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Let $H_d:=\mathbb{C}_d[x,y,z]$ denote the space of homogeneous degree $d$ polynomials in $x$, $y$, $z$ with complex coefficients. I'd like to show that every $f\in H_3$ can be represented as $$ f=\det\begin{pmatrix}q_1 & q_2\\ \ell_1 & \ell_2\end{pmatrix}, q_i\in H_2, \ell_i\in H_1, i=1,2. $$ This is different from the usual determinantal representations, where each matrix entry is a linear form.

I suspect that the answer is true, and it can be proved using technique from A.Beauville's "Determinantal hypersurfaces" - which is not easy to read. Is there an easier argument?

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Let $(\ell_1,\ell_2)$ be the ideal of a point in $V(f)$, so $f \in (\ell_1,\ell_2)$.

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    $\begingroup$ Thanks, this explains why the difficulty starts with symmetric representations, or ones of size bigger than 2x2... $\endgroup$
    – Dima Pasechnik
    Commented Jun 21, 2017 at 14:52
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    $\begingroup$ Wow, I was super terse this morning. In case any students come across this I will add bit of detail: If $V(\ell_1,\ell_2) \subset V(f)$, then $f \in (\ell_1,\ell_2)$. So we can write $f = q_1 \ell_1 + q_2 \ell_2$ for some $q_1,q_2$. By homogeneity, $q_1,q_2$ are quadratic forms. Then $$f = \det \begin{pmatrix} q_2 & -q_1 \\ \ell_1 & \ell_2 \end{pmatrix}$$ (minor rearrangement of indices). ... $\endgroup$
    – Zach Teitler
    Commented Jun 21, 2017 at 19:29
  • $\begingroup$ ... Explicitly, a general cubic curve can be written in Hesse normal form as $f = x^3+y^3+z^3+6hxyz$ for some $h$, after a linear change of coordinates. The point $[-1:1:0] = V(x+y,z) \in V(f)$. Then $f = \det \begin{pmatrix} z^2+6hxy & -(x^2-xy+y^2) \\ x-y & z \end{pmatrix}.$ This leaves some non-generic curves to be handled individually. $\endgroup$
    – Zach Teitler
    Commented Jun 21, 2017 at 19:32

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