant $1/4$ proves rational linear independence.