Analytic number theory and elliptic curves course of the University of Oxford (C9.1)

Source: Mathematics release, bundle Part C 2007 11.pdf, pages 365–371. The combined Mathematics and Mathematics and Philosophy Part C paper was sat on 5 June 2010, 2.30–5.30pm.

Bounded exact-paper availability checks found no matching answer advertisement. The general even-year policy establishes coverage back to at least 2016 and does not itself establish this 2010 paper. Both sections and all printed targets are covered below; the unlettered theorem requests in Questions 4 and 5 are included alongside their displayed subdivisions.

For $N=\lfloor x\rfloor$, interchange a finite sum and an integral:This proves the Abel's summation formula, including the term at $n=1$.

Apply it with $f(n)=1$, $F(t)=1/t$. Writing $\{t\}=t-\lfloor t\rfloor$ givesThe last two terms are $O(x^{-1})$, proving the first expansion. For the refinement let $B_1(t)=\{t\}-1/2$. It has mean zero on each unit interval, so its primitive $P(t)={\int }_0^t{B}_1(u)du$ is bounded and periodic. Integration by parts givesThus for integer $x$, where $\{x\}=0$,For the zeta representation, integrating each defining integral by parts yieldsSumming from $1$ to $N-1$ and substituting into the definition of $\zeta$ givesSince $0\le x-n\le 1$,This is uniform on compact subsets of $\mathrm{Re}s>0$, so the series is holomorphic there and limits at $s=1$ are legitimate after subtracting the pole. Expanding $N^{1-s}=1-(s-1)\log N+O((s-1{)}^2)$ givesThe final tail is $O(N^{-1})$, and the harmonic expansion sends the remaining difference to $\gamma$. Therefore

Use the convention $f(1)=1$ in the definition of a multiplicative arithmetic function. For a finite set of primes, multiplying the absolutely convergent local series gives the Dirichlet sum over integers supported on those primes, by unique prime factorization. If all primes up to $Y$ are included, any missing integer is larger than $Y$, so the difference from the full series is bounded by ${\sum }_{n>Y}|f(n)|n^{-\sigma }$, which tends to zero. This proves the Euler product in its absolute-convergence half-plane.

For coprime $m,n$, every divisor of $mn$ is uniquely a product of a divisor of $m$ and a divisor of $n$. Hence ${\sigma }_{\alpha }(mn)={\sigma }_{\alpha }(m){\sigma }_{\alpha }(n)$. The squared modulus is also multiplicative when $\alpha =i\theta$.

At a prime put $z=p^{-s}$ and $a=p^{i\theta }$, so $|a|=1$. Summing by the two divisor exponents givesFor example, expand the square as ${\sum }_{j,l\le k}{a}^{j-l}$ and sum $z^k$ from $k=\max (j,l)$ to obtain this rational expression. It remains valid for $a=1$. Absolute convergence for $\sigma >1$ follows from $|{\sigma }_{i\theta }(p^k)|\le k+1$ and convergence of ${\sum }_p{p}^{-\sigma }$. Multiplying the factors therefore provesSuppose now that $\zeta (1+i\theta )=0$. Necessarily $\theta \ne 0$, since $s=1$ is a pole. Complex conjugation gives a zero also at $1-i\theta$. In the numerator, the double pole at $s=1$ is canceled by those two zeros in the shifted factors. At $s=1\pm i\theta$, the possible simple pole of a shifted factor is canceled by the double zero of $\zeta (s{)}^2$. Thus on $\mathrm{Re}s>0$ the only possible remaining poles come from zeros of $\zeta (2s)$. Since the Euler product excludes zeros for $\mathrm{Re}(2s)>1$, these possible poles lie in $0<\mathrm{Re}s\le 1/2$. In particular $f$ is regular for $\mathrm{Re}s>1/2$.

The assumed nonnegative-series principle now makes its Dirichlet series converge and equal $f$ there. Its first coefficient is one and all coefficients are nonnegative, so $f(\sigma )\ge 1$ for real $\sigma >1/2$. But as $\sigma \downarrow 1/2$, the numerator remains finite while $\zeta (2\sigma )\to +\infty$, forcing $f(\sigma )\to 0$. This contradiction provesAt the real point $s=1$ the statement is understood as a pole rather than a finite nonzero value. The pole-location argument did not assume the zero-free boundary before proving it.

The vertical integral is absolutely convergent because the denominator is quadratic in the imaginary part. If $x<1$, move its line to the right: no poles are crossed, and the integral on $\mathrm{Re}s=R$ is $O(x^R/R)$, tending to zero. If $x>1$, move left across the simple poles at $0,-1$. Their residues are $1$ and $-x^{-1}$, and the far-left integral tends to zero. Horizontal segments vanish by their $O(|\mathrm{Im}s{|}^{-2})$ decay for a fixed pair of vertical lines. At $x=1$, dominated convergence on the original line gives the limiting value zero. Hence the kernel is $0$ below one and $1-x^{-1}$ at or above one.

On a line $c>2$, absolute convergence permits interchanging the totient Dirichlet series and this integral. The $n$th term is $x\varphi (n)$ times the kernel at $x/n$, giving exactly $\varphi (n)(x-n)$ for $n\le x$. ThusWe may take $c=5/2$ and shift to $\mathrm{Re}s=3/2$. The denominator has no zeros in this strip; the only pole crossed is $s=2$, where $\zeta (s-1)$ has residue one. Its residue in the integrand isOn the new line, the supplied estimates bound the integrand by $O(x^{5/2}|t{|}^{-3/2}{\log }^8|t|)$ for $|t|\ge 2$, an integrable function. The compact portion is bounded as well. The same estimates, uniformly across the strip, make the horizontal integrals tend to zero. ConsequentlyFor the Tauberian step let $A(x)={\sum }_{n\le x}\varphi (n)$. It is nondecreasing because the summands are nonnegative, and $S(x)={\int }_0^{x}A(t)dt$. For fixed $0<ϵ<1$,After division by $x^2$ and passage to lower and upper limits, the asymptotic for $S$ gives bounds $(3-3ϵ+{ϵ}^2)/{\pi }^2$ and $(3+3ϵ+{ϵ}^2)/{\pi }^2$. Letting $ϵ\downarrow 0$ squeezes them together. ThereforeThis uses monotonicity of the summatory function; differentiating an asymptotic formula without such control would not justify the conclusion.

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,

absolute convergence depends only on $\sigma =\mathrm{Re}s$, since $|a_n{n}^{-s}|=|a_n|n^{-\sigma }$. Let $A$ be the set of real $\sigma$ for which $\sum |a_n|n^{-\sigma }$ converges. Comparison with a convergent member shows that $A$ is upward closed. The hypotheses give a member of $A$ and a real number outside $A$, so its infimum ${\sigma }_0$ is finite. If $\sigma >{\sigma }_0$, some $\tau \in A$ lies below $\sigma$, giving convergence by comparison; if $\sigma <{\sigma }_0$, convergence would contradict the definition of the infimum. Thus ${\sigma }_0=\inf A$ is the abscissa of absolute convergence of a Dirichlet series. This argument leaves convergence on the boundary open.

Partition $1,\dots ,n$ according to $d=n/\gcd (k,n)$. Writing $k=(n/d)j$ identifies the part indexed by $d\mid n$ with the integers $1\le j\le d$ coprime to $d$. There are $\varphi (d)$ of them. Counting the entire partition proves the totient divisor-sum identity

We use the multiplication theorem for absolutely convergent Dirichlet series: their product is the Dirichlet series of the Dirichlet convolution of their coefficients. absolute convergence justifies rearrangement of the double sum. Here $0<\varphi (n)\le n$, so its series converges absolutely for $\sigma >2$, andThe Euler product implies $\zeta (s)\ne 0$ for $\sigma >1$. Division gives

We finally use the simple pole of the Riemann zeta function at one, of residue one, and $0<\zeta (2)<\infty$. The quotient therefore tends to $+\infty$ as real $s\downarrow 2$. If $\sum \varphi (n)n^{-2}$ converged, its nonnegative terms would dominate the series at every real $s>2$, giving a uniform finite bound and a contradiction. Thus the series diverges at two; comparison then excludes absolute convergence at every $\sigma <2$. Combined with the convergence already proved, the abscissa is exactly two. This is the totient Dirichlet series and its absolute convergence boundary.

summation by parts, with $S(x)={\sum }_{n\le x}\chi (n)$, gives for an integer $N$The boundary term tends to zero when $\mathrm{Re}s>0$ because $S$ is bounded. In the original region of absolute convergence this provesFor a compact subset of $\mathrm{Re}s>0$, choose $\delta >0$ below all its real parts and bound $|s|$ there. The integral tail is bounded by a constant times ${\int }_N^{\infty }{x}^{-1-\delta }dx$, uniformly on that compact set. Its truncated integrals are holomorphic function in $s$, so locally uniform convergence proves a analytic continuation throughout $\mathrm{Re}s>0$. This is holomorphy of a Dirichlet series from bounded partial sums.

The Euler product, valid for $\mathrm{Re}s>1$, isExpand the logarithms of these absolutely convergent Euler products. Writing $\theta =mt\log p$, the logarithm of the product in the desired inequality isFor odd $p$, $\chi (p{)}^m$ is $1$ or $-1$. In the first case the numerator is the given nonnegative trigonometric polynomial. In the second replace $\theta$ by $\theta +\pi$, which changes the sign of $\cos \theta$ and leaves $\cos 2\theta$ unchanged. At $p=2$ the numerator is $3+\cos 2\theta \ge 2$. Every summand is nonnegative. Exponentiation provesThis is the three-four-one inequality for Euler products applied to the real character modulo four.

Suppose $t\ne 0$ and $L(1+it)=0$. Holomorphy gives $L(\sigma +it)=O(\sigma -1)$ as $\sigma \downarrow 1$. The only pole of the Riemann zeta function is at one, so $\zeta (\sigma +2it)$ remains bounded, while $\zeta (\sigma {)}^3=O((\sigma -1{)}^{-3})$. The displayed product is then $O(\sigma -1)$, contradicting its lower bound. Hence $L(1+it)\ne 0$ for $t\ne 0$.

At $t=0$ the last zeta factor also has a pole, so that contradiction no longer follows from the inequality. Instead evaluate the convergent alternating series:For rigor, integrate the finite geometric identity for $(1+x^2{)}^{-1}$; the remainder has absolute integral at most $1/(2N+3)$ and tends to zero. summation by parts above identifies this value with the analytic continuation at one.

Write $s=\sigma +it$; the source's rectangle notation $|\mathrm{Im}(t)|\le T$ is interpreted as the intended height restriction $|t|\le T$. We use the standard facts that $\zeta$ has no zeros on $\sigma \ge 1$, has a simple pole at one, and is meromorphic with no other pole. The function $(s-1)\zeta (s)$ is holomorphic function and nonzero on the compact set $\{1+it:|t|\le T\}$, including its removable value one at $s=1$. It remains nonzero in a neighborhood of this compact set. Together with nonvanishing to its right this gives some $0<{\alpha }_T<1$ such that the whole closed rectangle ${\alpha }_T\le \sigma \le 2$, $|t|\le T$ contains no zeta zero. The pole at one is harmless: $1/\zeta$ extends holomorphically there with value zero.

The linearly smoothed Perron kernel, for $y>0$ and $c>0$, isFor $y>1$ shift the contour left across the simple poles at zero and minus one, whose residues are $y$ and $-1$. For $y<1$ shift right, crossing no pole. At $y=1$ the integral is zero, also obtained by direct integration of $1/s-1/(s+1)$. The quadratic denominator gives absolute integrability on the original vertical line.

For $c>1$ use the absolutely convergent Dirichlet series $1/\zeta (s)=\sum \mu (n)n^{-s}$. Interchanging the sum and the contour integral is justified by $\sum |\mu (n)|n^{-c}<\infty$ and the integrable denominator. The term for $n$ is $n$ times the kernel at $y=x/n$, hence $(x-n)$ when $n<x$ and zero otherwise. Consequently

Here is a contour argument that needs only the stated bound, with no unproved uniform estimate left of one. Take $x>e$, $c=1+1/\log x$, and first hold $T\ge 2$ fixed. On the two tails $|t|>T$, the given bound and $x^{c+1}=e{x}^2$ giveShift the remaining segment to $\sigma ={\alpha }_T$. The integrand is holomorphic function inside the rectangle: the zero of $1/\zeta$ at one cancels the apparent zeta pole, and $0,-1$ lie outside. On the fixed left side its integral is $O_T(x^{1+{\alpha }_T})=o_T(x^2)$. On the two horizontal sides, $1/(\zeta (s)s(s+1))$ is bounded by some finite constant depending on $T$. Therefore their integrals are bounded byIt follows thatNow let $T\to \infty$; the integrable tail tends to zero. ThusThe order of limits matters: first $x\to \infty$ for each fixed rectangle, then $T\to \infty$. This is linearly smoothed Möbius cancellation from finite zero-free rectangles.

Work with the smooth projective cubic, including its points at infinity. For two points $P,Q$, draw their chord, using the tangent when they coincide, and let $R$ be its third intersection counted with intersection multiplicity. Draw the line through $R$ and $o$, again using the tangent if necessary, and let $S$ be its third intersection. Define $P+Q=S$. When $o$ is a flex, this is the familiar reflection of the first third-intersection point across the Weierstrass involution. The two-line construction also works when the chosen identity is not a flex.

To prove associativity without assuming it in the geometry, use divisor class. A nonsingular plane cubic has genus one. By the Riemann-Roch theorem, every divisor class of degree one contains exactly one effective divisor on an algebraic curve of degree one: for such a divisor on an algebraic curve $D$,and $\mathrm{deg}(K_C-D)=-1$, so $\ell (K_C-D)=0$. Thus $\ell (D)=1$, and the nonzero section is unique up to scalar. Over $K$, a degree-one effective divisor on an algebraic curve defined over $K$ is a $K$-rational point. This also proves uniqueness of that point. In particular the mapis injective and identifies these points with the corresponding degree-zero Picard group of a curve: to recover the point from a class, add $[o]$ and take its unique effective representative. This is the elliptic curve group law from Riemann-Roch.