Bézout's theorem for intersection of curves

I want to solve the following problem.

Verify Bézout's theorem for the curves $(X^{2}-Y)^{2}-X^{5}=0$ and $X^{4}+X^{3}Y-Y^{2}=0$.

Let $\mathcal{C}=\mathrm{V}(F)$ and $\mathcal{C}'=\mathrm{V}(G)$, with $F=(X^{2}-YZ)^{2}Z-X^{5}$ and $G=X^{4}+X^{3}Y-Y^{2}Z^{2}$. We compute the resultant: $$R_{F,G}^{X}=Y^{11}Z^{9}-8Y^{10}Z^{10}-16Y^{9}Z^{11}=$$$$=Y^{9}Z^{9}(Y^{2}-8YZ-16Z^{2})=Y^{9}Z^{9}(Y-(4+4\sqrt{2})Z)(Y-(4-4\sqrt{2})Z)$$

There are four intersection points and their multiplicities must add $20$.

Bézout's theorem says that $$\sum _{P\in P^2} \text{mult}_P(\mathcal C \cap \mathcal C ')=\text{degree}(F)\cdot \text{degree}(G)$$

So it verifies Bèzout's theorem:$$20=\sum _{P\in P^2} \text{mult}_P(\mathcal C \cap \mathcal C ')=\text{degree}(F)\cdot \text{degree}(G)=5\cdot 4 = 20$$

Edit: I computed the resultant in a wrong way, so the result didn't make sense. Now it is ok.

• The version of Bezout's theorem I know says that that equality holds for projective curves, but your curves aren't given by homogeneous equations. Perhaps I am also misunderstanding something. – Alfred Yerger Jan 16 '18 at 2:04
• Shouldn't the $X$-resultant a homogeneous polynomial in $Y,Z$? Clearly $X=Z=0, Y=1$ is a solution, not reflected in your resultant. – Mohan Jan 16 '18 at 2:37
• Yes, I can homogenize the two polynomials and I did it. – Miguel Mars Jan 16 '18 at 2:38
• I computed the resultant using Maple. Maybe I did it wrong. – Miguel Mars Jan 16 '18 at 2:39
• (Ok, mystery solved: I was evaluating in $Z=1$ by mistake. Sorry) – Miguel Mars Jan 16 '18 at 2:45