 Solution

The genus-degree formula givesbecause the nonsingular plane cubic is already its own normalization of an algebraic curve. If it were birational to the projective line, the two smooth projective curves would have isomorphic function fields. The uniqueness of the finite normalization model of a one-variable function field would then make them isomorphic. But the projective line has genus of a smooth projective curve zero. Therefore the cubic is not rational. This obstruction works in the stated positive characteristics as well as in characteristic zero.