Norm of powers of a maximal ideal (in residually finite rings) Let $A$ be a residually finite integral domain and $M$ a maximal ideal in $A$. Is this true that
$$|A/M^k|=|A/M|^k \quad (k\in\textbf{N}) \quad ?$$
In Hirano's article On Residually Finite Rings we can read in page 11/14 (proof of proposition 4) an argument working in "Asano order", and I have not the background to understand this proof (I don't know category and tensor product yet...). 
Can anyone help with this, or have an easier proof/counter-example of this fact ?
(This result is not true in a more general context, see Norm of powers of a maximal ideal.)
Many thanks and happy new year !
 A: Very much like the previous answers you got for a similar problem, I suggest you try to work out the case of $A=F[t^2,t^3]$ ($F$ a finite field, in which case $A$ is residually finite integral domain) and $M=(t^2,t^3)$. You should run into trouble already for $k=2$.
A: Let $A=\mathbb Z[\sqrt{-3}]$. 
The ring $A$ is residually finite since every non-zero ideal $I$ contains a non-zero integer (if $a+b\sqrt{-3}\in I$ then $n=a^2+3b^2\in I\cap\mathbb Z$), so $A/I\simeq \frac{A/nA}{I/nA}$ and $A/nA$ is finite.  
Set $M=(1+\sqrt{-3},1-\sqrt{-3})$. We have $N(M)=2$, $M^2=(2)M$, hence $N(M^k)=2^{2k-1}$ for all $k\ge1$. (Here $N(I)=|A/I|$ for $I$ a non-zero ideal.) 
Remark. The commutative analogous of the Asano orders are Dedekind domains, and for residually finite Dedekind domains we have $N(IJ)=N(I)N(J)$. This is why I've chosen $\mathbb Z[\sqrt{−3}]$ and not, for instance, $\mathbb Z[\sqrt{−5}]$. 

Let $A$ be a residually finite Dedekind domain, and $I,J$ ideals. Then $N(IJ)=N(I)N(J)$.

The proof reduces easily (via CRT) to show that $N(A/P^{m+n})=N(A/P^m)N(A/P^n)$ where $P$ is a non-zero prime ideal. But $A/P^i\simeq A_P/P^iA_P$, $A_P$ is a local PID, and then $N(A_P/P^iA_P)=N(A_P/PA_P)^i$.
