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.

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 $1-xy$).
