Implicitization of a cubic Let $\nu$ be the map $\mathbb{P}^1 \to \mathbb{P}^2$ defined by $$\nu([X_0,X_1])=[A_0(X_0,X_1),A_1(X_0,X_1),A_2(X_0,X_1)]$$ where the $(A_i)$ are homogenous polynomials of degree 3, without common zeros. 
To prove the image of this map is included in a cubic curve of $\mathbb{P}^2$, we can look at the pullback of the map $\nu$ that gives a linear map from the space of homogenous polynomials of degree 3 on $\mathbb{P}^2$ to the space of homogenous polynomials of degree 9 on $\mathbb{P}^1$ (this comes from a hint in an exercise of Harris' book Algebraic Geometry). It is then sufficient to prove that this pullback is not surjective since the two spaces have dimension 10.
I have some difficulty to prove that by a simple (meaning not using powerful theorems of elimination theory) method. Any hint about this hint ? (Harris seems to imply in his hint that it is quite obvious : perhaps there is something very simple to see that I am missing)
PS: I posted below a simple answer I just found. If someone can check it, I will be grateful !
 A: Well, after some more thinking, I found the following elementary solution :
1- First note that the image by the pullback is not changed by the action of $PGL(3)$ on the point $[A_0,A_1,A_2]$
2- Second, note that the three homogenous polynomials of degree 3 do not generate the space of degree 3 homogenous polynomials on the projective line, but they are supposed to be independant.
3- Then using a transformation of $PGL(3)$ applied to $[A_0,A_1,A_2]$, we can get a new triple of polynomials such as, if we consider the basis $(X_0^3,X_0^2X_1,X_0X_1^2,X_1^3)$, either for at least one element of such a basis, there will be only one polynomial having a non zero coordinate on this element (Jordan normal form) and at least a non zero coordinate on one other element of the basis, either we will get the basis minus one element of it.
4- Then it is obvious to see that the image of the pullback do not contain $X_0^9$ or $X_0^8X_1$ or $X_0X_1^8$ or $X_1^9$. 
I got the feeling this solution is a bit too adhoc, and lacks elegance. 
A: I think you can try with Bézout's theorem. 
To compute the degree of the image, you count the number of its intersection points with a generic line in the projective plane. Let $c_0, c_1, c_2$ be constants in the base field. We have to solve the equation 
$$c_0A_0(x,y)+c_1A_1(x,y)+c_2A_2(x,y)=0$$
in $\mathbb P^1$. It has clearly at most three solutions (unless the polynomial $c_0A_0+c_1A_1+c_2A_2$ is zero, which can't happen for general enough $c_i$'s). Send these solutions by $\nu$ to $\mathbb P^2$ and we see that there are at most three intersection points.
