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Let $f \in \mathbb{Q}[x,y]$ be a polynomial, define a curve $C = \{ f(x,y) = 0 \}$. Suppose the genus of $C$ is $0$ and $C(\mathbb{Q}) \neq \emptyset$. Then $C(\mathbb{Q})$ is birational to $\mathbb{A}^1_\mathbb{Q}$.

I have heard of this result, but I have not been able to find a reference; all the sources I have looked so far deals with algebraically closed fields. Could someone please suggest me a reference? Thank you.

Johnny T.
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    What do you get with $f(x,y) = x^2+y^2$? – reuns Apr 27 '23 at 18:43
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    Let $P$ be a rational point on $C$. By Riemann-Roch, $H^0(C,P)$ has dimension $2$. Choosing a basis $f_1, f_2$ for $H^0(C,P)$, we obtain a rational map $\varphi: C \dashrightarrow \mathbb{P}^1$ given by $Q \mapsto [f_1(Q) : f_2(Q)]$, and $\deg(\varphi) = \deg(P) = 1$, so it is a birational equivalence. It remains to show that we can choose $f_1, f_2$ in $K(C)$ (rather than just $\overline{K}(C)$). I think this can probably be shown using Galois cohomology. – Viktor Vaughn Apr 27 '23 at 19:13
  • @reuns I guess the exact statement I'm asking for is not correct. Is there a way to fix it? – Johnny T. Apr 28 '23 at 09:05
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    You want one rational point on the normalization ($\Bbb{Q}(x,i)$ doesn't have any rational place). So on the singular curve you want either a smooth rational point or infinitely many rational points (most being automatically smooth). Assuming that $f$ is absolutely irreducible may work too. – reuns Apr 28 '23 at 09:15
  • @reuns That makes more sense. thank you. would you know a reference which I can cite by chance? – Johnny T. Apr 28 '23 at 09:24
  • or do you know a reference I can use to reproduce the argument you mention? – Johnny T. Apr 30 '23 at 20:06
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    https://math.stackexchange.com/questions/1470969/can-a-conic-over-mathbbq-with-no-mathbbq-points-have-a-point-of-degree/1472591#1472591

    Answers your question I believe.

     – Alex Youcis May 01 '23 at 06:15

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