Showing that intersection multiplicity at a point is finite for prime divisors My question has two parts two it: one vaguely more elementary, one perhaps less so.
In Beauville (Complex Algebraic Surfaces), we define the multiplicity of intersection of two (irreducible, no common component) curves $C,D$ on a surface by first taking the intersection multiplicity of them at a point $x \in C \cap D$:
$$m_x(C,D) = \mbox{dim}_k \mathcal{O}_x/(f,g),$$
where $f, g$ are local equations for $C,D$ at $x$, and the dimension on the right is as a $k$-vector space.
Beauville claims this is finite dimensional from the Nullstellensatz, using that $f,g$ vanish on $x$ so that $M_x$ in $I(V((f,g))) = \sqrt {(f,g)}$, where $M_x$ is the maximal ideal defining $x$. This gives that $M_x^k \subset (f,g)$ for some $k$. I'm not sure how this then gives the answer: it doesn't seem far off, but I can't finish. Could anyone point me in the right direction on how to conclude?
Secondly, I'm trying to extend this to effective divisors on a surface, defined as a (nonsingular) dimension 2 projective scheme over a field $k$, as in Hartshorne. The same definition certainly works, as Hartshorne uses it: but he sidesteps showing finiteness by showing this agrees with his more technical previous definition.
Without the Nullstellensatz, I'm vaguely lost for even a starting point. Any pointers on how to begin to go about this?
Thanks!
 A: This is not a complete answer, I'd like to hear suggestions for completing it.
Let $(R,m)$ be a (regular) local $\mathbb{C}$-algebra (this is the case since in Beauville, $S$ is a smooth projective surface over $\mathbb{C}.$) We would like to show that $\dim_{\mathbb{C}}R/m^n < \infty$ for $n \geq 1.$ We do this by induction on $n.$ 
Here I think, we need to assume that $x \in S$ is a closed point. I'd like to hear suggestions for dropping this assumption.
For $n=1,$ in fact $R/m$ is a finite algebraic extension of $\mathbb{C},$ hence is $\cong \mathbb{C}.$
For $n=2,$ look at the following short exact sequence of $R$-modules, ($\mathbb{C}$-modules)
$$0 \to m/m^2 \to R/m^2 \to R/m \to 0$$
Since $\dim_{\mathbb{C}}$ is additive, we get $\dim_{\mathbb{C}}(R/m^2) < \infty.$ Note that $\dim_{\mathbb{C}}(m/m^2)$ is finite ($S$ is smooth.)
Now for $n=3,$ we have $0\to m^2/m^3 \to R/m^3 \to m/m^2 \to 0$ so it's enough to show that $m^2/m^3$ is finite dimensional. Look at the following SES 
$$0 \to m^2/m^3 \to m/m^3 \to m/m^2 \to 0$$
In fact, we have $m/m^3 \otimes_R R/m \cong \frac{m/m^3}{m(m/m^3)}\cong m/m^2$ which is also a $\mathbb{C}$-module isomorphism. I think, we can conclude that $\dim_{\mathbb{C}}m/m^3 < \infty.$ Therefore, $\dim_{\mathbb{C}}(R/m^3)< \infty.$ The same holds for $n$ with the required inductive steps. 
