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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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This is the most well-known example of Hasse (local-global) principle, which does not hold in higher genus. In your particular case, it is easy. Basically the proof is the same as finding all the rational solutions to $x^2+y^2=1$ as the following:

Assume $C$ is smooth. Then the genus formula says $f$ is of degree [at most] $2$. Use rational slope lines passing through the rational points to intersect the curve $f=0$, namely, solve equations in terms of the slope. Plugging in the linear curve into the quadratic equation. It factors into two solutions, one is the given rational point, the other gives you a parametrization.

If $C$ is not smooth, the degree might be higher. For example, $y^2=x^3-x^2$ is a rational curve. But you can still use this method, since the genus $0$ assumption basically allows you to cancel terms.

[Edit]: As reuns remarked, some extra assumptions are needed for the statement to be correct. One way to fix this is by restricting to the smooth case (apparently my careless writing skips that point):

If $C\subseteq \mathbb{A}^2$ is a smooth curve of genus $0$ defined over $\mathbb{Q}$ with a $\mathbb{Q}$-point, then $C$ is rational.

user unknown
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  • Thank you for this explanation. Would you know a reference by any chance? – Johnny T. Apr 28 '23 at 09:04
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    You're skipping over a lot here - for instance, you're upgrading from affine to projective without saying anything, then the final sentence is pretty vague. – KReiser Apr 28 '23 at 12:26
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    @JohnnyT. I don't know a reference, since this seems an elementary number theory question. You can look up Serre's A course in arithmetic for a more general version, which would be an overkill. Also, you may look up "Brauer group". – user unknown May 01 '23 at 05:57