Is ideal norm multiplicative in arbitrary order of number field? 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
 A: As Hurkyl mentioned, there are easy counterexamples. But one can say much more. Suppose that$\rm\,D\,$ is a finite norm domain, i.e. every ideal $\ne 0\,$ has finite norm $\rm\,|D/I|.\,$ Then it is easy to prove


*

*the norm is multiplicative for all nonzero ideals of $\rm\,D$ $\!\iff\!$ $\rm\,D\,$ is a Dedekind domain   


Thus norm multiplicativity fails for finite norm domains that are not integrally closed, as in the example in Hurkyl's answer. For the simple proof, and a related criterion, see
Butts, H. S.; Wade, L. I. $\ $ Two criteria for Dedekind domains.
Amer. Math. Monthly  73, 1966, 14-21.
A: 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:
$$\frac{\mathbb{Z}\oplus (3i \mathbb{Z})}{9\mathbb{Z}\oplus (9\cdot 3i\mathbb{Z})}\cong \frac{\mathbb{Z}}{9\mathbb{Z}}\oplus \frac{\mathbb{Z}}{9\mathbb{Z}}$$
A: Let $R$ be the ring $\mathbb{Z}[3i]$ with $\mathbb{Z}$-basis $(1,3i)$. This is an order in $\mathbb{Q}(i)$.
Let $I$ be the $R$-ideal $(3,3i)$ with norm $[R:I]=3$.  Note that $I$ is not a principal $R$-ideal.
We have $I^2 = (9,9i)$, which has norm $[R:I^2]=27$.
Now, $N(I) = 3$ and $N(I^2) = 27$, so the norm is not multiplicative.
The prime ideal $(3, 3i)$ is the only "singularity" of $R$. If $J$ is any ideal relatively prime to $(3,3i)$, then $N(J) = N(J \mathcal{O}_K)$, and so the norm is multiplicative on such ideals. A quick proof of this fact is:
$$ R / J \cong (R / J)[1/3] = R[1/3] / J = \mathcal{O}_K[1/3] / J = (\mathcal{O}_K / J)[1/3] = \mathcal{O}_K / J $$
However $\mathfrak{p} = (3, 3i)$ doesn't share a similar property; it "splits" into the principal prime ideal $(3) \subseteq \mathcal{O}_K$; we have $R / \mathfrak{p} \cong \mathbb{F}_3$, but $\mathcal{O}_K / (3) \cong \mathbb{F}_9$.
