An irreducible plane cubic has at most one singular point
= An irreducible plane cubic has at most one singular point
Two distinct singular points would impose at least two zeros each on the restriction of the cubic to their joining line. A nonzero homogeneous polynomial of degree three cannot have total zero multiplicity four, so that restriction vanishes identically. The line would be a component, contradicting the <irreducible polynomial> hypothesis. This argument works in every characteristic.