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Let $C$ be a smooth, projective, irreducible curve in $\mathbb C\mathbb P^n$ defined over $\mathbb Q$.

  1. Is it true that if its genus is $0$ then $C(\mathbb Q)$ is isomorphic with $\mathbb Q\mathbb P^1$ (over $\mathbb Q$)?

  2. I understand that Mordell's theorem says that if genus is $1$ then $C(\mathbb Q)$ is either empty or finitely generated, and

  3. Falting's theorem says that for $g\geq 2$, $C(\mathbb Q)$ is finite.

Is there a (fairly elementary) textbook discussing these statements?

Adam
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The curve $X^2+Y^2+Z^2=0$ in $\Bbb P^2$ is of genus zero but not isomorphic over $\Bbb Q$ to $\Bbb P^1(\Bbb Q)$ (since it has no points over $\Bbb Q$).

Good luck finding an elementary book discussing Faltings' theorem beyond simply mentioning it. (Although Bombieri found a simpler proof than Faltings' that's far from easy).

Angina Seng
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  • Thanks! I understand though that if $C$ of genus $0$ has a rational point then $C(\mathbb Q)= \mathbb P^1(\mathbb Q).$ Correct? Would there be a simple reference for that? – Adam Nov 08 '19 at 18:10
  • @Adam I discuss (the well-known) proof here: https://math.stackexchange.com/questions/3201863/how-many-solutions-of-the-equation-ax2-by2-1-are-there-with-x-y-%e2%88%88-ma/3202865#3202865 – Alex Youcis Nov 09 '19 at 14:26
  • @Adam PS, I don't know what your level is at, but there is at least a very famous heuristic for why the Mordell conjecture should be true. See this: https://mathoverflow.net/a/56016/38867 – Alex Youcis Nov 09 '19 at 14:28