Dimension of $\mathfrak{m}^k/\mathfrak{m}^{k+1}$?

Question

Let $C \subset \mathbb{P}_\mathbb{C}^2$ be a projective curve, and $p \in C$ a point of multiplicity $m$. If $\mathcal{O}_p(C)$ is the local ring of $C$ at $p$, and $\mathfrak{m} \subset \mathcal{O}_p(C)$ is its maximal ideal, what is $\text{dim}_\mathbb{C} \mathfrak{m}^k/\mathfrak{m}^{k+1}$?

Answer 1

We show that$$\dim_\mathbb{C} \mathfrak{m}^k/\mathfrak{m}^{k+1} = k+1$$if $k < m$.

Let $p = [a, b, c]$ (we can assume $c \neq 0$). Then

• we can identify $\mathcal{O}_p$ with $\mathbb{C}[x, y]/(f)$,
• we can identify $p$ with $(a/c, b/c) \in \mathbb{C}^2$,
• $f(a/c, b/c) = 0$.

In this case, we have$$\mathfrak{m} = \left\langle x - {a\over{c}}, y - {b\over{c}}\right\rangle$$$$\implies f = \sum_{i + j \ge m} a_{ij} \left(x - {a\over{c}}\right)^i \left( x - {b\over{c}}\right)^j g_{ij} (x, y)$$$$\implies \mathfrak{m}^k = \sum_{i + j = k} \mathcal{O}_p \left(x - {a\over{c}}\right)^i \left( x - {b\over{c}}\right)^j$$$$\implies \mathfrak{m}^{k+1} = \sum_{i + j = k+1} \mathcal{O}_p \left(x - {a\over{c}}\right)^i \left( x - {b\over{c}}\right)^j.$$It is clear that $\mathfrak{m}^k/\mathfrak{m}^{k+1}$ is spanned by $(x - a/c)^i(x - b/c)^j$ for $i + j = k$ over $\mathbb{C}$. The linear independence of these polynomials come from the fact that the minimal degree of homogeneous components of $f$ and the minimal degree of homogeneous compoenents of $g \in \mathfrak{m}^{k+1}$ are at least $m$. Hence, we have$$\text{dim}_\mathbb{C} \mathfrak{m}^k/\mathfrak{m}^{k+1} = k+1$$if $k < m$, as desired.