Multiplicity of point of algebraic plane curve

Question

I'm reading Fulton's Curves at the moment, and I'm stuck on Theorem 2 of section 3.2. He has defined the multiplicity $m_P(C)$ of a point $P = (0,0)$ of curve $C = V(F)$ (let's say $C$ is irreducible for now) as the degree of the lowest non-zero form comprising $F$. In the theorem I'm stuck on, he gives a more intrinsic notion of multiplicity.

Theorem. Let $P$ be a point on an irreducible curve $C$. Then for all sufficiently large $n$, $$m_P(C) = \dim_k{\mathfrak{m}^n / \mathfrak{m}^{n+1}}$$.

Here $\mathfrak{m}$ is the maximal ideal of the local ring $\mathcal{O}_P$. It's actually the first line of the proof I am stuck with:

From the exact sequence $$ 0 \to \mathfrak{m}^n/\mathfrak{m}^{n+1} \to \mathcal{O}/\mathfrak{m}^{n+1} \to \mathcal{O}/\mathfrak{m}^{n} \to 0$$ it follows that it is enough to show that $\dim_k{\mathcal{O}/\mathfrak{m}^n} = n m_P(C) + s$ for some constant $s$ and all $n \geq m_P(C)$.

He does give a reference to a previous theorem as to why this should follow, but the reference is just to the rank-nullity theorem. I guess the $s$ and $nm_P(C)$ somehow refer to the dimensions of $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ and $\mathcal{O}/\mathfrak{m}^{n+1}$, but I'm not sure how.

Thanks

Answer 1

For a short exact sequence, $0 \to A \to B \to C \to 0$, we have $dim(B) = dim(A) + dim(C)$. (Think of direct sum short exact sequence, $0 \to A \to A \oplus C \to C \to 0$.)

Rearranging, we have $dim(A) = dim(B) - dim(C)$, for a general short exact sequence. (Indeed, this is rank nullity. C is the image, and $A$ is the kernel.)

So if $dim_k \mathscr{O}/\mathscr{m}^n = n m_P(C) + s$,

we get $dim_k (m^{n+1}/m^n) = dim_k(\mathscr{O}/m^{n+1}) - dim_k ( \mathscr{O}/m^n) = (n+1) m_P(C) + s - (n m_P(C) + s) = m_P(C)$

Answer 2

All ideals of $\newcommand{\co}{\mathcal{O}}\co_P$ are vector spaces over $k$. If we have an exact sequence of $k$-vector spaces $$0\to U\to V\to W\to0$$ then $\dim_k(V)=\dim_k(U)+\dim_k(W)$; this is essentially the rank-nullity formula. Here we have $$\newcommand{\fm}{\mathfrak{m}} 0\to \fm^n/\fm^{n+1}\to \co_P/\fm^{n+1} \to \co_P/\fm^n\to0$$ and so by rank-nullity, $$d_{n+1}=c_n+d_n$$ where $d_n=\dim_k(\co_P/\fm^n)$ and $c_n=\dim_k(\fm^n/\fm^{n+1})$.

We need to prove that $c_n=m_P(C)$ for $n>N_0$ say. If we can prove that $d_n=n m_P(C)+s$ for $n>N_0$, then it follows that $$c_n=d_{n+1}-d_n=((n+1)m_p(C)+s)-(n m_P(C)+s)=m_P(C)$$ for $n>N_0$.