Crossposted from MO, where it is downvoted: https://mathoverflow.net/questions/309908/removing-singular-points-from-curves
Basically I am interested when can we find non-singular birationally equivalent curve.
Working over the rationals.
Q1: Let $C : f(x,y)=0$ be singular affine curve. When can we find non-singular curve $C_2$ birationally equivalent to $C$?
Q2: Let $C : f(x,y,z)=0$ be singular projective curve. When can we find non-singular curve $C_2$ birationally equivalent to $C$?
Partial results:
For the affine case looks like the answer is always via disequality constraints. $x \ne x0$.
Let $(x0,y0)$ be singular point on $C$. Add new variable $z$ and let $C_2 : f(x,y)=0, z(x-x0)=1$. This removes $(x0,y0)$ from the curve and doesn't introduce new singular points. Similar trick works if the singular locus is not zero dimensional.
The disequality constrain trick doesn't appear to work entirely for projective curves.
Since the answer to Q2 is likely negative, can we find projective $C_2$ with only one or few singular points?
From the comments: Per wikipedia, every curve $C$ over a field has a resolution of singularities, that is, a smooth curve $\widetilde{C}$ along with a proper birational map $\pi:\widetilde{C}\to C$. Thus every singular curve is birational to a smooth curve.