 Integral basis of an odd biquadratic field containing square root of two

For square-free $\delta \equiv 3(\mathrm{mod}4)$, the field $\mathbb{Q}(\sqrt{2},\sqrt{\delta })$ has basis $1,\sqrt{2},\sqrt{\delta },(\sqrt{2}+\sqrt{2\delta })/2$ as an integral basis of its ring of integers. Relative traces force all rational coefficients to be half-integers; the norm to $\mathbb{Q}(\sqrt{\delta })$ forces the constant and $\sqrt{\delta }$ coefficients to be integers and the two other numerators to have equal parity. The half-sum is integral because it satisfies $X^4-(1+\delta )X^2+(\delta -1{)}^2/4=0$.