= Solution
The three quadratic subfields are $\mathbb Q(\sqrt2)$, $\mathbb Q(\sqrt p)$ and $\mathbb Q(\sqrt{2p})$. Their integer rings are respectively $\mathbb Z[\sqrt2]$, $\mathbb Z[\sqrt p]$ and $\mathbb Z[\sqrt{2p}]$, since their squarefree radicands are $2,3,2$ modulo four. Write $\theta=r+s\sqrt2+t\sqrt p+u\sqrt{2p}$. Relative traces to these three fields are $2r+2s\sqrt2$, $2r+2t\sqrt p$ and $2r+2u\sqrt{2p}$. Traces of <algebraic integers> are <integral>, so all four doubled coefficients are integers. Thus
$$
\theta=\tfrac12(a+b\sqrt p)+\tfrac12(c+d\sqrt p)\sqrt2,\qquad a,b,c,d\in\mathbb Z.
$$
The PDF names the wrong quadratic subfield for its next displayed norm coefficients. They are obtained from $k=\mathbb Q(\sqrt p)$, not $\mathbb Q(\sqrt2)$. The relevant conjugation changes the sign of $\sqrt2$, giving
$$
N_{K/\mathbb Q(\sqrt p)}(\theta)=\frac{a^2+pb^2-2c^2-2pd^2+2(ab-2cd)\sqrt p}{4}.
$$
Its membership in $\mathbb Z[\sqrt p]$ proves the requested divisibilities. For comparison, the norm to the printed subfield is $[a^2-pb^2+2c^2-2pd^2+2(ac-pbd)\sqrt2]/4$, 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: \b[$a$ and $b$ are even]. Substituting this into the first divisibility modulo four shows $c^2+pd^2$ is even, hence \b[$c\equiv d\pmod2$]. Therefore every <algebraic integer> is an integer linear combination of
$$
\boxed{1,\quad\sqrt2,\quad\sqrt p,\quad w=\frac{(1+\sqrt p)\sqrt2}{2}.}
$$
Indeed the last two half-coefficients are $[(c-d)/2]\sqrt2+dw$. These four elements are rationally independent. They are all <integral>: for $w$, $w^2=(p+1)/2+\sqrt p$, and consequently
$$
w^4-(p+1)w^2+\frac{(p-1)^2}{4}=0
$$
is 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.
Back to article page
