Let $F$ be an irreducible polynomial in $k[x,y]$ with $k$ an algebraically closed field, and $P=(0,0)$ a zero of the polynomial. Let $(\mathcal O_P(F),\mathfrak m)$ be the local ring at $P$, and let $m_P$ be the multiplicity of the curve defined by $F$ at $P$. Show that $\dim_k(\mathfrak m^n / \mathfrak m^{n+1})=n+1$ whenever $0\leq n<m_P$.
For reference, this is exercise 3.13 from Fulton's Algebraic Curves.
I tried doing an induction: the case $n=0$ is easy. For the induction step, I tried replicating the Theorem 2 from earlier in the chapter:
There is a short exact sequence $$\mathfrak m^n / \mathfrak m^{n+1}\to\mathcal O_P(F) /\mathfrak m^{n+1}\to \mathcal O_P(F)/\mathfrak m^n$$
From which we get that $\dim_k(\mathfrak m^n/\mathfrak m^{n+1})=\dim_k(\mathcal O_P(F)/\mathfrak m^{n+1})-\dim_k(\mathcal O_P(F)/\mathfrak m^n)$. We can compute the latter term using the induction hypothesis, so I need to find $\dim_k(\mathcal O_P(F)/\mathfrak m^{n+1})$. Fulton uses the following series of identifications: Let $I=(x,y)$ be an ideal. Then we have that $\mathfrak m^n=I^n\mathcal O_P(F)$, so that:
$$\mathcal O_P(F)/\mathfrak m^{n+1}\cong \mathcal O_P(F)/I^{n+1}\mathcal O_P(F)\cong \mathcal O_P(\mathbb A^2)/(F,I^{n+1})\mathcal O_P(\mathbb A^2)\cong k[x,y]/(F,I^{n+1})$$
(This skips a lot of detail, but it is all justified). But now, I am having trouble computing the dimension of the latter term. In the book, Fulton uses the fact that $n\geq m_P$ to create a short exact sequence, but here it doesn't seem to work.
Best Answer
$F$ has multiplicity $m_P$ at $P$, so $F\in I^{m_P}$ and hence $(F,I^{n+1})=I^{n+1}$ for $n<m_P$. This immediately gives $\dim_k\mathfrak{m}^n/\mathfrak{m}^{n+1}=n+1$.