 Solution

The three quadratic subfields are $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{p})$ and $\mathbb{Q}(\sqrt{2p})$. Their integer rings are respectively $\mathbb{Z}[\sqrt{2}]$, $\mathbb{Z}[\sqrt{p}]$ and $\mathbb{Z}[\sqrt{2p}]$, since their squarefree radicands are $2,3,2$ modulo four. Write $\theta =r+s\sqrt{2}+t\sqrt{p}+u\sqrt{2p}$. Relative traces to these three fields are $2r+2s\sqrt{2}$, $2r+2t\sqrt{p}$ and $2r+2u\sqrt{2p}$. Traces of algebraic integers are integral, so all four doubled coefficients are integers. ThusThe PDF names the wrong quadratic subfield for its next displayed norm coefficients. They are obtained from $k=\mathbb{Q}(\sqrt{p})$, not $\mathbb{Q}(\sqrt{2})$. The relevant conjugation changes the sign of $\sqrt{2}$, givingIts membership in $\mathbb{Z}[\sqrt{p}]$ proves the requested divisibilities. For comparison, the norm to the printed subfield is $[a^2-p{b}^2+2c^2-2p{d}^2+2(ac-pbd)\sqrt{2}]/4$, a different expression.

The first divisibility modulo two makes $a,b$ have the same parity. The second makes $ab$ even, so they cannot both be odd: $a$ and $b$ are even. Substituting this into the first divisibility modulo four shows $c^2+p{d}^2$ is even, hence $c\equiv d(\mathrm{mod}2)$. Therefore every algebraic integer is an integer linear combination ofIndeed the last two half-coefficients are $[(c-d)/2]\sqrt{2}+dw$. These four elements are rationally independent. They are all integral: for $w$, $w^2=(p+1)/2+\sqrt{p}$, and consequentlyis a monic polynomial with integer coefficients. Their integer span consists of algebraic integers and contains every algebraic integer by the preceding argument, proving the integral basis of a two-prime biquadratic field. This repairs the subfield typo without changing the claimed basis.