Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 19H / d / Solution

We prove the conjugate product of a quadratic ideal identity $\mathfrak{a}\overline{\mathfrak{a}}=(N(\mathfrak{a}))$, which applies to the given two-generator ideal without assuming its displayed generators form an integral basis.

The quadratic ring of integers of a number field has a basis $1,\omega$ over $\mathbb{Z}$: the vector 1 is primitive, since a rational algebraic integer must be an integer, and extends to an integral basis. Write ${\omega }^2-T\omega +U=0$, with $T,U\in \mathbb{Z}$. Let $B$ be the least positive integer in $\mathfrak{a}$ and let $s>0$ generate the set of coefficients of $\omega$ in elements of $\mathfrak{a}$. Its lattice basis can be chosen as $B,r+s\omega$, with $r\in \mathbb{Z}$. Since $B\omega \in \mathfrak{a}$, comparison of coefficients shows $s\mid B$ and $s\mid r$. Write $B=sd$, $r=se$ and $\theta =e+\omega$. ThenThe scaled ideal $\mathfrak{j}=s^{-1}\mathfrak{a}$ remains an ${\mathcal{O}}_L$-ideal. From ${\theta }^2-T^{\prime }\theta +U^{\prime }=0$ we get $U^{\prime }\in \mathfrak{j}\cap \mathbb{Z}=d\mathbb{Z}$, so $d\mid U^{\prime }$. The product $\mathfrak{j}\bar{\mathfrak{j}}$ is generated as an ideal by $d^2,d\theta ,d\bar{\theta },U^{\prime }$ and is contained in $(d)$.

Moreover $\gcd (d,T^{\prime },U^{\prime }/d)=1$. If a prime $\ell$ divided all three, then $\ell \mid T^{\prime }$ and ${\ell }^2\mid U^{\prime }$, so $\theta /\ell$ would satisfy the monic integer polynomial $z^2-(T^{\prime }/\ell )z+U^{\prime }/{\ell }^2$. It would lie in ${\mathcal{O}}_L$, impossible because its $\omega$ coefficient in the integral basis is $1/\ell$. Bezout identity now expresses one as an integer combination of $d,T^{\prime },U^{\prime }/d$. Multiplying by $d$, the terms are $d^2$, $d{T}^{\prime }=d\theta +d\bar{\theta }$, and $U^{\prime }$, all in the product. Hence $d$ belongs to it, proving $\mathfrak{j}\bar{\mathfrak{j}}=(d)$.

The determinant of the lattice basis $sd,s(e+\omega )$ gives the ideal norm $N(\mathfrak{a})=s^{2}d$. ConsequentlyIn particular $⟨b,\alpha ⟩⟨b,\bar{\alpha }⟩$ is principal. This does not require the original $b$ to be nonzero or the ideal itself to be principal.