Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2009 (C9.1, DMAC 2482)

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.

Let $H$ denote the divisor class cut out by a line. The two lines in the chord-and-tangent group law giveTheir difference gives $P+Q\sim o+S$, henceAddition of divisor class is associative. Therefore both $(P+Q)+R$ and $P+(Q+R)$ map to $[P-o]+[Q-o]+[R-o]$. Injectivity of $\iota$ proves the chord-and-tangent group law is associative. The same argument gives identity $o$, commutativity and inverses, and covers tangencies through the use of intersection multiplicities. All constructions are defined over $K$.

The three suggested values areThere is an obstruction at $p=3$: every integer residue satisfies $x^3-x\equiv 0(\mathrm{mod}3)$, whereas $-1\equiv 2$ is not a square modulo three. Reduction of a solution in the p-adic integers would give an impossible solution modulo three.

At $p=2$, take $x=-2$. The odd 2-adic unit in a ring $-7$ is congruent to one modulo eight, so it is a square in ${\mathbb{Z}}_2$. We use the standard 2-adic square criterion that an odd unit in a ring is a square exactly when it is one modulo eight. Its sufficiency follows by successively adjusting an odd square root modulo $2^m$ by $2^{m-1}$ to obtain the next required square congruence. Thus $y^2=-7$ has a 2-adic integer solution.

For $p\ge 5$ other than $23$, the discriminant of $y^2=x^3-x-1$ isso this model has good reduction of an elliptic curve. Hasse's theorem on elliptic curves givesSince there is just one point at infinity, there is an affine point. Smoothness means that at least one partial derivative of $y^2-x^3+x+1$ is a unit in a ring modulo $p$, so Hensel's lemma lifts the affine point to ${\mathbb{Z}}_p^2$. The last exceptional odd prime is also soluble: at $p=23$, $h(-2)=-7\equiv 16$ has square roots $\pm 4$, which are simple roots of $Y^2+7$ modulo $23$, and lift by Hensel's lemma.

For small-prime checks the suggested values also give $x=0$ at $p=5$ and $x=3$ at $p=7$, where respectively $-1\equiv 2^2$ and $23\equiv 3^2$. ThereforeThe same prime list holds if one only asks for a solution modulo $p$, but the lifting above establishes the asserted p-adic integer solutions. This is local solubility of the cubic x cubed minus x minus one.

In the elliptic curve method, addition is attempted modulo the composite number. A denominator that is not a unit in a ring supplies a factor through its greatest common divisor with that number.

First double $P$ using the elliptic-curve addition formula. Its tangent slope isThese calculations are valid modulo $221$ since the only initial denominator is two, which is invertible. Thus $2P=(14,-53)$, equivalently $(14,168)$ modulo $221$.

To form $3P=P+2P$, the chord slope would requireThe denominator hasIt cannot be inverted, and thirteen is the requested proper factor. The failure has a geometric explanation: modulo thirteen, $2P=(1,-1)=-P$, so $3P$ is the point at infinity. Modulo seventeen that chord denominator remains nonzero. The differing behavior at the two primes is exactly what the elliptic curve method detects.

Choose the blocksIf $p^k\le n<p^{k+1}$, then $|x_n{|}_p=p^{-k}$, so $x_n\to 0$ in ${\mathbb{Q}}_p$. By the ultrametric inequality, for $N\ge M$,The partial sums are Cauchy sequence, and completeness of ${\mathbb{Q}}_p$ proves convergence of $\sum x_n$.

In the real series of p-adic absolute values, the block $p^k\le n<p^{k+1}$ contains $(p-1)p^k$ equal terms of size $p^{-k}$. Its sum is $p-1$, independently of $k$. Infinitely many such positive block sums force divergence. This exhibits p-adic convergence without summable absolute values: the criterion is that the terms tend to zero, rather than real absolute summability.

Put $U(T)=F_X(0,T)$. The formal group identity $F(X,0)=X$ gives $U(0)=1$, so $U$ is a unit in a ring in $R[[T]]$, even if $R$ has zero divisors.

A differential $h(T)dT$ is invariant under formal translation ifas an identity of formal power series. Setting $X=0$ gives $h(T)U(T)=h(0)$. Consequently every invariant differential of a formal group law, if it exists, must have the form $h(0)U(T{)}^{-1}dT$. In particular there can be only one with constant coefficient one.

Existence follows directly from associativity. Differentiatewith respect to $Z$ at zero, obtainingEvery displayed $U$ is a unit in a ring, soThis is precisely invariance of $U(T{)}^{-1}dT$. ThusNormalization means its constant coefficient is one. This proves both existence and uniqueness over an arbitrary commutative ring, without division by any integer.

Let ${\omega }_G=h_G(T)dT$. The pullback through the formal group homomorphism isThe identity $f(F(X,T))=G(f(X),f(T))$ and the chain rule show that this pullback is invariant under $F$: translation by $T$ before $f$ becomes translation by $f(T)$ afterward. By part (i) it is a scalar multiple of ${\omega }_F$. Its coefficient at zero is $h_G(0)f^{\prime }(0)=f^{\prime }(0)$, so

Apply this to the multiplication-by-$p$ endomorphism $q(T)=[p](T)$. Since the linear coefficient of $F$ is $X+Y$, repeated addition gives $q^{\prime }(0)=p$. If ${\omega }_F=h(T)dT$ with $h(0)=1$, the pullback formula givesbecause $h(q(T))$ is a unit in a ring. Write $q(T)={\sum }_{m\ge 1}{a}_m{T}^m$. Then $m{a}_m\in pR$ for every $m$. If $p\nmid m$, choose integers $u,v$ with $um+vp=1$. It follows thatChoose $b_m\in R$ with $a_m=p{b}_m$ for those $m$ not divisible by $p$, and defineCoefficient comparison now provesNo cancellation of $p$ was used, so this prime-multiplication coefficient splitting for formal groups remains valid when $p$ is a zero divisor in $R$.

The discriminant of $y^2=x^3-9$ is $-16\cdot 27\cdot 81=-2^4{3}^7$, so five and seven are primes of good reduction of an elliptic curve. Count solutions over their residue fields.

Modulo five the values of $x^3-9$ for $x=0,1,2,3,4$ are $1,2,4,3,0$. They yield $2,0,2,0,1$ affine points, respectively, and the point at infinity gives $\#E({\mathbb{F}}_5)=6$.

Modulo seven the values for $x=0,\dots ,6$ are $5,6,6,4,6,4,4$. Only the three values four are nonzero squares, giving six affine points and $\#E({\mathbb{F}}_7)=7$.

For a rational elliptic curve at an odd prime of good reduction of an elliptic curve, the kernel of reduction on ${\mathbb{Q}}_p$ has no torsion element: it is the formal group on $p{\mathbb{Z}}_p$, whose formal logarithm embeds it in the additive group $p{\mathbb{Z}}_p$. Thus the entire torsion subgroup injects into each of the two finite reduction groups. Its order divides both six and seven, hence is one:Equivalently one may use prime-to-residue-characteristic injection at both primes; it gives the same coprime bound. This is a two-prime bound on rational elliptic torsion.

Since neither $n$ nor $-n$ is a square, both cubics have exactly one rational root, zero. Thus each torsion point of an elliptic curve group has exactly one nonidentity 2-torsion, namely $T=(0,0)$.

Neither curve has a rational point of order four. Indeed on $E_m:y^2=x^3+mx$ the duplication formula givesIf $2P=T$, then $P$ is neither the identity nor a 2-torsion, so the denominator is nonzero and $x(P{)}^2=m$. Thus $m$ must be a rational square; an integer that is a rational square is an integer square. Applying this with $m=n$ or $m=-n$ contradicts the hypotheses.

If $3\nmid n$, both curves have good reduction of an elliptic curve at three and exactly four points over ${\mathbb{F}}_3$. To see this, $x^3=x$ there, so the equation is $y^2=(1+m)x$. For $m=1$ the right side runs through all residues, and for $m=-1$ it is always zero; either way there are three affine points. Injection of torsion point of an elliptic curve at this odd good prime bounds each torsion element order by four. An order-four torsion subgroup with just one nonzero 2-torsion would contain an order-four point, which was excluded. Therefore in this case both torsion subgroups equal $\{o,T\}$.

The remaining case is $3\mid n$. The literal hypothesis $15\nmid n$ then ensures $5\nmid n$, so both curves have good reduction of an elliptic curve at five. Direct square counting givesFor example the square residues are $0,1,4$; substituting $x=0,\dots ,4$ gives this table, including the point at infinity. The pair $m=n,-n$ has residues $(1,4)$ or $(2,3)$ in some order. In either pair at least one reduction group has order two or four. The same injection and absence of order-four points then force that curve's torsion point of an elliptic curve to have order two. ThusThis torsion alternative for opposite quartic-twist coefficients uses only $15\nmid n$; it does not strengthen the source hypothesis to coprimality with fifteen.

We use two-isogeny descent, stating the arithmetic results involved. Fordefine the two-torsion square-class homomorphism ${\alpha }_E:E(\mathbb{Q})\to {\mathbb{Q}}^{\times }/{\mathbb{Q}}^{\times 2}$ by ${\alpha }_E(o)=1$, ${\alpha }_E((0,0))=b$, and ${\alpha }_E((x,y))=x$ otherwise. Its kernel is the image of the dual two-isogeny formula; the analogous map ${\alpha }_{E^{\prime }}$ has kernel the image of the forward two-isogeny formula. The two-isogeny rank formula isThe images have signed squarefree representatives supported on primes dividing $b$ and $a^2-4b$, respectively. This prime-support bound in two-isogeny descent follows by valuation comparison: if a prime does not divide $b$, the valuations of $x$ at that prime must be even, whether positive or negative. A candidate squarefree class $d$ is represented precisely when its quartic covering in a two-isogeny descenthas the corresponding rational solution, apart from the separately included identity and two-torsion cases; clearing denominators permits coprime integers $U,V$.

Here $a=5,b=-5$, soOn $E$ the possible classes are $1,-1,5,-5$. They all occur: the identity gives one, $T=(0,0)$ gives $-5$, and $(-1,3)$ gives $-1$. Since the image is a group, their product also gives five. Thus its image has order four.

On $E^{\prime }$, the quadratic factor $(X-5{)}^2+20$ is strictly positive for real $X$. Every nonzero rational $X$-coordinate is therefore positive. The prime-support bound in two-isogeny descent restricts its image to $1,3,5,15$. The identity gives one and $(0,0)$ gives $45$, that is class five.

Class three is impossible. Its quartic covering in a two-isogeny descent would requireModulo three this is $N^2\equiv 2U^2{V}^2$, so three divides $N$ and at least one of $U,V$. They cannot both be divisible. If $3\mid U$ and $3\nmid V$, the right side has valuation exactly one, from $15V^4$. If $3\mid V$ and $3\nmid U$, it again has valuation exactly one, from $3U^4$. Either case contradicts its being a square. Class fifteen is then impossible too, since multiplying it by the known image class five would give class three.

Hence the second image is exactly $\{1,5\}$, of order two. The rank formula givesThis proves the rank-one elliptic curve with coefficients five and minus five calculation, including the local obstruction that removes the spurious extra descent classes.

For $E:y^2=x^3+px$ with $p>0$, the factor $x^2+p$ is strictly positive on the real line. Thus every nonzero rational $x$-coordinate is positive. In the two-isogeny descent map, prime support restricts its square class to $1$ or $p$. Both are realized by the identity and $(0,0)$, although only the upper bound is needed:The two-isogenous curve is $E^{\prime }:Y^2=X^3-4pX$. Its two-torsion square-class homomorphism has signed squarefree representatives supported on the primes dividing $4p$. For odd $p$ these are amongHence the second image has at most eight elements. At $p=2$ the list reduces to $1,-1,2,-2$, so the bound by eight remains valid. The two-isogeny rank formula therefore givesIn fact the same simple count yields $r\le 1$ when $p=2$. The positivity restriction on the first descent image is essential for the stated bound of two for all primes.

Use the two-isogeny formula formula with $a=\alpha$ and $b=-\alpha$. Its companion isBoth curves are nonsingular because $\alpha \ne 0,-4$. With $b^{\prime }=\alpha (\alpha +4)$, chooseEach map extends across its exceptional affine points, and $\hat{\varphi }\varphi =[2]$. On $D$, the first coordinate of the dual can also be written $v^2/(4u^2)$.

Let $P=(1,1)$ and solve $\hat{\varphi }(u,v)=P$. The first coordinate gives $v^2=4u^2$. Since $u\ne 0$, the equation of $D$ then givesThe second coordinate selects the signs. The two rational preimages are exactlySubstituting into the dual formula checks that both map to $(1,1)$. Their coordinates are nonexceptional under the assumptions, and the degree-two map has no further preimages.

The two-torsion square-class homomorphism ${\alpha }_D$ has kernel $\varphi (C(\mathbb{Q}))$, and on these two points it is their nonzero $u$-coordinate modulo squares. Thus $Q_1$ is in the image of $\varphi$ exactly when $\alpha$ is a rational square, and $Q_2$ is in that image exactly when $\alpha +4$ is a rational square.

If $P=2R$, then $\varphi (R)$ is one of the two preimages, because $\hat{\varphi }\varphi (R)=2R=P$. Conversely, if $Q_j=\varphi (R)$, applying the dual shows $P=2R$. ThereforeFinally, if an integer equals $(a/b{)}^2$ in lowest terms, then $b^2\mid a^2$ forces $b=1$. Thus rational squareness of these integer parameters is exactly integer squareness, andThis is the two-isogeny halving criterion for the alpha family.