Proof for "any $p \in \operatorname{Sym^3(\mathbb C^2)}$ can be written as a product of linear factors?"

Question

Here is the paper I am trying to understand:

I need a proof that "any $p \in \operatorname{Sym^3(\mathbb C^2)}$ can be written as a product of linear factors $(a_1 x + b_1 y)(a_2 x + b_2 y)(a_3 x + b_3 y)$" please?

Intuitively I agree on this statement because the basis elements $(x^3, x^2y, xy^2, y^3)$ can be obtained from these linear factors but this is not a rigorous proof.

Also, I do not understand the statement: what if $\theta$ is the cube of the inverse of the determinant character of $G'.$ This will descend to a character of $G.$

Could anyone explain this for me please?

Answer 1

You have a polynomial $$p(x,y)=ax^3+bx^2y+cxy^2+dy^3=y^3\Big(a\Big(\frac{x}{y}\Big)^3+b\Big(\frac{x}{y}\Big)^2+c\Big(\frac{x}{y}\Big)+d\Big)=y^3q\Big(\frac{x}{y}\Big)$$ where $q(t)=at^3+bt^2+ct+d$. Since $\mathbb{C}$ is algebraically closed, $q$ can be factor into linear terms : $q(t)=(a_1t+b_1)(a_2t+b_2)(a_3t+b_3)$. Thus $$p(x,y)=y^3\Big(a_1\frac{x}{y}+b_1\Big)\Big(a_2\frac{x}{y}+b_2\Big)\Big(a_3\frac{x}{y}+b_3\Big)=(a_1x+b_1y)(a_2x+b_2y)(a_3x+b_3y)$$ You have an equality of polynomials which holds for any $y\neq 0$, thus which hold for any $y$.