= Integral basis of an odd biquadratic field containing square root of two
{title2=$1,\sqrt2,\sqrt\delta,(\sqrt2+\sqrt{2\delta})/2$}
For square-free $\delta\equiv3\pmod4$, the field $\mathbb Q(\sqrt2,\sqrt\delta)$ has basis $1,\sqrt2,\sqrt\delta,(\sqrt2+\sqrt{2\delta})/2$ 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 $X^4-(1+\delta)X^2+(\delta-1)^2/4=0$.
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