= Integral basis of a pure quartic field with radicand three modulo four
{title2=$\mathcal O_{\mathbb Q(\sqrt[4]d)}=\mathbb Z[\sqrt[4]d]$}
For a nonzero squarefree integer $d\equiv3\pmod4$, set $\theta=\sqrt[4]d$. The <power basis> $1,\theta,\theta^2,\theta^3$ is an <integral basis>, with <discriminant> $-256d^3$. The polynomial is Eisenstein at every prime dividing $d$; for $d=-1$, shift $X^4+1$ by one to obtain an Eisenstein polynomial at two. The <discriminant-index formula for an integral lattice> restricts any power-basis index prime to $2d$. An Eisenstein prime cannot divide the index: the norm of $r\theta^i/p$ has valuation $i-4<0$ for $p\nmid r$ and $i<4$, while $\theta^4/p$ is integral; multiplying a putative denominator-$p$ integral element successively by powers of $\theta$ forces each numerator coefficient divisible by $p$.
It remains to exclude two. A putative integral element $(a+b\theta+c\theta^2+e\theta^3)/2$ has relative norm to $\mathbb Q(\sqrt d)$ with coefficients
$$
\frac{a^2+dc^2-2dbe}{4},\qquad\frac{2ac-b^2-de^2}{4}.
$$
The quadratic field has <ring of integers> $\mathbb Z[\sqrt d]$, so both coefficients are integers. The first congruence forces $a,c$ to have the same parity. If they are odd, the second requires $b^2+3e^2\equiv2\pmod4$, impossible. If they are even, the second forces $b,e$ to have the same parity and the first forces $be$ even, so both are even. Thus no nontrivial denominator-two element exists, proving index one.
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