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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!

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  • $\begingroup$ What do you mean by number of intersections? Do you mean the number of common zeros of $F$ and $G$? It is true that there are $12$ points of intersection, when counted with multiplicities, but in your case those $12$ points are not distinct. $\endgroup$ – Nils Matthes Apr 27 '13 at 8:16
  • $\begingroup$ My final objetive is to find how many points are in this intersection and then their multiplicities. I need this to apply Hurwitz's Formula and calculate the genus of the Klein curve. $\endgroup$ – Marra Apr 27 '13 at 13:41
  • $\begingroup$ @GustavoMarra Dear Gustavo, I don't know if you receive any notifications after an answer is updated, but in any case, since you said in the above comment that you need the multiplicities of the points of intersection, at the suggestion of Georges Elencwajg I added the computations of the multiplicities to my previous answer. $\endgroup$ – Adrián Barquero Apr 28 '13 at 18:00
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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.

enter image description here

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 $1$.

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    $\begingroup$ Congratulations for getting your hands dirty with actual computations, dear Adrián: +1 $\endgroup$ – Georges Elencwajg Apr 27 '13 at 17:54
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    $\begingroup$ ...and your answer yould be even better if you gave a succinct proof that the multiplicity of intersection is $2$ at infinity, $3$ at the origin $[0,0,1]$ and $1$ at each of the remaining $7$ points. $\endgroup$ – Georges Elencwajg Apr 27 '13 at 17:57
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    $\begingroup$ Dear Adrián, your calculation is perfect and I much prefer detailed , explicit and honest computations to enigmatic pronouncements. Let's leave conciseness to haikus! $\endgroup$ – Georges Elencwajg Apr 28 '13 at 18:03
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    $\begingroup$ @AdriánBarquero I think you have a mistake: $ x + z^3 + zx^3 - \frac{1}{3}(3z^2 + x^3)$ is not $x - x^3 /3 + zx^3$. So it is not so simple $\endgroup$ – user128245 Feb 13 '18 at 0:36
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    $\begingroup$ @Goodnighyn Dear Goodnighyn, thanks for spotting that mistake. You are indeed right, the intended calculation had been $x + z^3 + zx^3 - \frac{z}{3}(3z^2 + x^3)$, which hopefully I got right this time. The rest of the calculation remained essentially the same, so no extra complications arose out of my mistake ;-) $\endgroup$ – Adrián Barquero Feb 13 '18 at 5:36
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By Bezout's theorem the curves will intersect in $\text{deg}(F)\cdot \text{deg}(G)=12$ points when counted with multiplicity.

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