Let $R=C[x_1,...,x_n]$ and $M$ be a maximal ideal of $R$ such that $R/M$ is a finite dimensional $C-$algebra. Can we deduce that $R/M^n$ for n>1 is also finite dimensional $C$-algebra?
We know that $M^n\subset M$ so $R/M \subset R/M^n$. However we cannot conclude that $R/M^n$ is also finite dimensional.
Also we know that since $M$ is maximal, $M=(x_1-a_1,...,x_n-a_n)$, so $R/M$ is isomorphic to $C$. I don't know if we can conclude that $R/M^n$ is also finite dimensional.
Let $R=\mathbb{C}[X_1, \cdots, X_n]$ and $m=(x_1-a_1, \cdots x_n-a_n)$ be a maximal ideal of $R$. $R/m = \mathbb{C}$.
$m^k$ is generated by $\{(x_1-a_1)^{e_1}\cdots(x_n-a_n)^{e_n} \colon \sum_j e_j = k\}$ and $R/m^k$ is generated by $\{(x_1-a_1)^{i_1}\cdots (x_n-a_n)^{i_n}\colon \sum_j i_j < k\}$. Then $R/m^k$ is finite dimensional.