If $\mathcal{O}$ is the ring of integers of some number field and $I$ is some nonzero ideal in $\mathcal{O}$, the ideal norm $N(I)$ can be defined via $$N(I):=[\mathcal{O}:I]$$ It can be shown that $N$ is multiplicative: $N(IJ)=N(I)N(J)$ for any two ideals $I$ and $J$. The only proof for multiplicativity I know uses the fact that $\mathcal{O}$ admits unique factorization into prime ideals, i.e. that $\mathcal{O}$ is a Dedekind domain.
My Question: Is the fact that we have factorization into prime ideals necessary for proving this or is it provable without using it (and also not using the Dedekind property in any other way)? In other words: If we define the ideal norm in the above way for any integral domain $R$ with the property that all the residue rings $R/I$ are finite when $I\neq 0$, is the norm always multiplicative or is there a counterexample?
The natural idea to find a counterexample is to take a non maximal order. Let $R=\mathbb{Z}[i\sqrt{3}]$. Set $\alpha=i\sqrt{3}$, so $\alpha^2=-3$.
Fact 1. If $I\neq 0$, $R/I$ is finite.
Assume that $I\neq 0$, and let $a$ be a nonzero element of $I$. Then $R/(a)\to R/I$ is surjective, so it is enough to prove that $R/(a)$ is nonzero. But since $R$ is a free abelian grop of rank $2$ and $a\neq 0$, $(a)=Ra$ is free abelian of rank $2$. It is known that case that the quotient is finite (take a basis of $R$ adapted to $aR$).
Fact 2. Let $I=(2,1+\alpha)$. Then $I^2=2I$. Indeed, $$I^2=(4,2(1+\alpha), 4+2\alpha)=(4,2(1+\alpha),-2+2\alpha),$$ so $$I^2=(4,2(1+\alpha),-4+2(1+\alpha))=(4,2(1+\alpha))=2I.$$
So assuming multiplicativity of the norm, we would get $N(I)=N(2)$.
Now $R/I$ has 2 elements (morally speaking, you identify $i\sqrt{3}$ to $-1$, and then you mod out by $2$). But $R/(2)$ has $4$ elements since $R/(2)\simeq\mathbb{F}_2[X]/(X^2+3)$ and $X^2+3$ has degree $2$.
All in all, $N(I)=2,$ but $N(2)=4$, so $N$ is not multiplicative.