# A problem of four curves

This is a generalization of my previous question, a problem of a cubic and six conics.

Let a curve $(K)$ of degree $m$ and three curves $(C_i)$ of degree $n$, for $i=1,2,3$. Let $(C_1)$ meets $(K)$ at $mn$ points. Let $(C_2)$ meets $(K)$ at $mn$ points. Let $(C_3)$ meets $(K)$ at $mn$ points.

Let $(C_1), (C_2)$ and $K$ have $d$ common points. Let points $P_1, P_2, …,P_{mn-d}$ lie on $(C_1)$ and $(K)$ but don’t lie on $(C_2)$.

Let $(C_2), (C_3)$ and $K$ have $mn-d$ common points. Let points $Q_1, Q_2, Q_3,….,Q_d$ lie on $C_3$ and $(K)$ but don’t lie on $C_2$.

The problem: show that the $mn$ points $P_1, P_2, …,P_{mn-d}, Q_1, Q_2, Q_3,….,Q_d$ lie on a curve of degree $n$, where $mn-d < \frac{n^2+3n}{2}$ and $d < \frac{n^2+3n}{2}$.

The first figure I, make with $m=3$, $n=2$ and $d=2, mn-d=4$ The second figure I, make with $m=4$, $n=2$ and $d=4, mn-d=4$. In this case 8 points lie on a new conic.

The third figure, I make with $m=4$, $n=3$ and $d=6, mn-d=12-6=6$. In this case, I make curve $C_1, C_2, C_3$ = three cubics, one cubic = three line, the curve $(K)$ = $4$ line, the new curve is a conic through $8$ points and a line share a same line of cubic 1.

• I thank to dear PhD. @Allen Knutson for your edit. – Oai Thanh Đào Feb 29 '16 at 17:17
• Please do not use the tag geometry. It is deprecated. – user9072 Mar 13 '16 at 16:15

This is a very easy exercise if $K$ is nonsingular. Experts will probably tell what does/does not go wrong if singularities are present.
Under the assumptions, $m<n+3$, hence $n-m>-3$ and $H^1(\mathbb{P}^2;\mathcal{O}_{\mathbb{P}^2}(n-m))=0$. From the classical exact sequence $$0\to\mathcal{O}_{\mathbb{P}^2}(n-m)\to\mathcal{O}_{\mathbb{P}^2}(n)\to\mathcal{O}_{K}(n)\to0,$$ the restriction map $H^0(\mathbb{P}^2;\mathcal{O}_{\mathbb{P}^2}(n))\to H^0(K;\mathcal{O}_{K}(n))$ is surjective. On the other hand, your $mn$ points are in the linear system $|C_1+C_3-C_2|$ (restricted to $K$), which represents precisely the sections of $\mathcal{O}_{K}(n)$.
• It is true if $K$ is nonsingular. I think it is always true, but then you should be careful with counting points and deciding which ones are common. Also, after another thought, I don't see why you need the degree inequalities, as $H^1(\mathbb{P}^2;\mathcal{O}_{\mathbb{P}^2}(d))=0$ for any $d$. – Alex Degtyarev Apr 9 '16 at 8:10