Dimension of ring by $\dim_k (m^i/m^{i +1})$ for all $i(2≦i＜∞)$ is the same as the embedding dimension?

Question

Let $A$ be a Noetherian local ring. Define $m$ be it's maximal ideal and $A/m$ be residue field. Then we can define embedding dimension of $A$ by $\dim_k(m/m^2)$, here $\dim_k m/m^2$ is dimension of $m/m^2$ as vector space over $k$.

But I think we can also define $\dim_k (m^i/m^{i +1})$ for fixed $i$ $(2\le i＜∞)$ in the same way because the action is well-defined in the same way and we can regard $m^i/m^{i +1}$ as $k$-vector space.

Then, $\dim_k (m/m^2) ＝ \dim_k (m^i/m^{i +1})$ for all $i$ $(2\le i＜∞)$ ? If sometimes differ by $I$, then it is meaningless to define new dimension by $\dim_k (m^i/m^{i +1})$?

Thank you in advance.

Answer 1

Let $A = k[X,Y]/(X^2, Y^2) = \mathrm{span}_k\{1,X,Y,XY\}$. Then $\mathfrak{m} = \mathrm{span}_k\{X,Y, XY\}$, $\mathfrak{m}^2 = \mathrm{span}_k\{XY\}$ and $\mathfrak{m}^i = 0$ for $i>2$. So $$\dim(\mathfrak{m}/\mathfrak{m}^2) = 2, \\ \dim(\mathfrak{m}^2/\mathfrak{m}^3) = 1 \text{ and} \\ \dim(\mathfrak{m}^i/\mathfrak{m}^{i+1}) = 0 \text{ for } i>2.$$