# Algebraic curves, intersection

Given the two plane curves, $$F=2X_0^2X_2-4X_0X_1^2+X_0X_1X_2+X_1^2X_2$$ $$G=4X_0^2X_2-4X_0X_1^2+X_0X_1X_2-X_1^2X_2$$ I want to calculate the multiplicity of the intersection at $(1:0:0)$, but I have no idea how to compute it. I think I understood the theory but when it comes to calculate an example I don't know how to start.

The answer should depend on your definition of intersection multiplicity. For curves, one of the workable definitions is $$\dim_k k[x,y]_{\mathscr p}/(f,g)$$ where $\mathscr p$ is the maximal ideal corresponding to the point we're interested in.
Since your curve is projective, and we are supposed to look at the point $(1:0:0)$, we look in the affine chart $U_0$ by setting $X_0=1$. Thus the relevant ideal is generated by $$f = 2x_2-4x_1^2+x_1x_2+x_1^2x_2$$ and $$g = 4x_2-4x_1^2+x_1x_2-x_1^2 x_2.$$ In this chart we are looking at their interesection multiplicity at the point $(0,0) \subset \mathbb A^2$. So we lokalize at the prime ideal $\mathscr p = (x,y)$. This means that we can divide by anything not in $\mathscr p$.
Note that $f-g = -2x_2+2x_1^2x_2$ and that we always have $(f,g)=(f,f-g)$. Also note that $f-g=x_2(2x_1^2-2)$. The second factor is a unit in our localized ring, ring so that $(f,g)=(f,x_2)$. This allows for simplifying $f$ by subtracting all terms involving $x_2$.
We find that $(f,g)=(x_1^2,x_2)$. This implies that $$\dim_k k[x,y]_{(x,y)}/(f,g) = 2.$$