Is ideal norm multiplicative in arbitrary order of number field?

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 $\mathcal{O}_K / (3) \cong \mathbb{F}_9$.

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.

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:

$$\frac{\mathbb{Z}\oplus (3i \mathbb{Z})}{9\mathbb{Z}\oplus (9\cdot 3i\mathbb{Z})}\cong \frac{\mathbb{Z}}{9\mathbb{Z}}\oplus \frac{\mathbb{Z}}{9\mathbb{Z}}$$