Calculating intersection number of $(x^2+y^2)^3-4x^2y^2=0 $ and $x=0$ at $(0,0)$ I am trying to find the intersection number of  $(x^2+y^2)^3-4x^2y^2=0 $ and $x=0$ at $(0,0)$. The intersection number of $F$ and $G$ is defined to be $dim_k(O_p/(F,G))$(Here $O_p$ is the local ring at P). I do not know how to calculate this. If someone can help it would be great. Thanks.
 A: This is one of those things where everyone expects you to grok some feature of commutative algebra without ever teaching or explaining it to you.
In particular, suppose we have a ring $R$, an ideal $I$, and a prime ideal $\mathfrak{p} < R$.  Then by definition we have an exact sequence
$$0 \to I \to R \to R/I \to 0.$$
Tensoring by $R_\mathfrak{p}$, we have an exact sequence
$$0 \to I \otimes_R R_\mathfrak{p} \to R_\mathfrak{p} \to (R/I) \otimes_R R_\mathfrak{p} \to 0$$
of $R_\mathfrak{p}$-modules, since localization is exact.  In other words,
$$\frac{R_\mathfrak{p}}{I \otimes_R R_\mathfrak{p}} \cong \frac{R}{I} \otimes_R R_\mathfrak{p}.$$
Suppose $I$ is a finitely generated ideal $I = (f_1, \ldots, f_n)$.  This is the same as saying that
$$I = \sum_{i=1}^n f_i R,$$
and moreover we have
$$I \otimes_R R_\mathfrak{p} = \sum_{i=1}^n f_i R_\mathfrak{p}.$$
So in this case the equation above says that
$$\frac{R_\mathfrak{p}}{f_1 R_\mathfrak{p} + \cdots + f_n R_\mathfrak{p}} \cong \frac{R}{f_1 R + \cdots + f_n R} \otimes_R R_\mathfrak{p}.$$
This should let you use your knowledge of $k[x,y]/(F,G)$ to understand $\mathcal{O}_P/(F,G)$.
