Let $k$ be an algebraically closed field, $\operatorname{char}k=0$, $F$ be an irreducible homogeneous polynomial of degree$>1$ in $k[X,Y,Z]$, and $H=\det\left(\begin{array}{ccc}F_{xx}&F_{xy}&F_{xz}\\F_{yx}&F_{yy}&F_{yz}\\F_{zx}&F_{zy}&F_{zz}\end{array}\right)$. Make more clear, in this setting, that $H\neq 0$ is always true.

Why is $H$ not 0? Is there a pure algebraic proof of this ?


  • 8
    $\begingroup$ The question is to prove that, for any $F$, this is not identically zero. As I assume wxu knows, the geometric explanation is that (in characteristic zero) it is impossible to have a plane curve of degree $>1$ where every point is a flex. I don't know an algebraic proof, though. $\endgroup$ Jul 30, 2011 at 0:44
  • 3
    $\begingroup$ It seems that this was a conjecture of Hesse, proven by Gordan and Noether, so I suppose it's not trivial. Here is a recent paper on the topic: arxiv.org/abs/0802.0959 $\endgroup$
    – John M
    Aug 7, 2011 at 13:48
  • 1
    $\begingroup$ @wxu : Notice that \det is a standard operator name in $\TeX$. I changed \mathrm{Det} to \det. $\endgroup$ May 23, 2012 at 16:15

1 Answer 1


The question makes sense in any number of variables $n$. As pointed out by John M in the comments, Hesse conjectured that $H\equiv 0$ if and only if the integral variety $V(F)$ defined by $F$ is a cone (i.e. after a suitable linear transformation, $F$ depends only on $n-1$ variables). This condition is clearly sufficient (note that for $n\le 3$, $F$ irreducible implies that $V(F)$ can't be a cone unless $\deg F=1$). Gordan and Noether proved the conjecture for $n\le 4$ and showed it is false for $n\ge 5$.

For $n=3$ (your situation), there is a down-to-earth and purely algebraic proof in a paper of Christoph Lossen http://www.mathematik.uni-kl.de/~lossen/download/Lossen003/Lossen003.ps.gz (the idea of dual variety appears in the proof but is not explicitely named). In the preprint pointed out by John M (now published in Connect. Math. 60 (2009)), there is a more geometric proof (easy consequence of Proposition 1.6 attributed to F. Zak).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.