# Why is the Hessian of an irreducible polynomial not zero?

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 ?

Thanks.

• 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. Jul 30, 2011 at 0:44
• 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 Aug 7, 2011 at 13:48
• @wxu : Notice that \det is a standard operator name in $\TeX$. I changed \mathrm{Det} to \det. May 23, 2012 at 16:15

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).