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An algebraic curve (in this question) is the zero set   $C = f^{-1}(X\ Y)$ of any polynomial   $f\in\mathbb R[X\ Y]$;   we say then that   $f$   represents   $C$.   An algebraic curve   $C$   is non-rational $\ \Leftrightarrow\ $ there does not exist any polynomial   $f\in \mathbb Q[X\ Y]$   which represents   $C$. An algebraic curve   $C$   is irreducible $\ \Leftrightarrow\ $ it is not a union of any two curves, different from   $C$.   The following problem is open to me:

Question: does there exist a non-rational irreducible curve which contains infinitely many rational points   (i.e. when   $C\cap\mathbb Q^2$   is infinite)?


COMMENT: Curve   $(X-\frac 1{\sqrt 2})^2 + (Y-\frac 1{\sqrt 2})^2 = 1$   has exactly one rational point. This promises a taste for trying the rational points of non-rational curves, and of the geometric-combinatorial considerations related to them (other fields and dimensions are possible too).

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Are you thinking of curves as subsets of $\mathbb{R}^2$ or subsets of $\mathbb{C}^2$? The reason I ask is to make sure you have in mind the latter, since if you meant the former then every irreducible nonempty curve would be a single point. –  Michael Zieve Sep 6 '13 at 19:59
    
Thank you, @Michael. I quickly added the addition (correction) about the sub-curves being different also from the total. Will this help the real case? –  Włodzimierz Holsztyński Sep 6 '13 at 20:13
    
Yes, that fixes things. –  Michael Zieve Sep 6 '13 at 20:56
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Since $C\cap\mathbb Q^2$ is infinite and $C$ is irreducible, this set is Zariski dense in $C$. Now take the ideal $I$ in $\mathbb Q[X,Y]$ of polynomials vanishing at $C\cap \mathbb Q^2$. This ideal contains the ideal $I_C$ of $C$, so $V(I)\subset V(I_C)=C$. On the other hand, $V(I)$ contains $C\cap\mathbb Q^2$, hence its Zariski closure, so that $V(I)$ contains $C$. Finally, $C$ is defined by an ideal in $\mathbb Q[X,Y]$, so is rational. (NB. Rational is an adjective which applies to curves, and by Faltings, besides elliptic curves, non-rational curves have only finitely many rational point!) –  ACL Sep 6 '13 at 21:16
    
@ACL, thank you for your answer and a comment about (my amateurish) terminology. –  Włodzimierz Holsztyński Sep 6 '13 at 22:58

1 Answer 1

Using a basis for $\mathbb R$ over $\mathbb Q$ (or just the sub-vector space generated by the coefficients of $f$), we can write the equation for a point being on $C$, $f(x,y)=0$, as a set of equations with rational coefficients - we view $f(x,y)$ as a vector, and set the entries to $0$ indepedently.

This gives a list of rational polynomial equations that must be satisfied by all the rational points. If they have no common factor, they have finitely many common roots, by Bezout's theorem. If they do have a common factor, it is also a factor of $f$, so $C$ is either reducible or rational.

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