Why is the Hessian of an irreducible polynomial not zero?

Solution 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).