Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2008 / B / 5 / iii

The reduction has equation $F=x^3+y^3=0$. At the origin both first partial derivatives vanish, so the point is singular, also when $p=3$. A lift reducing to the origin would require $x,y\in p{\mathbb{Z}}_p$. Its two cubes would then lie in $p^3{\mathbb{Z}}_p$, whereas the right side of the original equation has valuation one. This is impossible.

For an example which does lift, chooseIt has the same reduction and the rational point $(p,0)$ reduces to the origin. Over ${\mathbb{Q}}_p$, which has characteristic zero, a singular affine point would require $x=y=0$, which does not lie on this curve. Its projective closure is also smooth: the simultaneous partial derivatives of $X^3+Y^3-p^3{Z}^3$ cannot vanish at a projective point. Thus this is a nonsingular curve for every prime $p$, including three. A singular point in the reduction can therefore fail to lift or can lift, depending on the original equation.