Firstly, I apologise that this question is specific to two polynomials. I understand that this post will not help a lot of people, and for that I am sorry. I will make it up to you all by anwsering lots more questions in the future! However, as my lecturers enjoy time off on the weekend, I am unable to ask this to anyone else who would know what the heck I'm talking about. Sorry if my notation is not widely used, I can elaborate more if required. Please excuse my poor Latex too.
Let:
$B = Y^2 - X^3 + X$, and
$F = (X^2 + Y^2)^3 - 4X^2Y^2$.
I wish to find the intersection number of $B$ and $F$ at $P = (0,0)$
i.e $I(P,F \cap B). $
However, I want to use the following proposition to do this:
If $P$ is simple on $B$ then $I(P,F \cap B) = ord_P^B(F) $.
I will omit proving that $P$ is simple on $B$ as I hope it is clear to see it is.
I know that $I(P,F \cap B) > M_p(F) * M_p(B) = 1 * 4 $, where $M_P(G)$ is the multiplicity of $P$ on $G$.
Now, consider the Image of $X$ and $Y$:
$K[X,Y] \longrightarrow \frac{K[X,Y]}{Y^2 - X^3 + X} \longrightarrow O_P(Y^2 - X^3 + X)$
$X \longrightarrow x$
$Y \longrightarrow y$
Where lower case denotes the image.
Then $ord_P^{Y^2 - X^3 + X}((X^2 + Y^2)^3 - 4X^2Y^2)$
Can be found by substituting $y^2 = x^3 - x$ into $(x^2 + y^2)^3 - 4x^2y^2$
Which gives:
$ord_P^{Y^2 - X^3 + X}(x^9+3x^8 -5x^6 -4x^5 +3x^4 +3x^3)=3$
i.e $I(P,F \cap B) = 3$, a contradiction to $I(P,F \cap B) > 4$ as above.
The correct answer I know is 6, so where have I gone wrong?
The key to the solution is that the curve $C=V(B)$ is indeed smooth at $P$ and that a uniformizing parameter there is $y$, since $\frac {\partial B}{\partial x}(P)\neq0$ .
On the other hand the function $x$ satisfies $ord_P(x)=2$ since $x=-y^2+x^3$ and moreover $x^2=y^4+\cdots$, where $\cdots$ are higher order terms.
Then the order of $(x^2 + y^2)^3 - 4x^2y^2=(y^4+\cdots+y^2)^3 -4(y^4+\cdots)y^2=-3y^6+\cdots$ is $6$, just as you seemed to know.