Checking the intersection multiplicity for two curves Hi guys I am using Fulton's book on algebraic curves as a main material to study. I am trying to check the intersection multiplicity of $p=(0,0)$ of $g(x,y)=y^2-x^3-x^2$ and $h(x,y)= y^2-x^3+x$
My approach is inspired by chapter 3.3 of Fulton. 
First $Mult_p(g)=2$ because the we are looking at the origin and thus the multiplicity is the power of the smallest order term of the polynomial. Similarly $Mult_p(h)=1$.
Now I make the following sequence:
$\mathbb{C}[x,y]/I^2 \times \mathbb{C}[x,y]/I \rightarrow \mathbb{C}[x,y]/I^3 \rightarrow \mathbb{C}[x,y]/(g,h,I^3)$.
Where $I=\langle x,y \rangle$
Where we can call the first arrow $\psi(A,B)=Ag+Bh$ and the second to be the projection map $\pi$. Now we notice that the $\ker \pi = \dim \mathbb{C}[x,y]/I^2 + \dim \mathbb{C}[x,y]/I$ iff and only if $\psi$ is injective.
So that is what we aim to show. Notice that by Fulton's algebraic curves $I_p(p; F \cap G)= I_p(p; F \cap G+ AF$ and $I_p(p; F \cap G)=I_p(p; G \cap F)$. Thus we can simplify both g and h and doing so I got $g'= g-h=x^2+x$ and $h'= h+xg'-g'=y^2$
Thus we want to show $\ker p= \{Ag' +Bh'\}= I^3$.
My question is this correct so far, am I on the right track. Also is there a overall better way to find the intersection multiplicity.
 A: Here are two ways of computing the intersection number using Fulton's approach.  Fulton lists 7 axioms that the intersection number satisfies on pp. 36-37.
The quick way: Axiom (5) states that $I_P(g \cap h) \geq m_P(g) m_P(h)$ with equality if $g$ and $h$ have no common tangents at $P$.  Examining the homogeneous terms of lowest order, we find
$$
y^2 - x^2 = (y-x)(y+x) \implies g \text{ has tangents } y-x, y+x\\
$$
and $h$ has the single tangent $x$ at $P$.  Using the multiplicities you've already computed, we find $I_P(g \cap h) = 2 \cdot 1 = 2$.
Another way: We can use axiom (7), as you suggested.  Note that
$$h - g = y^2 - x^3 + x - (y^2 - x^3 - x^2) = x^2 + x \, .$$
Using axioms (7) and (6), we find
\begin{align*}
I_P(g \cap h) &= I_P(y^2 - x^3 - x^2 \cap y^2 - x^3 + x) \overset{(7)}{=} I_P(y^2 - x^3 - x^2 \cap x^2 + x)\\
&\overset{(7)}{=} I_P(y^2 \cap x(x+1)) \overset{(6)}{=} I_P(y^2 \cap x) + I_P(y^2 \cap x + 1) \overset{(2)}{=} I_P(y^2 \cap x) = 2 \, .
\end{align*}
(Since $P$ doesn't lie on the intersection of $y^2$ and $x+1$, which consists only of the point $(-1,0)$, then $I_P(y^2 \cap x+1) = 0$ by axiom (2).)
