[Math] Every smooth cubic curve has a flex point

Question

I want to show that every smooth irreducible plane cubic $C$ has a flex point, i.e. a point $P$ with $i_P(C, T_C(P)) = 3)$ (where $T_C(P)$ is the tangent to $C$ at $P$). I know how to do this in characteristic zero – take the Hessian and use Bezout's theorem – but don't know how to proceed in characteristic $p$. Can anyone provide a proof or point me to a resource where this is shown? I'm fine with chracteristic $\neq 2,3$.

Answer 1

For characteristic $\not=2,3$ you can always put the curve in short Weierstrass form and choose the point at infinity. I'm not sure about char=$2,3$.

EDIT: For characteristic 2 or 3 we can put the curve in Weierstrass form, and again the only point at infinity will be $[0:1:0]$, with tangent equal to $Z$, and the tangent will intersect the curve three times in that point, making it a flex.