TITLE: Is ideal norm multiplicative in arbitrary order of number field?
QUESTION [6 upvotes]: I'm wondering if is true that in any order $\mathcal{O}$ of a number field $K$ the ideal norm (defined by $N(I) = | \mathcal{O}/I|$ is multiplicative. I'm reading Cox's book Prime numbers of the form $x^2 +ny^2$ and he proves it for proper ideals of quadratic fields (i.e. ideals $I$ such that $\mathcal{O} = \{ \beta \in K | \beta I \subset I\}$) which happen to be the invertible ideals. His proof really seems to use all the hypotheses so my (mostly random though) guess would be that it is false (the multiplicativity) in general.
If it is indeed false is it for true for special classes of ideals like for example propers ideals in arbitrary number fields or invertible ideals etc...
Thanks

REPLY [2 votes]: Here a counterexample:
Let R be a number ring and $I\in R$ its ideal, I want to show that the norm of an ideal, defined as $[R:I]$ is not in general multiplicative. If we consider the ring $\mathbb{Z}[\sqrt{-19}]$ and its ideal $I=(2,1+\sqrt{-19})$ we easily notice that : $I^2=2I=(4,2+2\sqrt{-19})$. Since $N(I)=\sharp\frac{R}{I}$, if we compute:
$$\frac{\mathbb{Z}[\sqrt{-19}]}{(2,1+\sqrt{-19})}\cong \frac{\mathbb{Z}[x]}{(2,1+x,x^2+19)}\cong\frac{\mathbb{F}_{2}[x]}{(1+x,x^2+1)}$$
And sending $x \to -1$ we get:
$$\frac{\mathbb{F}_{2}[x]}{(1+x,x^2+1)}\cong \mathbb{F}_{2}$$
Therefore, $N(I)=2$ and, if the norm is multiplicative, we should have $4=N(I)N(I)=N(I^2)=N(2I)=N(2)N(I)$ so what we need is to compute $N(2)$:
$$\frac{\mathbb{Z}[\sqrt{-19}]}{(2)}\cong\frac{\mathbb{Z}[x]}{(2,x^2+19)}\cong \frac{\mathbb{F}_2[x]}{(x^2+1)}$$
Whose cardinality is 4 and then $N(2)N(I)=4\cdot 2\neq4$
The example of $\mathbb{Z}[3i]$ is wrong because $N((9))=81$ since:
$$\frac{\mathbb{Z}[3i]}{(9)}\cong \frac{\mathbb{Z}[x]}{(9,x^2+9)}$$
But $(9,x^2+9)=(9,x^2)$ and so:
$$\frac{\mathbb{Z}[x]}{(9,x^2)}\cong \frac{\frac{\mathbb{Z}}{9\mathbb{Z}}}{x^2}$$
which has 81 element. Another way to see this is to consider:
$$\frac{\mathbb{Z}\oplus (3i \mathbb{Z})}{9\mathbb{Z}\oplus (9\cdot 3i\mathbb{Z})}\cong \frac{\mathbb{Z}}{9\mathbb{Z}}\oplus \frac{\mathbb{Z}}{9\mathbb{Z}}$$