Suppose $k$ is an algebraically closed field and $C,C'$ two affine curves defined by respectively $F,G\in k[X,Y]$ which don't have common factors. Suppose $x=(a_1,a_2)\in C\cap C'$. Note $I_x=( X-a_1,Y-a_2)$ the maximal ideal defining $x$. We define $\mathcal{O}_{\mathbb{A}^2,x}$ as the localization of $k[X,Y]$ with respect to the multiplicative set $S=k[X,Y]- I_x$.
Every reference I looked defines the multiplicity of the intersection of $C$ and $C'$ to be $\dim_k \mathcal{O}_{\mathbb{A}^2,x}/( F,G)$. My question is: what is precisely $(F,G)$ in this writing?
If $(F,G)$ is simply the ideal generated by $F,G$ in $k[X,Y]$ everything is ok because it's a sub-$k$-vector space of $\mathcal{O}_{\mathbb{A}^2,x}$ and we can happily count the dimension of the quotient space which is nearly always infinite in my opinion.
More sense would be to replace the ideal $(F,G)$ of $k[X,Y]$ with its localisation with respect to the set $S$. This would make our quotient a $k$-algebra and moreover we can use the fact that $S^{-1}M/S^{-1}N\cong S^{-1}(M/N)$ and we see that actually the quotient is actually $\mathcal{O}_{C\cap C',x}$ which we can define as a localisation of $k[X,Y]/(F,G)$ with respect to the set $k[X,Y]/(F,G)-I_x/(F,G)$.
So, is this problem only a question of definition and the two versions are isomorphic? Or the authors simply omit this detail?
Let $C, C'$ be two curves in $\mathbb A^2$ given by the polynomials $F, G \in k[x,y]$ and let $P \in \mathbb A^2.$ We define the intersection multiplicity of $C,C'$ at the point $P$ to be $I(P, C, C'):= \dim_k\mathcal O_{\mathbb A^2, P}/(F,G)\mathcal O_{\mathbb A^2, P}.$ (Here $(F,G)\mathcal O_{\mathbb A^2, P}$ means the ideal in $\mathcal O_{\mathbb A^2, P}$ generated by the images of the polynomials $F,G$ in $\mathcal O_{\mathbb A^2, P}.$)
Fact: $I(P, C, C') = \infty$ if and only if $P$ lies in a common component of $C$ and $C'.$
In this case, since it is given that there is no common component of $C$ and $C',$ we will only prove that $I(P, C, C') < \infty.$ Any prime ideal $\mathfrak p$ in $\mathcal O_{\mathbb A^2, P}$ containing $F$ and $G$ (here we are using same notation for $F,G$ and their images in $\mathcal O_{\mathbb A^2, P}$) must contain an irreducible factor of each of these polynomials, say $f$ and $g$ corresponding to $F$ and $G$ respectively. Now both $f$ and $g$ generates non-zero prime ideals contained in $\mathfrak p$ and these two prime ideals are distinct (otherwise the curves $C$ and $C'$ will have a common component). This shows that height of $\mathfrak p$ is at least two. On the other hand, Krull-dimension of $\mathcal O_{\mathbb A^2, P}$ is two. Consequently $\mathfrak p$ is a maximal ideal. So the local ring $\mathcal O_{\mathbb A^2, P}/(F,G)\mathcal O_{\mathbb A^2, P}$ has a unique prime ideal, namely the maximal ideal itself. Hence it is Artinian and thus a finite dimensional vector space over $k.$