Intersection of smooth projective plane curves

Question

I need to calculate the number of intersections of the smooth projective plane curves defined by the zero locus of the homogeneous polynomials $$ F(x,y,z)=xy^3+yz^3+zx^3\text{ (its zero locus is called the }Klein\text{ }Curve\text{)} $$ and $$ G(x,y,z)=\dfrac{\partial F}{\partial z}=3z^2y+x^3 $$ It is easily seen that both are homogeneous and non-singular (therefore irredutible). I believe that the number of intersections is 4, but I have no base in algebraic curves theory to prove that!

Answer 1

I presume you're working over the complex numbers, and in that case, as has already been mentioned, by Bezout's theorem, there are 12 intersection points, counted with their multiplicities.

But in this particular case, it is not that hard to find the actual points of intersection as follows. First we find the points at infinity, i.e., the points $[X, Y, 0] \in \mathbb{P}^2$.

Points at Infinity

Let $Z = 0$, then your equations become

$$ \begin{cases} XY^3 &= 0\\ X^3 &= 0 \end{cases} \implies X = 0,\, Y\neq 0 $$ Thus we get only one intersection point at infinity, namely the point $[0, Y, 0] = [0, 1, 0]$.

Points in the affine plane $\mathbb{A}^2 = \{ [X, Y, Z] \in \mathbb{P}^2 \mid Z \neq 0 \}$

Now, suppose that $Z \neq 0$, so that we can dehomogenize the equation with respect to $Z$. Hence we consider the system

$$ \begin{cases} f(x, y) = F(x, y, 1) = xy^3 + y + x^3 &= 0\\ g(x, y) = G(x, y, 1) = 3y + x^3 &= 0 \end{cases} $$

The from the second equation you can solve for $y$ in terms of $x$, then plug that in the first equation to get a degree 10 equation in $x$, which luckily is easy to solve. In the end, assuming that I didn't make any mistakes, you get the "affine" points $[0, 0, 1]$ and

$$ \left[ \sqrt[7]{18} \zeta_7^k, -\frac{\sqrt[7]{18^3} \zeta_7^{3k} }{3}, 1 \right] $$

where $k = 0, 1, 2, 3, 4, 5, 6$ and $\zeta_7 = e^{2\pi i /7}$ is a primitive seventh root of unity.

Thus, if everything is correct, there are $9$ different points of intersection.

The following plot shows the real intersection points $(0, 0)$ and $\left(\sqrt[7]{18}, -\frac{\sqrt[7]{18^3}}{3} \right)$. It appears that the origin $(0, 0)$ is an inflection point, so it's multiplicity would be at least 3.

Computation of the multiplicities

We will compute the multiplicities of the $9$ points of intersection listed above. As Georges Elencwajg comments below, it will turn out that the multiplicity of the origin $[0, 0, 1]$ is $3$, the multiplicity of the point at infinity $[0, 1, 0]$ is $2$ and the multiplicity of the remaining points $P_k := \left[ \sqrt[7]{18} \zeta_7^k, -\frac{\sqrt[7]{18^3} \zeta_7^{3k} }{3}, 1 \right]$ is $1$.

First recall that the multiplicity of the intersection of the curves $C_F: F = 0$ and $C_G: G = 0$, denoted by $I_P(C_F, C_G)$, at a point $P$ is given by the dimension as a $\mathbb{C}$-vector space of the following quotient ring

$$ I_P(C_F, C_G) = \dim_\mathbb{C}{ \mathcal{O}_{ \mathbb{P}^2, P} } / \langle F, G \rangle $$

where $\mathcal{O}_{ \mathbb{P}^2, P}$ is the local ring of $\mathbb{P}^2$ at $P$.

Multiplicity of the origin $[0, 0, 1]$

An affine neighborhood of the origin is obtained by dehomogenizing with respect to $Z$. The corresponding point in the affine neighborhood is $(0, 0)$, so that the local ring is $\mathcal{O}_{ \mathbb{P}^2, [0, 0, 1]} = \mathbb{C}[x, y]_{(x, y)} = \left \{ \frac{a(x, y)}{b(x, y)} \in \mathbb{C}(x, y) \mid b(0, 0) \neq 0 \right \}$. Now we must quotient by the ideal $\langle xy^3 + y + x^3, 3y + x^3 \rangle$ in $\mathbb{C}[x, y]_{(x, y)}$. Observe that

$$ \begin{array} (\langle xy^3 + y + x^3, 3y + x^3 \rangle &= \langle xy^3 + y + x^3 - (3y+ x^3), 3y + x^3 \rangle = \langle xy^3 - 2y, 3y + x^3 \rangle\\ & = \langle y\underbrace{(xy^2 - 2)}_{\text{unit in $\mathbb{C}[x, y]_{(x, y)}$}}, 3y + x^3 \rangle = \langle y, 3y + x^3 \rangle = \langle y, x^3 \rangle \end{array} $$

Hence the elements $1, x, x^2$ form a basis of the $\mathbb{C}$-vector space $\mathcal{O}_{ \mathbb{P}^2, [0, 0, 1]} / \langle y, x^3 \rangle$, so that the intersection multiplicity $I_{[0, 0, 1]} (C_F, C_G) = 3$.

Multiplicity of the point at infinity $[0, 1, 0]$

Now an affine neighborhood can be obtained by dehomogenizing with respect to $Y$. The corresponding point in the affine neighborhood is $(0, 0)$ again, but now the local ring $\mathcal{O}_{ \mathbb{P}^2, [0, 1, 0]} = \mathbb{C}[x, z]_{(x, z)}$. The nonhomogeneous polynomials are now $h(x, z) = F(x, 1, z) = x + z^3 + zx^3$ and $t(x, z) = G(x, 1, z) = 3z^2 + x^3$.

Then the ideal $\langle h, t \rangle$ can be simplified as follows.

$$ \begin{array} (\langle h, t \rangle &= \langle x + z^3 + zx^3 , 3z^2 + x^3 \rangle = \langle x + z^3 + zx^3 - \frac{z}{3}(3z^2 + x^3), 3z^2 + x^3 \rangle \\ &= \langle x + \frac{2}{3}zx^3 , 3z^2 + x^3 \rangle = \langle x\underbrace{( 1 + \frac{2}{3}zx^2)}_{\text{unit in $\mathbb{C}[x, z]_{(x, z)}$}} , 3z^2 + x^3 \rangle \\ &= \langle x, 3z^2 + x^3 \rangle = \langle x, z^2 \rangle \end{array} $$

Therefore a $\mathbb{C}$-basis for the vector space ${ \mathcal{O}_{ \mathbb{P}^2, [0, 1, 0]} } / \langle x, z^2 \rangle$ is given by the elements $1, z$, so that its dimension is $2$ and hence the intersection multiplicity is $I_{[0, 1, 0]} (C_F, C_G) = 2$.

Multiplicities of the points $P_k$

Finally, since by Bezout's theorem we know that

$$ \sum_{P \in C_F \cap C_G} I_P(C_F, C_G) = 12 $$

and we have

$$ \begin{array} .\sum_{P \in C_F \cap C_G} I_P(C_F, C_G) = I_{[0, 0, 1]}(C_F, C_G) + I_{[0, 1, 0]}(C_F, C_G) + \sum_{k = 0}^{6} I_{P_k}(C_F, C_G)\\ = 3 + 2 + \sum_{k = 0}^{6} I_{P_k}(C_F, C_G) \end{array} $$

then this means that each of the seven points $P_k$ has multiplicity $1$.

Answer 2

By Bezout's theorem the curves will intersect in $\text{deg}(F)\cdot \text{deg}(G)=12$ points when counted with multiplicity.