Projective cubic curve passing 9 points

This is a problem from our school's algebraic curve course midterm exam.

Let $p_1,\dots,p_9$ be nine distinct points on $\mathbb{CP}^{2}$. None of 3 of them lie on same line, and none of 7 of them lie on same conic.

(1) Prove that there exist a conic $C$ containing all these 9 points. (2) Prove that $C$ is irreducible. (3) Is $C$ unique?

Actually I proved (1) (using 9$\times$10 matrix) and (2) (using Bezout's theorem), but I can't solve (3). Intuitively, if we consider arbitrary two projective cubic curve, then they will meet at nine points and they will satisfy the condition of problem. But I cannot find explicit example.

I also tried to show that such $C$ is unique. Suppose not, then we can find linearly inedependent homogeneous cubic polynomials $P,Q$. Let $R=aP+bQ$, then $R\neq 0$ for any $(a,b)\neq (0,0)$ and we can choose $a,b$ so that $Z(R)$ contains $p_i$'s and $q$ which is different from $p_i$'s. But I cannot find $q$ which gives contradiction.

Is $C$ is unique or not? Thanks in advance.

• You're right: the intersection of two cubics in $\mathbf P^2$ shows that $C$ need not be unique. I don't know a nice way to choose two explicit cubics which will guarantee that no 3 of the points lie on a line, and no 7 lie on a cubic, but certainly such cubics exist. – Nefertiti Nov 2 '16 at 11:41

If you pick smooth cubic $C\subset \mathbb{P}^2$ and 8 general points $p_1,\ldots,p_8\in C$ and pick the 9th point $p_9\in C$ such that $p_1+\cdots+p_9\sim 3L$, where $L$ is the class of a line, then this should be an example. The fact that $p_1+\cdots+p_9\sim 3L$ means exactly that there is another cubic passing through these 9 points.
We know such a point $p_9$ exists and is unique by Riemann-Roch for curves (or fix a basepoint and use the group law on an elliptic curve). To see that the points satisfy the stated conditions, I claim that in general no 3 lie on a line and no 6 lie on a conic.
We should be able to pick the 8 points $p_1,\ldots,p_8\subset C\times\cdots \times C$ within some Zariski dense subset, because having 3 lying on a line or 6 on a conic is a closed condition (for example by using the group law on an elliptic curve).
Put more concretely, after choosing a base point on $C$, we have a map from $\mathbb{C}/\Lambda$ to the complex points of $C$ that respects the group law on $C$. Then, the condition that, for example, $p_1+p_2+p_3\in \Lambda$ cuts out a codimension 1 locus in the complex manifold $(\mathbb{C}/\Lambda)^8$. We repeat this for all subsets of 3 and 6 points of $p_1,\ldots,p_9$ (where $p_9=-p_1-\cdots-p_8$), generate a finite number of bad loci, and throw them out.