Fix an algebraically closed field $k$ of characteristic $3$. Consider a curve in $\mathbb{P}^2_{k}$ given by $$xy^2+yz^2+zx^2=0.$$ It seems to be a cubic with a cusp at $[1:1:1]$, whose Hessian $(x+y+z)^3$ meets it only at the cusp. Hence this cubic has no inflection point.
On the other hand, e.g. Appendix A of Silverman's The Arithmetic of Elliptic curves shows that every cuspidal cubic in $\mathbb{P}^2_{k}$ is projectively equivalent to $x^3=y^2 z$, so every point should be an iflection point.
What am I missing here? If I got Silverman wrong, and there is more than one cuspidal cubic on $\mathbb{P}^2_{k}$, could you give some reference where these cubics are classified? Also, do similar pathologies occur in characteristic two?
EDIT: After some computation, I arrived to a following conclusion: a cubic $C\subseteq \mathbb{P}^{2}_{k}$ over an algebraically closed field $k$ has an inflection point, unless $\mathrm{char} k=3$ and $C$ is projectively equivalent to the above curve. Therefore, I would like to ask now:
- Is there any reference for this fact?
- It is known (see e.g. Hartshorne's Example 6.11.4) that for a cuspidal cubic $C=\{y^3=x^2z\}$, we have $\mathrm{Pic}^0(C^{\mathrm{reg}})\cong\mathbb{G}_{a}$ as group schemes. Is there an analogous way to define a group structure on the smooth locus of the above curve? (Clearly, $\mathbb{G}_{a}$ won't work since $3p\neq 0$ for every point $p$, but maybe $C^{\mathrm{reg}}$ is still a torsor over something?)