= Conjugate product of a quadratic ideal
{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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