Intersection multiplicity 
Let $f=y^2-x^3$ and $g=y^3-x^7$. Calculate the intersection multiplicity of $f$ and $g$ at $(0,0)$. 

I know the general technique for this (passing to the local ring) but I having difficulty with the fact that $3,7,2$ have no common factors.
 A: Using the properties of intersection number can save much time for the
calculation of intersection numbers.
If $P$ is a point, two affine plane curves $F$ and $G$ have no common
components, then: (see section 3.3 in Fulton's algebraic curves)


*

*$I(P,F\cap G)=I(F\cap (G+AF))$ for any $A\in k[X,Y]$;

*$I(P,F\cap G)=m_p(F)m_P(G)$, when $F$ and $G$ have not tangent
lines in common at $P$. 

*$I(P,F\cap GH)=I(P,F\cap G)+I(P,F\cap H)$ for $H\in k[X,Y]$.


Now, $P=(0,0)$, 
$I(P,y^2-x^3\cap y^3-x^7)=I(P, y^2-x^3\cap
((y^3-x^7)-(y^2-x^3)x^4))$
$\quad =I(P,y^2-x^3\cap y^2)+I(P,y^2-x^3\cap y-x^4).$
And $I(P, y^2-x^3\cap y^2)=I(P,x^3\cap y^2)=3\times 2=6$.
$I(P, y^2-x^3\cap y-x^4)=I(P, y^2-x^3\cap
((y-x^4)-(y^2-x^3)x))$
$\quad =I(P,y^2-x^3\cap y(1-xy))=3$. 
Thus $9=6+3$ is the
desired number.

$I(P,y^2-x^3\cap y(1-xy))=I(P,y^2-x^3\cap y)+I(P,y^2-x^3\cap (1-xy))$,
$I(P,y^2-x^3\cap y)=I(P,-x^3\cap y)=3$ and $I(P,y^2-x^3\cap (1-xy))=0$ ($P$ is not the zero of $1-xy$).
A: I think you no longer need an answer, but I’d like to write something here since it may help some other students.
We need four lemmas here(You can find them in lecture 2 of Andreas Gathmann’s notes in Plane Algebraic Curves):
Lemma (1): For any three plane curves F, G, H, $$\mu_p(F, G)=\mu_p(F, G+FH)$$
Lemma (2): For any three plane curves F, G, H, $$\mu_p(F, GH)=\mu_p(F, G)+\mu_p(F, H)$$
Lemma (3): $\mu_p(F, G)\ge 1$ iff $P\in F\cap G$.
Lemma (4): $\mu_p(F, G)=1$ iff $<F, G>=I_p$, where $I_p =\{f/g: f, g\in K[x, y], f(P)=0, g(P)\neq 0\}$.
Now we can compute $\mu_0(y^2-x^3, y^3-x^7)$.
$\begin{align}\mu_0(y^2-x^3, y^3-x^7)\overset{by (1)}{=}\mu_0(y^2-x^3, y^3-x^7-y(y^2-x^3))=\mu_0(y^2-x^3, x^3y-x^7)\overset{by (2)}{=}\mu_0(y^2-x^3, x^3)+\mu_0(y^2-x^3, y-x^4)\overset{by (1)}{=}\mu_0(y^2-x^3+x^3, x^3)+\mu_0(y^2-x^3-y(y-x^4), y-x^4))=\mu_0(y^2, x^3)+\mu_0(x^4y-x^3, y-x^4)\overset{by (2), (2), respectively}{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}6\mu_0(y, x)+\mu_0(x^3, y-x^4)+\mu_0(xy-1, y-x^4)\overset{by (4), (1), (3), respectively}{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}6+\mu_0(x^3, y-x^4+x(x^3))+0=6+\mu_0(x^3, y)\overset{by (2)}{=}6+3\mu_0(x, y)\overset{by (4)}{=}6+3=9\end{align}.$
