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I am trying to verify Bezout's theorem for the two following curves :

C: $y^2=x^3-x^2$ and

D:$y=x^2+x$

I found the intersection points: (0,0), ($ \beta_{\pm}$,-2) where $\beta_\pm=\frac{-1 \pm \sqrt 7 i}{2}$.

1) How do I find the intersection at the infinity line? I need to "dehomogenize" but I'm not sure what this means or how to do it.

2) I Calculated $I( (0,0), C \cap D)= I(y^2-x^3+x^2, y-x^2-x)=2 + I(2+y,y-x^2-x)$. I have been told $I(2+y,y-x^2-x)=0, $but I dont know why.

thanks,

1 Answers 1

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You can homogenise the equations to be $$y^2z=x^3-x^2z$$ and $$yz=x^2+xz$$ setting $z=1$ you have found the solutions. Setting $z=0$ we get $$x^3=0$$and $$x^2=0$$ so there is one solution at infinity $[0,1,0]$.

For the solution at the origin $(0,0)$ we see that the tangents at the origin are given by the lowest order terms in the nonhomogeneous equations, which give us $$y^2+x^2=0$$ or $$x+iy=0\text{and }x-iy=0$$ two tangent lines for the first equations and one tangent line $x-y=0$ for the second equation. None of these lines are equal and thus the intersection at the origin is $2$. To find the intersection multiplicity at the point at infinity we can set $y=1$ and then apply your technique to the equations

$$z=x^3-x^2z$$ and $$z=x^2+xz$$

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    How did you know where to put the z's in order to homogenize? i.e. how do you know that it becomes $y^2 z =x^3-x^2 z$ and $yz=x^2+xz$? Also, I understand how you found the tangent lines $(x+iy), (x-iy)$ and $(x-y)$, but I dont see how the intersection at the origin is 2 and why does it mean $I(2+y,y−x^2−x)=0?$ – 2017-01-21
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    I just use z to bump up the degrees to be all the same. That last thing is zero cause the origin is not a point of intersection. – 2017-01-21