# factorizing a quartic plane curve as $f_3f_1-f_2^2$

Let $C$ be a quartic plane curve. Suppose that for a given coordinate system $C=(F_4(x_0,x_1,x_2)=0)$ where there the polynomial $F_4$ factorizes as
$$F_4(x_0,x_1,x_2) = F_3(x_0,x_1,x_2)F_1(x_0,x_1,x_2)-F_2(x_0,x_1,x_2)^2$$ If the coordinate system is fixed. How many factorization are there?

More generally, Let $Hilb_d \cong \mathbb{P}^{Nd}$ be the Hilbert scheme of curves of degree $d$ in $\mathbb{P}^2$. We can define the map $$Hilb_3 \times Hilb_1 \times Hilb_2 \to Hilb_4$$ as $$\{ F_3(x_0,x_1,x_2), F_2(x_0,x_1,x_2), F_1(x_0,x_1,x_2) \} \to F_3F_1-F_2^2$$ What can I say about the fibers of the map?

I noticed that if $F_1(x_0,x_1,x_2)$ is a bitangent of the curve $C$, then I can start playing around... However, this seems to be an elementary question, and maybe I am missing something obvious.

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By the way, the apparent Hilbert-scheme map is actually not well-defined, because changing $F_1,F_2,F_3$ to scalar multiples in general changes the curve $F_1 F_3 = F_2^2$. –  Noam D. Elkies Feb 24 '14 at 4:55

Yes, if the quartic $C : F_4=0$ is smooth then the line $l : f_1=0$ must be one of the $28$ bitangents (because the restriction $F_4|_l$ is the square of $f_2|_l$), and then $f_2$ restricted to $l$ is one of the square roots of $F_4|_l$, and each of the lifts of $f_2|_l$ lets you solve for $f_3$.

Note that writing $F_4$ as $f_1 f_3 - f_2^2$ is not a factorization, and a parameter count would already indicate that there should be infinitely many such expressions: there are $3+6+10 = 19$ undetermined coefficients in $f_1,f_2,f_3$, and only $15$ coefficients to match in $F_4$. But there's also a $4$-dimensional group acting: multiply $f_1,f_3$ by $\lambda,\lambda^{-1}$, or change $f_2,f_3$ to $f_2 + g f_1$ and $f_3 + 2 g f_2 + g^2 f_1$ for any linear form $g$. So we expect finitely many solutions up to these transformations, and indeed we get the $28$ pairs described in the first paragraph.

(Likewise for $F_4 = b^2-ac$ with $a,b,c$ each of degree $2$ we have $2^6 - 1 = 63$ three-dimensional families, indexed by the nontrivial $2$-torsion divisor classes $D$ in the Jacobian of $C$; the function $a/b=b/c$ on $F_4$ is the quotient of two sections of $D+K$ where $K$ is the canonical divisor.)

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thanks for answering such an ill posed question. –  tkjs Feb 24 '14 at 21:33

If we are given one particular 'factorization': $F_4 = F_3F_1-F_2^2$, we can proceed like this:- Express $F_2$ as product of irreducible polynomials, say, $F_2 = P_1^{m_1}P_2^{m_2}\cdots P_k^{m_k}$. Now define $G_1$ as $\prod_j P_j^{a_j}$ with $a_j\leq 2m_j$, and $G_2$ as $F_2^2/G_1$. Now $F_4=0$ is the same variety as $F_2^2-G_1G_2=0$. So we get $(1+2m_1)(1+2m_2)\cdots(1+2m_k)$ choices for a fixed $F_2$.

Now take any polynomial $F$ and check if $F_4-F^2$ is reducible, if so express as a product of two polynomials: they play the role of $F_1, F_3$ and $F$ will play the role of $F_2$.

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