Norm of powers of a maximal ideal

Question

Let $A$ be a integral domain and $M$ a maximal ideal in $A$ such that the quotient $A/M$ is a finite ring (and thus a finite field). Is it true, in general, that $$|A/M^k|=|A/M|^k \quad (k\in\textbf{N})\ ?$$

Edit. (Counter-example in the answers, thanks to Jendrik Stelzner and Bib-lost). Nevertheless, I have the feeling that this fact happen when $A$ is assumed to be residually finite, i.e. for each non zero ideal of $A$, $A/I$ is finite (which is not the case of $\textbf{F}_p[X,Y]$). See my new question Norm of powers of a maximal ideal (in residually finite rings).

Many thanks and happy new year !

Answer 1

What about $\mathbb{F}_p[X,Y]$ and $(X,Y) \subseteq A$. We have $\mathbb{F}_p[X,Y]/(X,Y) = \mathbb{F}_p$. But if I am not mistaken we have $(X,Y)^2 = (X^2, XY, Y^2)$ and $\mathbb{F}_p[X,Y]/(X^2,XY,Y^2)$ should be three dimensional as an $\mathbb{F}_p$ vector space (a basis is given by the residue classes of $1$, $X$ and $Y$), so $$ |\mathbb{F}_p[X,Y]/(X,Y)^2| = p^3 \neq p^2 = |\mathbb{F}_p[X,Y]/(X,Y)|^2. $$

Answer 2

I don't think this is true. Consider $A = \mathbb{F}_{p}[X, Y]$, the polynomial ring in two variables over the finite field with $p$ elements. Let $M = (X, Y)$. Then $A/M \simeq \mathbb{F}_{p}$ (which contains $p$ elements), but $A/M^{2}$ is generated as an $\mathbb{F}_{p}$-vectorspace by $1$, $\overline{X}$ and $\overline{Y}$, thus containing $p^3$ elements.