What is the simplest example (or perhaps best reference) for the fact that there are genus $1$ curves (over a field of your choice --- or if you wish, over $\mathbb{Q}$, to make it more exciting) with no points of degree less than $n$? Brian Conrad gave a slick answer here: <https://math216.wordpress.com/2011/04/22/fourteenth-post/> but I wonder if there is something that can be explained more simply. My expectation is that this isn't possible, because describing genus one curves extrinsically is hard, but there may be a clever trick I am missing.



For each $n$: show there is a genus $1$ curve over some field $k$ with no points of degree less than $n$, (simple argument / best reference)?
How about a universal example?


Let $E$ be an elliptic curve over any field $k\_0$ and let $L$ be a degree $n$ line bundle on $E$. Then the actions of $n$-torsion points of $E$ by translation preserves $L$, and hence these automrophisms act by projective linear automorphisms of $H^0(E,L)$, giving a map $E[n] \to PGL\_n$. Let $k= k\_0( PGL\_n / E[n])$ and let $E'$ be the generic fiber of $(E \times PGL\_n) / E[n]$ over $PGL\_n/ E[n]$, a genus one curve.


I claim $E'$ has no $k$-points of degree $d<n$. Suppose it has one, then there is a degree $d$ divisor on $E$ defined over $k\_0(PGL\_n)$ such that acting by an automorphism in $E[n] \subseteq PGL\_n$ and translating by the same point in $E[n]$ returns the original point. However because $PGL\_n$ is rational, $k\_0(PGL\_n)$ is a rational function field, so every $k\_0$-point of $E$ is a $k$-point of $E$. So in fact there is a degree $d$ divisor invariant by translation by $n$-torsion points, which is impossible unless $d$ is a multiple of $n$.


