Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2010 / B (Elliptic curves)

The integer $n=-3$ gives $n^3-5=-32$, whose 2-adic norm is $2^{-5}<2^{-3}$. For the 5-adic condition, if $5\nmid n$ then $n^3-5$ is a unit, while if $5\mid n$ its valuation is exactly one. Neither case has norm below $5^{-3}$. The Newton construction for the remaining requests follows.

Let $d=|f^{\prime }(a_0)|>0$ and $q_0=|f(a_0)|/d^2<1$. At any step before reaching a root, setAssuming inductively that $|f^{\prime }(a_i)|=d$ and $q_i=|f(a_i)|/d^2<1$, this givesAll derivatives of a polynomial over $R$ are integral on $R$, so $d\le 1$. Hence $|h_i|<1$, and the ultrametric inequality keeps $a_{i+1}$ in $R$. If a root is reached, retain that value for subsequent terms, and the asserted weak step inequality still holds.

Taylor expansion of $f^{\prime }$ about $a_i$ has integral coefficients, soIn a Non-Archimedean absolute value, adding an error of norm strictly smaller than $d$ cannot change a quantity of norm $d$. ThereforeThis simultaneously verifies the derivative assumption used at the next step of part (i).

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.

We first prove all three general torsion assertions before treating the lettered cases. If a finite point has $|x{|}_p>1$, the cubic term dominates the integral lower terms, so $v_p(x)=-2m$, $v_p(y)=-3m$ for an integer $m\ge 1$. Its parameter $z=-x/y$ then lies in $p{\mathbb{Z}}_p$, with norm at most $p^{-1}$. The integral Weierstrass model defines its formal group at infinity over ${\mathbb{Z}}_p$: using $w=-1/y$ gives $w=z^3+Az{w}^2+B{w}^3$, an integral power series starting in degree three.

The supplied torsion estimate excludes a nonzero torsion parameter for $p\ge 3$, since $p^{-1}<p^{-1/(p-1)}$, and the lower bound increases for higher $p$-powers. For $p=2$ it excludes orders $2^n$ with $n\ge 2$, because their bound is strictly above $1/2$. The remaining possibility is order two, which would imply the finite point equals its negative and hence $y=0$. A nonintegral $x$ cannot solve the monic cubic with integral coefficients, so this is impossible too. Thus every finite torsion point has integral $x$, and the equation then forces integral $y$.

For good reduction the reduction map is a group homomorphism whose kernel is this formal group at infinity. Its torsion-free kernel proves injectivity on torsion, giving a subgroup of the finite reduced group. Applying local integrality to a rational torsion point at every prime gives $x,y\in \mathbb{Z}$.

If $y\ne 0$, the double of a rational torsion point is finite and integral. The doubling formulatherefore implies $y^2\mid (3x^2+a{)}^2$. Also $y^2=x^3+ax+b$. Evaluating the supplied polynomial identity at the integral $x$ now givesThis establishes the needed Nagell–Lutz theorem conclusions without assuming them in the lettered applications.

Nonsingularity with $b=a^2$ forces $a\ne 0$. TakeIt lies on the curve. Its tangent slope is $a/(2a)=1/2$, so the doubling formula gives $x(2P)=1/4$. If $P$ were torsion, $2P$ would be torsion as well, contradicting the integrality just proved. Therefore $P$ has infinite order.

For $y^2=x^3+x+2$, the point $P=(1,2)$ doubles to $(-1,0)$: its tangent slope is $(3+1)/(2\cdot 2)=1$, giving $x(2P)=-1$ and $y(2P)=0$. Thus $P$ has order four.

The curve has good reduction at $3$, since $4+27\cdot 4=112$ is nonzero modulo three. Its reduced points are the point at infinity, two points above $x=1$, and one above $x=2$, so there are four in total. Injectivity of torsion reduction bounds the rational torsion order by four. The known four points exhaust it:

Modulo five the curve is $y^2=x^3+x$, with good reduction because $4a^3+27b^2\equiv 4(\mathrm{mod}5)$. It has one point above each of $x=0,2,3$ and no others, plus infinity, so its group has order four. Rational torsion therefore has order dividing four.

A nonzero rational point of order two would have $y=0$ and integral $x$, so would give an integer root of $x^3+ax+b$. But modulo two this polynomial is $x^3+x+1$, which is one at both residues. Thus there is no rational 2-torsion. Every nontrivial finite group of order dividing four contains an element of order two, so

Complete the square-class map at the identity by $q(O)=1$; at the finite point $T=(0,0)$ its value is the class of $b$. For another rational point writeThese denominator powers follow because a negative valuation of $x$ forces the leading cubic term to dominate, so that valuation is even and the valuation of $y$ is three halves of it. The equation givesIf a prime $\ell \nmid b$ divides $r$, it does not divide the second factor, since $\ell \nmid s$. Thus $v_{\ell }(r)$ is even. All denominator valuations are even too. Every image class therefore has a representative which is a signed squarefree divisor of $b$, proving that $\mathrm{im}q$ is finite.

Use the standard 2-isogeny with kernel $\{O,T\}$ and dual isogeny satisfying $\hat{\varphi }\varphi =[2]$, $\varphi \hat{\varphi }=[2]$. Away from the exceptional points one may takeApplying the analogous isogeny to $E^{\prime }$ gives a curve with coefficients $4a,16b$; scaling its coordinates by $1/4,1/8$ gives the dual map back to $E$. The analogous descent map $q^{\prime }$ on $D$ has kernel $\varphi (C)$ and finite image. The sequenceis exact at the last two terms. Its outside groups are finite, identified with the two finite descent images, so $C/2C$ is finite.

For $E_p:y^2=x(x^2-p^2)$, the rational 2-torsion is $O,(0,0),(p,0),(-p,0)$, of order four. The first descent image is exactly $\{1,-1,p,-p\}$: these classes occur on the 2-torsion, and the divisor-support bound permits no others. The isogenous curve is $E_p^{\prime }:y^2=x(x^2+4p^2)$. Its real points have $x\ge 0$, so the second image lies in the four positive classes $\{1,2,p,2p\}$.

Here the kernel $\{O,(0,0)\}$ of the dual is contained in $\varphi (C)$, since $\varphi (\pm p,0)=(0,0)$. Thus the displayed map $D/\varphi (C)\to C/2C$ is injective, and exactness givesThe allowed rank formula makes the left side $2^{R+2}$, soNow consider the quartic relation. Normalize a nontrivial integer solution so that $p$ does not divide both $x,y$, by dividing them by $p$ and $z$ by $p^2$ whenever necessary. If $p\nmid x$, reduction modulo $p$ gives $x^4=2z^2$, with $z\ne 0$ modulo $p$, making $2$ a quadratic residue. If $p\mid x$, then $p\nmid y$ and $p\mid z$; write $x=pX$, $z=pZ$, divide by $p^2$, and reduce to get $y^4=2Z^2$ modulo $p$. Again $2$ is a square. ThereforeIf the class $2$ occurred in $\mathrm{im}{q}^{\prime }$, write the corresponding coordinate as $x^{\prime }=2u^2/v^2$. The curve equation then givesHere $u,v$ are nonzero coprime integers and $z$ is rational. Since $2z^2$ is an integer, its reduced denominator must be one, so $z$ is an integer too. For $p\equiv 3,5(\mathrm{mod}8)$ this contradicts the quartic obstruction. Hence class $2$ is absent. A subgroup of $\{1,2,p,2p\}$ omitting $2$ has order at most two, so $2^R\le 2$ and $R\le 1$.

For $p=5$, direct substitution verifies the nonintegral pointA rational torsion point would have integral coordinates by the preceding theorem, so this point has infinite order. Thus $R\ge 1$, and together with the upper bound,