Past exam of the number theory course of the University of Oxford / 2010 / B / 4 / Solution

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.