Why is the intersection of four generic cubics in $\mathbb{P}^2$ six points?

Question

I read today that a cubic surface can be constructed by blowing up 6 points in general position on $\mathbb{P}^2$. The crux of the proof depends on the fact that taking four general cubics $f_1,\ldots,f_4$, they give a rational morphism $\mathbb{P}^2 \to \mathbb{P}^3$ by $$ [x:y:z] \mapsto [f_1([x:y:z]): \cdots : f_4([x:y:z])] $$ why do these four polynomials have a vanishing locus of six points? Also, how does this generalize to other degree/dimensional varieties? This may be a really nice avenue for computing chow rings.

Answer 1

It is not true that a genral quartuple of cubic polynomials has 6 common zero points. In fact, they have none, and this can be easily seen geometrically.

Consider the triple Veronese (= anticanonical) embedding $v_3:P^2 \to P^9$. Then a choice of 4 cubic polynomials on $P^2$ is equivalent to a choice of 4 linear functions on $P^9$, i.e., to a choice of a $P^5 \subset P^9$. The zero locus of the polynomials is the intersection $v_3(P^2) \cap P^5$ in $P^9$ and it is empty for a general $P^5$ by dimension reasons.

On the other hand, if you want to get the situation, corresponding to a cubic surface, you can act as follows. Choose a general point on $S_9 = v_3(P_2)$ and let $S_8$ be its projection from this point. Then choose a general point on $S_8$ and let $S_7$ be its projection, e.t.c. The sequence $S_9 \subset P^9$, $S_8 \subset P^8$, \dots, $S_3 \subset P^3$ is a sequence of (anticanonically embedded) del Pezzo surfaces of the corresponding degrees. The surface $S_d$ is isomorphic to the blowup of a general set of $9-d$ points on $P^2$.