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

For $p=3$, a root would have valuation zero because five is a unit. It would therefore reduce to a unit cube modulo nine. The only cube residue classes modulo nine are $0,1,-1$, none equal to five. Thus no root exists in ${\mathbb{Q}}_3$.

For $p=5$, the valuation equation would be $3v_5(x)=1$, impossible because the valuation of a nonzero element of ${\mathbb{Q}}_5$ is an integer. Thus no root exists in ${\mathbb{Q}}_5$.

For $p=13$, $7^3\equiv 5(\mathrm{mod}13)$ and the derivative $3\cdot 7^2\equiv 4\not\equiv 0(\mathrm{mod}13)$. Hensel's lemma lifts this simple root to a root in ${\mathbb{Z}}_{13}\subseteq {\mathbb{Q}}_{13}$. Thus a root exists in ${\mathbb{Q}}_{13}$.

The expansion isThe notation has the units digit before the comma and successive positive powers afterwards, as defined in the source. To check it, the geometric series converges in the 3-adic norm and its sum is $1+(3+2\cdot 3^2)/(1-3^2)=1-21/8=-13/8$.

For the general repeating block, $|p^2{|}_p<1$ givesBoth numerator and denominator are integers, and the denominator is a $p$-adic unit.

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.

Let $\alpha ,\beta ,\gamma$ be the nonzero reductions of $a,b,c$. The square set in ${\mathbb{F}}_p$ has $(p+1)/2$ members, including zero. Multiplication by $\alpha$, and separately the transformation $z\mapsto \gamma -\beta z$, preserve this cardinality. Hence the two sets in the hint have total cardinality $p+1>p$ and must intersect. This gives a solution $(x_0,y_0)$ modulo $p$.

The solution cannot have both coordinates zero since $\gamma \ne 0$. If $x_0\ne 0$, choose any integral lift of $y_0$ and apply Hensel's lemma to the polynomial $a{X}^2+b{y}_0^2-c$ in $X$: its derivative at the reduced root is $2\alpha x_0\ne 0$. If $y_0\ne 0$, use the other variable instead. The nonzero derivative uses $p\ne 2$. Thus a solution lifts to $\boxed{x,y\in {\mathbb{Z}}_p}$ as required.