Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2010 / B / 4 / iii / Solution

The constant and linear Taylor terms cancel in the Newton step, leavingThus residuals strictly decrease whenever the current residual is nonzero, and the steps tend to zero doubly exponentially. By the ultrametric inequality, the distance between any two sufficiently late iterates is bounded by the largest intervening step. The sequence is Cauchy and converges in the complete valuation ring to $a$. Continuity gives $f(a)=0$, andThe literal infinite chain of strict inequalities needs the root-hit exception: if $f(a_0)=0$, or a later iterate is already a root, it cannot keep strictly decreasing below zero. The construction is eventually constant in that case; all existence and distance conclusions remain valid.

For the equality case, suppose $|f(a_0)|=d^2$ with $0<d<1$. The first Newton step has norm $|h_0|=d$. Write the Taylor coefficient of $h^2$ as $c_2(a_0)=f^{\prime \prime }(a_0)/2$. Its norm is strictly below one by hypothesis; all higher coefficients have norm at most one. HenceThe derivative changes by $2c_2(a_0)h_0+O(h_0^2)$, of norm strictly below $d$, so it still has norm $d$. The strict Hensel argument applies from this first iterate onward, and the root remains within distance $d=|f(a_0)|/d$ of $a_0$. If characteristic two is admitted, interpret the displayed half-second-derivative as the second Taylor coefficient ${\sum }_j\left(\genfrac{}{}{0ex}{}{j}{2}\right)c_j{b}^{j-2}$; this is the division-free form of the required hypothesis.

For the final local equation put $F(t)=t^3+3t$. Its values modulo nine are $0,4,5$, so the sum of two such values is confined to $0,1,4,5,8$. Conversely choose initial pairs for these five classes asrespectively. Fix the second coordinate $y_0$ and apply the equality-case Hensel result to $g(X)=F(X)+F(y_0)-n$ at the first coordinate $x_0$. Its residual is divisible by nine, $g^{\prime }(x_0)=3(x_0^2+1)$ has 3-adic norm $1/3$, and $g^{\prime \prime }(b)/2=3b$ has norm below one for every $b\in {\mathbb{Z}}_3$. The strict or equality case therefore supplies a root in ${\mathbb{Z}}_3$. Thusis the necessary and sufficient condition.