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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.

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This answer is maybe more than you asked for, but try to do all the computations yourself. Try to stop reading at some point, then guess the next step.

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 $$ \dim_k k[x,y]_{(x,y)}/(f,g) = 2. $$

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