A smooth cubic is not rational

Question

We consider projective curves over the closed field $\mathbb{k}$. It can be proven that the curve is rational iff its genus $g=0$. Also the curve is birationally equivalent to a nonsigular cubic iff its genus $g=1$. This means, in particular, that smooth cubics are not rational. But this proof is not so elementary, I think: it uses the notion of genus, that is rather difficult to introduce (divisors, spaces $L(D)$...). How the statement can be proven without divisors and genus?