Missing finitely many points by rational curve with parametrization by rational functions and rational curve self intersection I have two questions:


*

*Can a irreducible rational curve have infinitely self intersections? 

*If $f$ has rational coefficients and the solutions for $0=f(x,y)\in k[x,y]$ are parametrized by rational functions with rational coefficients of some parameter $t$, then the image of this parametrization over the rationals miss only finitely many rational points.
It is not clear to me why only finitely many points were missed. My first guess is some what related to rational points are dense. Can someone elaborate a bit geometric intuition on 2 and how to prove it(I think hint will suffice)?
 A: *

*I think it not possible for any plane algebraic curve $C$ to have infinitely many self-intersections. If $p$ is a self-intersection then locally around $p$ curve looks like $f=f_1f_2=0$, thus $df|p=0$. So, infinitely many points of self-intersection give us infinitely many points where $df=0$ or infinitely many zeroes of the system of equations $f=0$, $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$. But if two polynomials of two variables have infinitely many common zeroes they have a common irreducible component. Therefore, $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$ along $C$ a and any point of $C$ is singular. That is not possible because singular points form a closed subset of a variety.

*If you get such parametrization you curve is necessarily rational. This is because existence of rational map from $\mathbb{P}^1$ to the curve tell us that the curve is unirational. On the other hand Luroth's theorem tells us that any unirational curve is rational. 

A: For the second question the idea of @user45765 was correct.
Denote by $C$ the curve given by $f(x,y)=0$. Then, if there is a birrational map $\varphi:\mathbb{A}^1\rightarrow C$ (that is, a parametrization by rational functions) there are open sets $U\subseteq \mathbb{A}^1$ and $V\subseteq C$ such that $\varphi_{|U}:U\rightarrow V$ is an isomorphism. As $\varphi_{|U}$ is defined over $\mathbb{Q}$ it's inverse must be defined over $\mathbb{Q}$ as well (see here). Hence, $\varphi$ gives a bijection between the rational points in $U$ and $V$. As $C\setminus V$ is finite (for example, because of the description of the Zariski topology on a curve) we get the result above.
