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 $\theta=e+\omega$. Then
$$
\mathfrak a=s\langle d,\theta\rangle_{\mathbb Z},
\qquad \theta+\bar\theta=T'=T+2e,\quad
\theta\bar\theta=U'=e^2+eT+U.
$$
The scaled <ideal> $\mathfrak j=s^{-1}\mathfrak a$ remains an $\mathcal O_L$-ideal. From $\theta^2-T'\theta+U'=0$ we get $U'\in\mathfrak j\cap\mathbb Z=d\mathbb Z$, so $d\mid U'$. The product $\mathfrak j\bar{\mathfrak j}$ is generated as an <ideal> by $d^2,d\theta,d\bar\theta,U'$ and is contained in $(d)$.

Moreover $\gcd(d,T',U'/d)=1$. If a prime $\ell$ divided all three, then $\ell\mid T'$ and $\ell^2\mid U'$, so $\theta/\ell$ would satisfy the monic integer <polynomial> $z^2-(T'/\ell)z+U'/\ell^2$. 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 $d,T',U'/d$. Multiplying by $d$, the terms are $d^2$, $dT'=d\theta+d\bar\theta$, and $U'$, 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> $N(\mathfrak a)=s^2d$. Consequently
$$
\boxed{\mathfrak a\bar{\mathfrak a}=(s^2d)=(N(\mathfrak a))}.
$$
In particular $\langle b,\alpha\rangle\langle b,\bar\alpha\rangle$ is principal. This does not require the original $b$ to be nonzero or the <ideal> itself to be principal.

Solution

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

{title2=$\mathfrak a\bar{\mathfrak a}=(N(\mathfrak a))$}

For a nonzero integral ideal $\mathfrak a$ of the full <ring of integers of a number field> of degree two,
$$
\mathfrak a\overline{\mathfrak a}=(N(\mathfrak a)).
$$
Here conjugation is the nonidentity field automorphism and $N(\mathfrak a)$ is the additive index of the ideal. A direct proof writes $\mathfrak a=s\langle d,e+\omega\rangle_{\mathbb Z}$ in an <integral basis> $1,\omega$. Put $\theta=e+\omega$, $T=\theta+\bar\theta$, $U=\theta\bar\theta$. The ideal property gives $d\mid U$. Moreover $\gcd(d,T,U/d)=1$: a prime dividing all three would make $\theta/\ell$ an <algebraic integer>, contradicting its nonintegral coefficient of $\omega$. The product of $\langle d,\theta\rangle$ with its conjugate contains $d^2,dT,U$, so <Bezout identity> puts $d$ in it; all its generators lie in $(d)$. Scaling back gives $(s^2d)$, and the lattice determinant identifies $s^2d=N(\mathfrak a)$. The full ring of integers is essential to the integrality argument.

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 Incoming links: Conjugate product of a quadratic ideal