Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2007 / B

The strict inequality means $v_3(x^2+2)>2$. Taking $x=5$ gives $x^2+2=27$, soFor the other expression, if $3\nmid x$ then $x^2+3\not\equiv 0(\mathrm{mod}3)$ and its P-adic valuation is zero. If $3\mid x$, then $x^2+3=3(3(x/3{)}^2+1)$ has P-adic valuation exactly one. It can never have valuation greater than two. Thus no such integer exists for the second strict inequality.

In the P-adic numbers, ${\sum }_{j\ge 0}{p}^{2j}$ converges to $1/(1-p^2)$. ConsequentlySince $p\ne 2$, the digits one and two are both between zero and $p-1$, so no carrying is required. These are its p-adic digits, with the unique negative-power digit at exponent minus one.

No prime works. The P-adic valuation induces a well-defined homomorphismThe class of $p$ maps to one, whereas the class of $p^p$ maps to zero. They therefore cannot be equal. Equivalently, $p$ cannot be a $p$th power because its valuation is not divisible by $p$.

Choose a nonzero integer $D$ with $D\alpha ,D\beta \in \mathbb{Z}$. By the Chinese remainder theorem, choose $z_n\in \mathbb{Z}$ satisfying $z_n\equiv D\alpha (\mathrm{mod}{q}^n)$ and $z_n\equiv D\beta (\mathrm{mod}{r}^n)$. Put $x_n=z_n/D$. ThenThis proves the requested simultaneous approximation, even when the rational targets have denominators divisible by either prime.

A rational sequence converging to zero at every standard absolute value is:Its ordinary absolute value is at most $1/n!$, hence tends to zero. For a fixed prime $p$ and all sufficiently large $n$, the denominator is congruent to one modulo $p$, so $v_p(y_n)=v_p(n!)\to \infty$. Therefore $|y_n{|}_p\to 0$ for each prime as well. The assertion concerns each fixed prime separately, rather than uniform convergence over all primes.

If $p=0$ as an element of $K$, the stated lower bound is zero and is immediate. Otherwise set ${\rho }_1=|p{|}^{1/(p-1)}$. The supplied formal group expansion has $g(0)=0$, because $[p](0)=0$. For $z\in \mathcal{M}$, the non-Archimedean estimates giveIf $0<|z|<{\rho }_1$, then $|z{|}^p<|p||z|$, so the two terms cannot cancel and $|[p](z)|=|p||z|>0$. The same strict inequality persists under iteration, so no such $z$ can have finite $p$-power order. Every nonzero $p$-power torsion point therefore satisfies $|z|\ge {\rho }_1$.

For a point of order $p$, $[p](z)=0$ itself forces $|p||z|\le |z{|}^p$, proving the bound for $n=1$. For $n\ge 2$, put $w=[p](z)$, of order $p^{n-1}$. Since $|z|\ge {\rho }_1$, the preceding maximum is $|z{|}^p$. By induction,This proves the formal torsion radius for prime power order. Taking the positive $p$th root gives

For a short Weierstrass equation of an elliptic curve with coefficients in the p-adic integers, a point with nonintegral $x$ has valuations $v_p(x)=-2m$ and $v_p(y)=-3m$ for some integer $m\ge 1$: the term $x^3$ strictly dominates $Ax+B$ in the equation, so $2v_p(y)=3v_p(x)$. Its formal parameter $T=-x/y$ therefore has valuation $m\ge 1$. These points, together with the identity, form the formal neighborhood $E_1({\mathbb{Q}}_p)$ of the identity, with parameter in $p{\mathbb{Z}}_p$ and the elliptic formal group law over the p-adic integers. For a point with $x\in {\mathbb{Z}}_p$, the equation gives $y\in {\mathbb{Z}}_p$ too.

The formal neighborhood has no nontrivial torsion. Multiplication by an integer coprime to $p$ has series $[m](T)=mT+O(T^2)$ and preserves $|T{|}_p$, so any torsion there must have $p$-power order. If $p\ge 3$, part (i) gives $|T{|}_p\ge p^{-1/(p-1)}>p^{-1}$, contradicting $T\in p{\mathbb{Z}}_p$.

The printed hypotheses also allow $p=2$. In that case, a nontrivial torsion point in the formal neighborhood would have order a power of two and hence yield a point of order two in that neighborhood. But every nonidentity point of order two on a short Weierstrass equation of an elliptic curve has $y=0$ and $x$ a root of the monic cubic over the p-adic integers $X^3+AX+B$. Such a root lies in the p-adic integers, so it cannot belong to the formal neighborhood, whose nonidentity points have nonintegral coordinates. This excludes the remaining case. This is the torsion-free formal subgroup for a short Weierstrass equation. Thus every affine torsion point satisfiesWith good reduction, the reduction homomorphism $E({\mathbb{Q}}_p)\to \tilde{E}({\mathbb{F}}_p)$ has kernel exactly this torsion-free formal neighborhood. Its restriction to the torsion subgroup is therefore injective, proving the required subgroup isomorphism. The short model over the p-adic integers is essential to the prime-two argument; no blanket torsion-injectivity assertion for arbitrary models is needed.

The discriminant is $-64D^3$, which is nonzero modulo three because $D\equiv 2(\mathrm{mod}3)$. Thus there is good reduction at three. For every $x\in {\mathbb{F}}_3$, the reduced right side is $x^3-x=0$, so its only affine points are $(0,0),(1,0),(2,0)$. Together with the identity they form $C_2\times C_2$.

By part (ii), the rational torsion subgroup injects into this four-element group and hence has exponent two. Over $\mathbb{Q}$, the two-torsion points have $Y=0$, requiring $X(X^2+D)=0$. Since $D>0$, the only rational root is zero. Therefore

In the typical case, the line through $P,Q$, or their tangent when they coincide, is $v=mu+c$. Its intersections with $D$ have first coordinates given byTheir product is $c^2$. The third intersection is $-(P+Q)$, which has the same first coordinate as $P+Q$. ConsequentlyPassing to the quotient by rational squares gives $q(P)q(Q)q(P+Q)=1$. Every square class is its own inverse, soThe excluded cases have their specified special values; the source asks only for this chord-and-tangent argument away from them.

For an affine point of $C$ with $x\ne 0$, the first coordinate of its image is $(y/x{)}^2$. Thus its image has trivial square class whenever that coordinate is nonzero. The identity and $T=(0,0)$ map to the identity. If a point maps to $T_1=(0,0)$, it has $y=0$ and $x^2+ax+b=0$, forcing $b_1=a^2-4b$ to be a rational square; in that case $q(T_1)=1$ too. Hence $\varphi (C(\mathbb{Q}))\subseteq \ker q$.

Conversely, let $(u,v)\in D(\mathbb{Q})$ have $q(u,v)=1$ and $u\ne 0$. Write $u=t^2$ with $t\in {\mathbb{Q}}^{\ast }$. The equation of $D$ givesSet $x=(u-a+v/t)/2$ and $y=tx$. Then $x$ solves $x^2+(a-u)x+b=0$ and is nonzero because $b\ne 0$. Hence $y^2=u{x}^2=x(x^2+ax+b)$, so $(x,y)\in C(\mathbb{Q})$. Also $x-b/x=v/t$, giving $\varphi (x,y)=(u,v)$.

The identity is an image of the identity. Finally, $T_1\in \ker q$ exactly when $b_1$ is a rational square; then either nonzero root $x=(-a\pm \sqrt{b_1})/2$ gives $(x,0)\in C(\mathbb{Q})$ mapping to $T_1$. This covers all exceptional points and proves

Here $a=1$, $b=-2$ and the two-isogenous curve is $D:v^2=u(u^2-2u+9)$. Let ${\delta }_C$ and ${\delta }_D$ be the first-coordinate square-class maps, using the coefficient $b$ or $b_1$ at the point $(0,0)$. The same chord argument shows that both maps are homomorphisms.

For an equation $y^2=x(x^2+ax+b)$ with integer coefficients, its first-coordinate square class has a square-free representative dividing $b$, including a possible sign. Indeed, at a prime not dividing $b$, a negative valuation of $x$ must be even because $x^3$ dominates; a positive valuation must be even because $x^2+ax+b$ is a unit. Thus only primes dividing $b$ can occur to odd order. For $C$, the possibilities are $1,-1,2,-2$. The classes $1,-2$ occur at the identity and $(0,0)$. If class two occurred, write $x=2M^2/e^2$ with coprime integers $M,e$. Substituting in the equation would give a rational square, necessarily an integer square,Modulo three, the right side is two whenever $M,e$ are not both divisible by three: this holds whether just one or both are units. That is impossible. Since the image is a subgroup containing $-2$, exclusion of two also excludes minus one. Therefore $|{\delta }_C(C(\mathbb{Q}))|=2$.

For $D$, the quadratic $u^2-2u+9=(u-1{)}^2+8$ is positive, so every nonzero first coordinate is positive. Its square class can only be one or three. Both occur, with three supplied by $(3,6)$, so $|{\delta }_D(D(\mathbb{Q}))|=2$.

Let $\psi :D\to C$ be the dual two-isogeny, with $\psi \varphi =[2]$. The kernel identities of part (ii), applied to both curves and the usual coordinate scaling, give $[C:\psi D]=2$ and $[D:\varphi C]=2$. Moreover $\ker \psi =\{O,T_1\}$ lies in $\varphi C$, since $b_1=9$ is a square. Thus $[\psi D:2C]=[D:\varphi C]=2$, andThis is a rank-zero two-isogeny descent for roots zero one and minus two. The rational two-torsion has four points, since the cubic factors as $x(x-1)(x+2)$. By the Mordell-Weil theorem, the last index is $2^r|C(\mathbb{Q})[2]|=2^{r+2}$. Therefore

The doubling slope at $P=(2,1)$ is $(3\cdot 2^2-1)/(2\cdot 1)=11/2$, givingThese denominators are invertible modulo $N$. To add $P$ again, the difference of first coordinates is $105/4-2=97/4$, so the new slope would be $-1083/194$. This requires inversion of a number divisible by 97. The elliptic curve method instead computesIt is a proper factor. The failure of the group-law division modulo the composite integer reveals different behavior modulo its prime factors.

Write $x=c/d$ in lowest terms and put $H=\max \{|c|,|d|\}\ge 1$. For $P\ne O,(0,0)$,Before reduction to lowest terms, both numerator and denominator are bounded in magnitude by $(1+|a|+|b|)H^2$. Cancellation can only decrease the height. Hence one may chooseThe duplication formula for $C$ isThe numerator has magnitude at most $(1+|b|{)}^2{H}^4$, and the denominator at most $4(1+|a|+|b|)H^4$. Thussatisfies the second bound. At the identity or any point whose relevant image is the identity, the image height is zero and the same nonnegative constants work directly. Thus the formulas cover every rational point, including the exceptional points of the displayed coordinate expressions.

If $p\equiv 1(\mathrm{mod}4)$, choose $i\in {\mathbb{F}}_p$ with $i^2=-1$. The cubic has the three distinct roots $0,im,-im$, so all four two-torsion points are rational over ${\mathbb{F}}_p$. They form a subgroup of order four, hence $4\mid \#E({\mathbb{F}}_p)$ by Lagrange's theorem.

If $p\equiv 3(\mathrm{mod}4)$, let $\chi$ be the Legendre symbol with $\chi (0)=0$. Since $f(x)=x(x^2+m^2)$ is odd and $\chi (-1)=-1$, the terms for $x$ and $-x$ cancel in ${\sum }_x\chi (f(x))$. Counting the possible second coordinates givesThus in both casesThe assumptions $p\ne 2$ and $m\ne 0$ ensure nonsingularity and keep the displayed two-torsion roots distinct when they exist.