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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?

Mikolay
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    Have a look at Hartshorne, Chapter I.6, Exercise 2. – Nils Matthes Oct 12 '14 at 05:36
  • @NilsMatthes Your pointer allowed me to complete the last missing step in an elementary proof of elliptic curve associativity, which led to the paper https://arxiv.org/abs/2302.10640 Thanks! – Junyan Xu Feb 22 '23 at 06:10

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