The [uniformity conjecture](https://mathoverflow.net/questions/223913/what-is-the-exact-statement-about-uniform-boundedness-of-rational-points-on-curv)
basically states that the number of rational points on a smooth curve of genus $g >1$ over number field is bounded.


If we drop smoothness, there is counterexample coming from unbounded number
of singular points.


Can we drop smoothness if we don't count singular points?


i.e., The number of rational points which are not singular is bounded in term
of the genus $g > 1$ and the number field?



Can we drop smoothness in uniformity conjecture if we don't count singular points?
Yes. Consider the smooth projective model $\tilde{C}$ of your (geometrically irreducible) curve $C$ of (geometric) genus $g \ge 2$ over a number field $k$. Then the nonsingular points on $C$ are in bijection with a Zariski-open subset of $\tilde{C}$, and this bijection preserves $k$-rational points (because the canonical morphism $\tilde{C} \to C$ is defined over $k$). So $C$ can have at most as many non-singular $k$-rational points as $\tilde{C}$ has $k$-rational points, and the latter number is conjecturally bounded in terms of $g$ (which is the genus of $\tilde{C}$) and the field $k$ (or just its degree over $\mathbb Q$).


(Since the arithmetic genus cannot be smaller than the geometric genus, the statement is also valid for the arithmetic genus.)


