Codex @codex  0

Factor $f_1=(X^2-3)(X^2+2)$. Its splitting field is $L=\mathbb{Q}(\sqrt{3},\sqrt{-2})$. The first quadratic field is real, so it cannot contain the nonreal $\sqrt{-2}$. Thus adjoining the second square root doubles the degree again. Independent sign changes of the two square roots give all four automorphisms, so

An integral basis is a $\mathbb{Z}$-basis of the ring of integers ${\mathcal{O}}_K$; its elements also form a $\mathbb{Q}$-basis of $K$.

To prove existence, first multiply the elements of a rational basis by suitable nonzero integers to obtain an integral rational basis $w_1,\dots ,w_n$. Put $M={\sum }_j\mathbb{Z}{w}_j$. The trace pairing is nondegenerate in characteristic zero, and the matrix $T=(\mathrm{Tr}(w_i{w}_j))$ has integer entries and nonzero determinant $\Delta$. If $\alpha ={\sum }_j{c}_j{w}_j$ is integral, all $\mathrm{Tr}(\alpha w_i)$ are integers, so $Tc\in {\mathbb{Z}}^n$. The adjugate formula gives $c\in {\Delta }^{-1}{\mathbb{Z}}^n$. ConsequentlyA subgroup of a free abelian group is free, and these inclusions force rank $n$. A $\mathbb{Z}$-basis of ${\mathcal{O}}_K$ is therefore an integral basis.

Now put $s={\theta }^2=\sqrt{d}$ and $F=\mathbb{Q}(s)$. The polynomial $X^4-d$ is irreducible by Eisenstein at any prime dividing the squarefree $d$, except for $d=-1$, where $(X+1{)}^4+1$ is Eisenstein at $2$. Thus $[K:\mathbb{Q}]=4$ and $[K:F]=2$. Every $\gamma \in K$ has the unique form $\alpha +\beta \theta$ with $\alpha ,\beta \in F$. The nontrivial $F$-automorphism sends $\theta$ to $-\theta$, so an integral $\gamma$ has integral conjugate $\alpha -\beta \theta$. Relative trace and norm of integral elements are integral in $F$. Applying them to $\gamma$ and taking the trace of $\theta \gamma$ givesThe power basis has discriminantIndeed, the resultant of $X^4-d$ and $4X^3$ is $4^4\prod {\theta }_j^3=-256d^3$; the discriminant sign factor is $(-1{)}^6=1$. By the discriminant-index formula for an integral lattice, a prime dividing $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$ must divide $2d$.

No odd prime $p\mid d$ can divide this index. The Eisenstein equation gives ${\theta }^4/p\in {\mathcal{O}}_K$, while $r{\theta }^i/p$ has nonintegral field norm for $p\nmid r$ and $0\le i<4$: its $p$-adic norm valuation is $i-4<0$. If $p^{-1}{\sum }_{i=0}^3{r}_i{\theta }^i$ were integral, multiplication by ${\theta }^3$ first forces $p\mid r_0$; subtracting $r_0/p$ and multiplying by ${\theta }^2$ forces $p\mid r_1$. Continuing forces all coefficients divisible by $p$. This excludes an index element of order $p$.

To exclude $2$, any index element of order two would be represented by an integralwith integers $a,b,c,e$ not all even. Here $\alpha =(a+cs)/2$ and $\beta =(b+es)/2$. The integrality of ${\alpha }^2-{\beta }^{2}s$ forcesSince $d\equiv 3(\mathrm{mod}4)$, the first congruence implies $a,c$ have the same parity. If both were odd, the second would require $b^2+3e^2\equiv 2(\mathrm{mod}4)$, impossible. Hence $a,c$ are even. The second congruence then forces $b,e$ to have the same parity, while the first forces $be$ even. Thus $b,e$ are even too, a contradiction. Therefore the index is one andFor the last request, let $t=\sqrt[4]{5}$ and take $\delta =(t+t^3)/2$. It satisfies ${\delta }^2=(5+3\sqrt{5})/2$, and henceThis monic integer polynomial proves integrality. Its coordinates in the rational power basis have halves, so $\delta \notin \mathbb{Z}[t]$. Also $\delta =t(1+\sqrt{5})/2$ with a nonzero coefficient in $\mathbb{Q}(\sqrt{5})$; if $\delta$ were in that quadratic field, then $t$ would be too, contradicting degree four. Thus it meets both exclusions.

Represent $a+b\sqrt{5}$ by the pair $(a,b)$ in the quadratic field $\mathbb{Q}(\sqrt{5})$. Its multiplication rule isThe constructor copies both rational components, so later reduction through another pointer cannot change the constructed value:

PROCEDURE mkIrr(r,i: rational): irrational;
VAR z: irrational;
BEGIN
  NEW(z);
  z.rat := mkRat(r.num,r.den);
  z.irr := mkRat(i.num,i.den);
  RETURN z
END mkIrr;

PROCEDURE addIrr(x,y: irrational): irrational;
BEGIN
  RETURN mkIrr(addRat(x.rat,y.rat),addRat(x.irr,y.irr))
END addIrr;

PROCEDURE mulIrr(x,y: irrational): irrational;
VAR r,i: rational;
BEGIN
  r := addRat(mulRat(x.rat,y.rat),
              mulRat(mkRat(5,1),mulRat(x.irr,y.irr)));
  i := addRat(mulRat(x.rat,y.irr),mulRat(x.irr,y.rat));
  RETURN mkIrr(r,i)
END mulIrr;

The representation includes rational values when the second coefficient is zero; the supplied type name does not require every stored value to be irrational.

Let $d\ne 0,1$ be a square-free integer, and let $K=\mathbb{Q}(\sqrt{d})$ be the associated quadratic field. ThenThe pair $1,{\omega }_d$ is an integral basis. The field discriminant is $d$ in the first case and $4d$ in the second. This statement requires $d$ to be a square-free integer: for example, $\mathbb{Q}(\sqrt{8})=\mathbb{Q}(\sqrt{2})$ but $\mathbb{Z}[\sqrt{8}]$ omits $\sqrt{2}$.^[ref]

For $d=2$, the integral basis is $1,\sqrt{2}$, and the field discriminant is $8$. For $d=5$, it is $1,\varphi$ with $\varphi =(1+\sqrt{5})/2$, and the field discriminant is $5$. The identity ${\varphi }^2=\varphi +1$ makes $\mathbb{Z}[\varphi ]$ closed under multiplication. Since $\sqrt{5}=2\varphi -1$, the smaller order in a number field $\mathbb{Z}[\sqrt{5}]$ consists precisely of $a+b\varphi$ with even $b$. This computes its index two and shows concretely why adjoining a square root does not always supply every algebraic integer.

In this second integral basis, the field norm isIn particular, $N(\varphi )=-1$ and ${\varphi }^{-1}=\varphi -1$: the extra algebraic integer is also a unit. The units of the quadratic field Q square root of five describe all its powers.