Ideal norm

For a nonzero integral ideal $I\subseteq {\mathcal{O}}_K$, its norm is the finite index $N(I)=|{\mathcal{O}}_K/I|$. Norms multiply under ideal multiplication.

For a nonzero integral ideal $\mathfrak{a}$ of the full ring of integers of a number field of degree two,Here conjugation is the nonidentity field automorphism and $N(\mathfrak{a})$ is the additive index of the ideal. A direct proof writes $\mathfrak{a}=s⟨d,e+\omega {⟩}_{\mathbb{Z}}$ 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 $⟨d,\theta ⟩$ with its conjugate contains $d^2,dT,U$, so Bezout identity puts $d$ in it; all its generators lie in $(d)$. Scaling back gives $(s^{2}d)$, and the lattice determinant identifies $s^{2}d=N(\mathfrak{a})$. The full ring of integers is essential to the integrality argument.

For a fractional ideal $I$ of ${\mathcal{O}}_L$, the relative ideal norm is the fractional ideal of ${\mathcal{O}}_K$ obtained multiplicatively from prime-ideal norms. Equivalently, its local valuation is the determinant valuation of $I$ as an ${\mathcal{O}}_K$-lattice relative to ${\mathcal{O}}_L$.