multiplicity of the intersections between $p=x_0x_1^2+x_1x_2^2+x_2x_0^2\,\,$ and $\,\,q=-8(x_0^3+x_1^3+x_2^3)+24x_0x_1x_2$ in $\mathbb{P^2(K)}$

Question

in $\mathbb{P^2(K)}$ where $\mathbb{K}$ is an algebraically closed field and $[x_0,x_1,x_2]$ the homogeneous coordinates, consider the following (homogeneous) polynomials:

$p=x_0x_1^2+x_1x_2^2+x_2x_0^2\,\,$ and $\,\,q=-8(x_0^3+x_1^3+x_2^3)+24x_0x_1x_2$

find the number and the multiplicity of the intersections between $p\,$ and $q$.

For the Bézout's theorem the number of intersections counted with their multiplicity is the product of the degrees of the two polynomials, which is $9$. Now I want to use the main theorem of elimination theory to eliminate one of the coordinates, in order to work on a single polynomial of degree $9$ in two variables. But I can't apply it because $p(1,0,0)=p(0,1,0)=p(0,0,1)=0$ and I need that one of those points doesn't lie on both of the curves to eliminate the corresponding coordinate.

Any suggestions? Maybe an appropriate automorphism of $\mathbb{P^2(K)}$ to workaround.

I computed the elimination of $x_0$ and the polynomial I obtained has the right degree: $-64x_1^9+192x_1^6x_2^3+1536x_1^3x_2^6+64x_2^3 \, = \, -64(x_1^3-3x_1x_2^2-x_2^3)(x_1^6+3x_1^4x_2^2-2x_1^3x_2^3+9x_1^2x_2^4-3x_1x_2^5+x_2^6)$

but $x_1=0$,$\,\,x_2=0$ is a partial solution and it doesn't correspond to any solution of the system.

Answer 1

First, suppose $x_0=0$. Then $p(0,x_1,x_2)=x_1x_2^2$ and $q(0,x_1,x_2)=-8(x_1^3+x_2^3)$, which intersect at $x_1=x_2=0$. Thus any intersection must occur on the affine chart $\{x_0\neq0\}$. In fact, one can generalize by symmetry: the intersections lie in the chart $\{x_0x_1x_2\neq0\}$. For this reason, I think your analysis of $p(1,0,0)$ etc. does not actually obstruct the main theorem. But converting multiple equations to a single high-degree polynomial usually makes things worse, not better.

Instead, follow Avnish Singh's advice to divide both equations by $x_0x_1x_2$. Then we have \begin{gather*} 0=\frac{x_1}{x_2}+\frac{x_2}{x_0}+\frac{x_0}{x_1} \\ 0=8\left(3-\left(\frac{x_0}{x_1}\frac{x_0}{x_2}+\frac{x_1}{x_0}\frac{x_1}{x_2}+\frac{x_2}{x_0}\frac{x_2}{x_1}\right)\right) \end{gather*} Now change variables: let $(\alpha,\beta,\gamma)=\left(\frac{x_0}{x_1},\frac{x_1}{x_2},\frac{x_2}{x_0}\right)$. Then \begin{gather*} 1=\alpha\beta\gamma \\ 0=\alpha+\beta+\gamma \\ 3=\frac{\alpha}{\gamma}+\frac{\beta}{\alpha}+\frac{\gamma}{\beta} \end{gather*}

Now let $(l_0,l_1,l_2)=\left(\frac{\alpha}{\gamma},\frac{\beta}{\alpha},\frac{\gamma}{\beta}\right)$. Then since $\gamma\neq0$, \begin{gather*} 1=l_0l_1l_2 \\ 1=l_0^2l_1\gamma^3 \\ 0=l_0+l_1l_0+1 \\ 3=l_0+l_1+l_2 \end{gather*} which are easy enough to solve by hand.

Answer 2

Equating both sides, and then bringing 24*x$_0*$x$_1*$x$_2$ to the left side and then dividing both sides by x$_0$x$_1$x$_2$ gives us

$24- [\frac{x_0}{x_1} + \frac{x_1}{x_2} + \frac{x_2}{x_0}] = 8[\frac{x^2_0}{x_1*x_2} + \frac{x^2_1}{x_0*x_2} + \frac{x^2_2}{x_0*x_1}]$

Consider the first bracket terms as m and the second brackets term as n.

Now, 24 - m = 8n

Also, using AM $\ge$ GM, we can find out the minimum values of m and n to be 3.

Now take m = 3 + i where i $\ge$ 0 and n =3 + j where j $\ge$ 0.

Putting these values in the above equation, we get

$24 - 3 - i= 8(3+j) $

$\implies -i -8j= 3$

Now, where i and j are positive values , the above equation has no solutions.

Thus the solutions exist only when we can't equate AM $\ge$ GM that is when $x_1$ or $x_2$ or $x_3$ = $0$