Irreducibility for cubics and existence of inflection points

Consider a nonzero homogenuous degree three polynomial $P\in k[X,Y,Z]$ in $3$ variables with coefficients in a field $k$ of characteristic $\neq 2, 3$.

1. How can one check whether $P$ is irreducible? Is there an analogue for cubics of the discriminant of a quadratic form that decides this question for conics?
2. In the first paragraph here it is claimed that an irreducible cubic either has an inflection point or a singular point, and that this follows from Bezout's theorem. Could someone provide details?

I'm starting to learn about elliptic curves : these are some elementary questions I couldn't find answers to online. Regarding the first point, all I have found is this fact which implies that the equation of an elliptic curve in Weierstrass form is irreducible as a polynomial $\in k[X,Y]$. I'm not quite sure whether this fact is answers my first question. Certainly irreducibility of the Weierstrass normal form should imply irreducibility of the homogeneous cubic $P$.

Reducing the equation of an elliptic curve to Weierstrass form requires as far as I understand, sending the flex point and its tangent line to the line at infinity, thus the existence of flex points is important.

• For 2. see this question and this pdf on Jan Stevens' homepage – Jan-Magnus Økland Apr 21 '18 at 15:53
• Bezout's theorem also implies that a reducible cubic defines a curve with a singular point. This is because the points of intersection of the curves determined by the factors are singular (their coordinates may be in an extension field of $k$ though). – Jyrki Lahtonen Apr 24 '18 at 2:24
• Olivier, I don't think so. Consider the (homogenization of) Foil of Descartes: $X^3+Y^3=3XYZ$. That cubic has a singularity at the origin, but it is irreducible. For if that cubic were a product of a linear and a quadratic the curve would be a union of a line and a conic which it manifestly isn't. – Jyrki Lahtonen Apr 24 '18 at 8:01
• Yes, that's equivalent to smoothness. In higher dimensions (with several constraint equations) you need the Jacobian to have full rank. IIRC smoothness is often defined a bit differently, but it comes to that rank condition for the Jacobian. In general you get one to the Jacobian for each generator of the ideal of polynomials vanishing on the variety. – Jyrki Lahtonen Apr 24 '18 at 8:06
• Sorry about not making this clear. The Foil of Descartes has a singularity at the origin of the affine chart $Z=1,x=X/Z,y=Y/Z$. In other words, at the point $[X:Y:Z]=[0:0:1]$ = the origin of that affine chart. The singularity is obvious in the plot of that real plane curve. Smoothness at a point of a projective curve can always be tested in an affine chart containing it. – Jyrki Lahtonen Apr 24 '18 at 8:18