Curve with a rational point but no new points in number fields of low degree Given an integer $d\geq 2$ is there an algebraic curve $C/\mathbb{Q}$ with $C(\mathbb{Q})\neq\emptyset$ and the natural map $C(\mathbb{Q})\to C(F)$ bijective for all number fields of degree at most $d$?
 A: Here's a family of examples that are geometrically irreducible. Let $p$ be a prime number and consider the modular curve $X_{1}(p)$. If $F$ is a number field, the points in $X_{1}(p)(F)$ are either non-cuspidal (and in bijection with elliptic curves $E/F$ that have a $F$-rational point of order $p$), or cuspidal. There are $p-1$ cusps of which $\frac{p-1}{2}$ are rational, and the other $\frac{p-1}{2}$ are in a single Galois orbit. (They are rational over $\mathbb{Q}(\zeta_{p} + \zeta_{p}^{-1})$.)
Merel's proof of the uniform boundedness conjecture implies that if $d$ is fixed, then there is a constant $C = C(d)$ so that $|E(F)_{{\rm tors}}| \leq C$ for all number fields $F$ of degree $d$.
If we choose a prime $p > \max \{2d+1, C \}$, then $X_{1}(p)(F)$ will have no noncuspidal points for any number field $F$ with $[F : \mathbb{Q}] \leq d$. Also, $X_{1}(p)(F)$ will contain no non-rational cuspidal points because these are defined over a number field of degree $\frac{p-1}{2} > d$.
A: In the case of affine algebraic curves, homogenizing a polynomial with no roots in a number field of degree up to $d$ should do. For example, $x^3 + y^2x + y^3 = 0$ has a rational point $(0, 0)$ and no other points in any quadratic number field.
