Norm in an order of a number field and Hilbert polynomial Let $O$ be an order in a number field (So $O$ is an one dimensional noetherian domain), I am curious about multiplication of norm function in $O$.
For every nonzero ideal $I \subseteq O$, let $N(I)=\#O/I$. The norm function need not to be multiplicative, for example take $O=\Bbb Z[2i], I=(2,2i),$ then $N(I^n)=2^{2n-1}$. Motivated by this example, I think $N(I^n)$ may look like $kq^n$ ($k,q$ depends on $I$) in general. Below are some of my observations.  
First of all, we have $O/I \cong \bigoplus_{p } O_p/IO_p$ where $p$ ranges over nonzero prime ideals (containing $I$) of $O$, as we know $O/I$ is Artin ($\text{dim}=0$ and Noetherian) hence is the product of its localization. Therefore $N(I)=\prod_p \#O_p/IO_p = \# \prod k(p)^{l(O_p/IO_p)}$ where $l()$ means the length of modules and $k(p)=O_p/pO_p=O/p$.
Then, we know that $N(IJ)=N(I)N(J)$ at least when one of $I,J$ is invertible. This is because $R/J \cong I/IJ=I \otimes R/J$, as locally they are isomorphic (Localization of $I$ is a principal ideal as projective module over local ring is free) hence they are isomorphic because $R/J$ is Artin.
By primary decomposition we know every nonzero ideal in $O$ is product of primary ideals (primary ideals with diffferent radical are coprime because $dim=1$ hence we can write intersection as product), so we need only analysis those ideals who's radical contains the conductor of $O$. In short, essentially we can analysis the problem locally. (A more precise decomposition is $I= \prod _p { I_p \cap O}$, and every $I_p \cap O$ contains some power of $p$)
For every $I$, we know that there exists $k,q \in \Bbb Q$ s.t $N(I^n)=kq^n, \forall n>>0$ because $l(O_p/I^nO_p)$ is eventually a polynomial of $deg=1$ by the general knowledge of Hilbert-Samuel function of notherian local rings. 
Here is my question: can we make sure that $N(I^n)=kq^n, \forall n>0$ ? If we write $k_I,q_I$ for $k,q$, do we have $q_{IJ}=q_Iq_J$ or $k_{IJ}=k_Ik_J$ or one-side inequalities? Can we describe $k_{I},q_{I}$ in a more concrete way? 
Edit: As examples in the answer below about the first part of the question shows, in general we cannot hope $N(I^n)=kq^n, \forall n>0$. However, does there exist $n_0 \in \Bbb N$ independent on $I$ such that $N(I^n)=k_Iq_I^n, \forall n>n_0, I \subseteq O$ ?
 A: This is not true. An easier type of ring to think about is a subring of $k[x]$, e.g. $R = k[x^3,x^4]$; we will deal with the number ring case later. For this type of ring, your question basically asks that $\dim R/I^n$ equals $an - b$ for some constants $a, b$. For $R = k[x^3,x^4]$, we get:


*

*$I = (x^3, x^4)$, so $\dim R/I = 1$ (only the constants are missing).

*$I^2 = (x^6,x^7,x^8)$, so $\dim R/I^2 = 3$ (the missing monomials are $1$, $x^3$, and $x^4$).

*$I^3 = (x^9,x^{10},x^{11},x^{12}) = (x^9,x^{10},x^{11})$, so $\dim R/I^3 = 6$ (the missing monomials are $1$, $x^3$, $x^4$, $x^6$, $x^7$, and $x^8$; note that $x^5 \not\in R$).


We conclude that the function $n \mapsto \dim R/I^n$ is not linear. (However, it is eventually linear.)
In a number ring, we can take $R = \mathbb Z[2\sqrt[3]{2}]$, with the maximal ideal $I = (2,2\sqrt[3]{2})$. We will write this as $R = \mathbb Z[x]/(x^3-16)$, with the ideal $I = (2,x)$. We get


*

*$|R/I| = 2$ since $I$ is maximal.

*$I^2 = (4,4\sqrt[3]{2},4\sqrt[3]{4})$, and
$$R/I^2 \cong \mathbb Z[x]/(x^3-16,4,2x,x^2) = (\mathbb Z/4)[x]/(2x,x^2),$$
which has $8$ elements (namely $a+bx$ for $a \in \{0,\ldots,4\}$, $b\in\{0,1\}$).

*$I^3 = (8,8\sqrt[3]{2},8\sqrt[3]{4},16) = (8,8\sqrt[3]{2},8\sqrt[3]{4})$, and
$$R/I^3 \cong \mathbb Z[x]/(x^3-16,8,4x,2x^2,x^3) = (\mathbb Z/8)[x]/(4x,2x^2,x^3),$$
which has $64$ elements (namely $a+bx+cx^2$ with $a \in \mathbb Z/8$, $b \in \mathbb Z/4$, and $c \in \mathbb Z/2$).


Again, we see that $|R/I^n|$ has size $2^1$, $2^3$, and $2^6$ for $n = 1$, $2$, $3$ respectively. This is not of the form $n \mapsto k q^n$ for any $k$ and $q$.
The precise relationship between these two examples is that the associated graded of $R$ with respect to the $2$-adic filtration is isomorphic to $\mathbb F_2[x^3,x^4]$, roughly because $\pi = \sqrt[3]{2}$ is a uniformiser in the normalisation, and we have taken the ideal $(\pi^3,\pi^4)$ inside the ring generated by $\pi^3$ and $\pi^4$. This explains why the dimension counts agree.
