Union of $\{[v^3]: v\in V\}$ and $\{[v^2w]:v,w\in V \text{ linearly independent}\}$ is a quartic surface in $\mathbb{P}^3$ I want to do Exercise 10.10 in J. Harris book "Algebraic Geometry", which is:
"Let $V$ be a 2-dimensional vector space over a field $K$ with char$(K)\neq 2,3$, let $C=\{[v^3]: v\in V\}$, $\Sigma = \{[v^2w]:v,w\in V \text{ linearly independent}\}$. Show that $C $ $\cup$ $\Sigma $ is a quartic surface in $\mathbb{P}^3$."
How can I solve this? I am relatively new to AG so I don't really know what I could try here. First of, if we have $\{x,y\}$ as a basis for $V$, we have a basis $\{x^3, 3x^2y, 3xy^2, y^3\}$ for Sym$^3(V)$. If $v = ax + by$, $w=cx+dy$, then $$v^3 = a^3x^3 + 3a^2bx^2y + 3ab^2xy^2 + b^3y^3$$ and $$v^2w = a^2cx^3 + (2abc + a^2d)x^2y + (b^2c +2abd)xy^2 + b^2dy^3$$ so we have:
$$C=\{[a^3, a^2b, ab^2, b^3]: (a,b) \in K^2 \setminus (0,0) \}$$
$$\Sigma = \{[a^2c, \frac{1}{3}(2abc + a^2d), \frac{1}{3}(b^2c +2abd), b^2d]: (a,b),(c,d) \in K^2 \setminus (0,0) \}$$
Did I calculate these sets correctly so far? Now I want to show that this union is a quartic surface, meaning that there exists a homogeneous polynomial of degree 4 that has precisely this union as its projective vanishing set. Is it possible to find such a polynomial? How can I find it? Or is there maybe some theorem that can be applied here? In the book, it is stated that $\Sigma$ is the tangent space of $C$, however we have not introduced them yet, so I cannot apply any tools regarding tangent spaces.
 A: Yes, everything's alright so far (except for a dropped linearly independent condition on your final math line describing $\Sigma$). The first thing to notice is that you can write $C\cup \Sigma$ as $[v^2w]$ with no linear independence conditions, so it suffices to find equations for that. I'll choose a slightly different basis than you: no division by 3 anywhere, which makes the set we're trying to find the ideal of $$\{[a^2c: 2abc+a^2d: 2abd+b^2c: b^2d] \mid (a,b),(c,d)\in k^2\setminus (0,0)\}.$$
The general method of finding the ideal of a parametrized variety like this belongs to an area called elimination theory. The way to do it is to declare new variables $x,y,z,w$ and compute the intersection of $(a^2c-x,2abc+a^2d-y,2abd+b^2c-z,b^2d-w)$ with $k[x,y,z,w]$ inside $k[a,b,c,d,x,y,z,w]$. This is called computing the elimination ideal, and this can be done with Grobner bases in a systematic fashion. While you can do it by hand, it can be very nice to get a computer to do it for you, and indeed the following code on the Sage math cell will give you the answer $(y^2z^2 - 4xz^3 - 4y^3w + 18xyzw - 27x^2w^2)$ :
R.<a,b,c,d,x,y,z,w> = PolynomialRing(QQ,8)
I = R*[ a^2*c-x, 2*a*b*c+a^2*d-y, b^2*c+2*a*b*d-z, b^2*d-w ]
J = I.elimination_ideal([a,b,c,d]); J

I wouldn't be surprised if there were more ad-hoc methods available for this specific example, but none spring to mind at the moment.
