Let $k$ be a global field and $C$ a smooth projective curve over $k$ which is not isotrivial. Then there is a well-known trichotomy:


* If $g(C) = 0$ and $C(k) \neq \emptyset$, then $C \cong \mathbb{P}^1$. In particular $C(k)$ is infinite.
* If $g(C) = 1$ and $C(k) \neq \emptyset$, then $C(k)$ is a finitely generated abelian group.
* If $g(C) \geq 2$ then $C(k)$ is finite (Mordell conjecture).


However, there exist projective *regular* curves $C$ over $k$ whenever $k$ is imperfect, which need not be smooth.


What happens with $C(k)$ in this case? Is there a similar structure theorem given by geometric invariants, which determine when $C(k)$ can be infinite?



Rational points on regular curves over global fields
*If the (geometrically integral, projective) curve $C$ over the global field $k$ is regular but not smooth over $k$, then $C(k)$ is finite*:


First of all $C$ is smooth over $k$ if and only if also the base change $C\_{\bar{k}}$ is regular.
This is equivalent to $C$ being conservative, i.e. $g(C)=g(C\_{\bar{k}})$, where $g$ is denotes the arithmetic genus of the normalization (or in function field terminology the genus of $k(C)/k$), see e.g. Theorem 2.5.1 in [1](https://arxiv.org/abs/2102.06941).


Finally, it is a consequence of a theorem of Jeong [2] that
for global fields $k$ (in fact for all $k$ finitely generated and of transcendence degree $1$ over a perfect field), $g(C)>g(C\_{\bar{k}})$ implies that $C(k)$ is finite, see Proposition 5.7 in [3].


[1](https://arxiv.org/abs/2102.06941) B. Poonen, Rational Points on Varieties
[2] S. Jeong. Rational points on algebraic curves that change genus. J. Number Theory 67(2):170–181, 1997
[3] <https://arxiv.org/abs/2102.06941>


