(i) Show that given any five points in $ℂℙ^2$ there is at least one conic passing through them.
Solution.
A conic is a curve of the form $a_1 x^2+a_2 x y+a_3 x z+a_4 y^2+a_5 y z+a_6 z^2$, and has 6 coefficients. Given 5 points in $ℙ^2$, each point specifies a linear constraint on the coefficients, and since there are 5 dimensions, by rank-nullity, dimension of image ≤5, so dimension of kernel≥1.
(ii) Let $C$ be a quartic curve (i.e. of degree 4), with four singular points. By choosing an appropriate conic $D$ and using the strong form of Bézout's theorem (involving intersection multiplicities $I_p(C, D)$) prove that $C$ must be reducible.
Solution.
Let $C$ be a projective curve of degree 4 in $ℙ^2$ with four singular points. By (i) exist a conic $D$ containing the 4 singular points and another point $p$ of $C$. By Bézout's theorem, if the two curves have no common component, then they have 8 points of intersection counting multiplicities. Since the 4 singular points have multiplicity greater than 1, and $C$ and $D$ have an additional point $p$ of intersection, hence the sum of the intersection multiplicities is at least 9. Thus $C$ and $D$ must have a common component, which implies $C$ is reducible.
(iii) Show that $y^4-4xzy^2-xz(z-x)^2=0$ is a quartic with three singular points.
Solution.
$P_x=-(3x^2+4y^2-4xz+z^2)z$
$P_y=4(y^2-2xz)y$
$P_z=-(x^2+4y^2-4xz+3z^2)x$
$P_x-P_z=(x - z) (x^2 - 6 x z + 4 y^2 + z^2)=0\implies z=x$ or $x^2 - 6 x z + 4 y^2 + z^2=0$
If $x=z$ then $P_x=0\implies y^2z=0\implies y=0$ [If $z=0$, by $P_y=0$, also $y=0$], solution is $[1,0,1]$.
If $x^2 - 6 x z + 4 y^2 + z^2=0$…(*), substitute $4y^2+z^2=-x^2+6xz$ into $P_x=0$
$$-(x+z)2xz=0$$
so $x=-z$ [If $xz=0$ then $P_y=0\implies y=0$, but $[1,0,0]$ or $[0,0,1]$ don't satisfy (*).]
$P_y=0\implies y=0$ or $±\sqrt{-2}x$, but $[1,0,-1]$ don't satisfy (*), so solution is $[1,±\sqrt{-2},-1]$.
So three singular points $[1,0,1],[1,±\sqrt{-2},-1]$.
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