Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 15 (Algebraic Geometry)

Choose an oriented basis ${\omega }_1,{\omega }_2$ of the period lattice, so $\mathrm{Im}({\omega }_2/{\omega }_1)>0$. The Weierstrass elliptic function is constructed by the normally convergent seriesOn compact sets away from the lattice the summands are $O(|\omega {|}^{-3})$, which is summable over a plane lattice. It is even, has double poles at the lattice with zero residues, and is periodic. To check periodicity without splitting divergent series, differentiate termwise: ${\wp }^{\prime }(z)=-2{\sum }_{\omega \in \Lambda }(z-\omega {)}^{-3}$ is invariant under lattice shifts. Hence $\wp (z+{\omega }_j)-\wp (z)$ is constant; evenness evaluated at $z=-{\omega }_j/2$ makes this constant zero.

The Weierstrass sigma function is the canonical productIts logarithmic tails are likewise $O(|\omega {|}^{-3})$, so it defines an entire function with simple zeros exactly at the lattice and ${\sigma }^{\prime }(0)=1$. Pairing opposite lattice factors makes it odd. Its logarithmic derivative is the Weierstrass zeta functionThus ${\eta }_j=\zeta (z+{\omega }_j)-\zeta (z)$ is constant. Integrating this identity and evaluating the sigma ratio at $z=-{\omega }_j/2$ givesIntegration of zeta around a fundamental cell, containing one pole of residue one, gives the Legendre relation for Weierstrass quasi-periods ${\eta }_1{\omega }_2-{\eta }_2{\omega }_1=2\pi i$. These formulas explain why sigma itself is not an elliptic function, but balanced products of translates can be.

For completeness, the basic algebraic property of wp is $({\wp }^{\prime }{)}^2=4{\wp }^3-g_2\wp -g_3$, where $g_2=60{\sum }_{\omega \ne 0}{\omega }^{-4}$ and $g_3=140{\sum }_{\omega \ne 0}{\omega }^{-6}$. The identity follows by canceling the principal parts using the Laurent expansion: the difference is an entire elliptic function, hence constant, and its constant term is zero. The discriminant is nonzero; equivalently the three half-period values are distinct. These properties lead to the cubic embedding in 2(ii).

Now let $F$ be a nonzero meromorphic function on the torus and lift it to a periodic meromorphic function on the plane. Its logarithmic derivative $g=F^{\prime }/F$ is periodic. Choose a fundamental parallelogram with no zero or pole on its boundary. The residue theorem applied to $g$ gives equality of the total zero and pole multiplicities. Applied to $zg$, it gives the difference of the sums of the zero and pole lifts. Pairing opposite edges explicitly yieldsEach integral of $F^{\prime }/F$ on the right is $2\pi i$ times an integer, since its endpoint values of $F$ coincide and the image path avoids zero. Therefore the zero sum and pole sum agree in $\mathbb{C}/\Lambda$.

Conversely, suppose the point sums agree modulo the lattice. Change one chosen lift of a point by a lattice element so the two sums agree exactly in $\mathbb{C}$; this changes no point of the torus. Thenhas the prescribed divisor. Its multiplier under either basis period is $\exp ({\eta }_j(\sum Q_i-\sum P_i))=1$, so it descends to the torus. Repetitions contribute their appropriate multiplicities. This proves the principality criterion on a complex elliptic curve:There is an implicit qualification in the wording about separate zero and pole divisors: their supports must be disjoint. If a point occurs on both lists, cancel common multiplicities first when prescribing the divisor difference. A function cannot simultaneously have a zero and a pole at the same point; for example identical one-point lists satisfy the sum condition but do not specify separate zero and pole divisors. For $n=0$, any nonzero constant supplies the empty divisor.

One form of the Theorem of the Cube is the cube principle: a line bundle $M$ on $E^3$ which is trivial on each of $E\times E\times \{O\}$, $E\times \{O\}\times E$, and $\{O\}\times E\times E$ is trivial. Here is a proof for a complex elliptic curve.

Since $E$ is a real two-dimensional torus, its integral cohomology has no torsion. The Künneth theorem decomposesEvery single-factor summand is detected on a face containing that factor, and every cross summand is detected on the corresponding two-factor face. Restriction of $M$ to those faces is trivial, so its First Chern class has zero component in every summand. Thus $c_1(M)=0$.

The holomorphic exponential sequence identifies the kernel of $c_1$ in the holomorphic Picard group with $H^1(E^3,\mathcal{O})/\mathrm{im}{H}^1(E^3,\mathbb{Z})$. First cohomology decomposes factorwise here: $H^1(E^3,\mathcal{O})={⨁}_j{p}_j^{\ast }{H}^1(E,\mathcal{O})$ and likewise for integral first cohomology. Analytically the former decomposition says that the harmonic representatives of $(0,1)$-cohomology on the flat torus are the constant forms $d{\overline{z}}_j$, one from each factor. Consequently this kernel is ${\mathrm{Pic}}^0(E{)}^3$, andfor degree-zero line bundles $M_j$ on $E$. Restrict to each coordinate axis. The hypothesis makes every $M_j$ trivial, so $M$ is trivial. This proves the cohomological proof of the cube theorem for elliptic curves; the holomorphic isomorphism is algebraic on these projective varieties by GAGA.

To state its usual addition-map consequence, write $m_I:E^3\to E$ for summation of the coordinates indexed by $I$ and $p_j=m_{\{j\}}$. For any line bundle $L$ on $E$,More canonically, include the inverse of the constant bundle with fibre $L_O$ in the alternating product. Its restriction to a zero-coordinate face has pairwise cancellations, leaving the trivial bundle. The cube principle proves its triviality, which gives the displayed isomorphism after choosing a trivialization of the constant fibre. Pullback along any triple of maps to $E$ gives the corresponding alternating identity for their seven nonempty sums.

A pointwise divisor version can also be seen directly from Question 1. For a divisor $D$ and fixed points $a,b,c$, the alternating translation divisorhas degree zero and point sum zero: translation by $t$ subtracts $\mathrm{deg}(D)t$ from its point sum, and all the alternating terms cancel. The principality criterion proves that this divisor is principal.

Over a number field $K$, use algebraic line bundles and maps defined over $K$, and use Galois-invariant divisors rather than arbitrary individual complex points. The geometric line-bundle cube identity itself is still exact over $K$. To see descent in this setting, first pass to an algebraic closure, or an embedding in $\mathbb{C}$. The alternating bundle becomes trivial there by the preceding proof. Choose its nowhere-zero section with value one in a chosen $K$-basis at $(O,O,O)$. Any Galois conjugate of that section differs by an invertible global function, hence a constant on the connected projective variety; its normalized value at the rational origin forces that constant to be one. The section therefore descends, giving a $K$-defined trivialization.

For the arithmetic formulation used with heights, the necessary modification is a uniformly bounded error. A chosen Weil height machine representative $h_L$ satisfiesuniformly for $P,Q,R\in E(K)$. Indeed, pullback and tensor product of line bundles correspond to composition and addition of heights only modulo bounded functions. This is the arithmetic cube identity for Weil heights. Arbitrary naive heights do not satisfy an exact numerical cube identity; the canonical height of an elliptic curve removes the bounded error, as explained in Question 4. This arithmetic qualification does not weaken the exact geometric theorem.

The Weierstrass elliptic function and its derivative give the maponto the Weierstrass equation of an elliptic curve $Y^{2}Z=4X^3-g_{2}X{Z}^2-g_3{Z}^3$. Near zero, multiplying these coordinates by $z^3$ gives $[z+O(z^5):-2+O(z^4):z^3]$, so the map extends holomorphically and is an immersion at the origin.

The meromorphic map $\wp :E\to {\mathbb{P}}^1$ has degree two, since its only pole on the torus has order two. Its evenness implies $\wp (u)=\wp (v)$ exactly when $u\equiv v$ or $u\equiv -v$. Oddness of ${\wp }^{\prime }$ distinguishes these two alternatives unless they are the same point class. At each nonzero half-period, ${\wp }^{\prime }=0$. These three points exhaust its zeros, since its only pole has order three. The three half-period values are distinct: equality of two of them would give two distinct zeros each of order at least two for $\wp -e$, contradicting its total pole order two. They are roots of $4X^3-g_{2}X-g_3$, so the polynomial has distinct roots and the cubic is smooth. At a half-period $h$ with value $e$, ${\wp }^{\prime \prime }(h)=2{\prod }_{e^{\prime }\ne e}(e-e^{\prime })\ne 0$. Elsewhere ${\wp }^{\prime }$ is nonzero. The coordinate pair therefore has nonzero derivative at every finite point, and the map is injective. Compactness and the smooth cubic equation then identify it as an embedding onto the cubic.

The hyperplane section at infinity has divisor $3[O]$. Thus the intersection divisor $D_{\ell }$ of any line with this cubic is linearly equivalent to $3[O]$. Applying the principality criterion on a complex elliptic curve to $D_{\ell }-3[O]$ shows that its three points, counted with intersection multiplicity, have sum zero in $E$.

Conversely, if the degree-three divisor $D=[u]+[v]+[w]$ has point sum zero, Question 1 gives a meromorphic function with divisor $D-3[O]$. A meromorphic function with no poles away from $O$ and pole order at most three lies inTo prove this description, subtract multiples of ${\wp }^{\prime }$ and $\wp$ to cancel the cubic and quadratic Laurent principal parts. The remaining possible simple-pole coefficient is a residue, which must be zero because it is the only pole on a compact torus. The remaining holomorphic elliptic function is constant. Thus the constructed function is a linear form in the projective coordinates, and its line has intersection divisor exactly $D$. This proves the chord-and-tangent collinearity criterion is $u+v+w=0$ in $E$, with multiplicities understood.

For three pairwise distinct point classes away from $O$, ordinary collinearity is equivalent to linear dependence of the coordinate columns, soThe original PDF has this three-column determinant; the TeX's extra copied column is a transcription defect.

There is also a genuine implicit distinctness condition in the printed determinant assertion. If $u=v={\omega }_1/10$ and $w={\omega }_1/5$, all entries are finite and the determinant is zero because two columns coincide, but $u+v+w=2{\omega }_1/5\notin \Lambda$. Thus the unrestricted literal equivalence is false. To specify repeated intersections, use the confluent elliptic collinearity determinant: for a double point at $u$ replace the second column by $c^{\prime }(u)$, where $c(z)=(1,\wp (z),{\wp }^{\prime }(z){)}^t$; for a triple point use $c(u),c^{\prime }(u),c^{\prime \prime }(u)$. These impose tangency or inflection, and recover the correct conditions $2u+w\in \Lambda$ or $3u\in \Lambda$. If a point is $O$, use the homogeneous coordinates or a local homogeneous lift, since the displayed affine entries have poles there.

Let $K$ be a number field of degree $d$, and normalize its absolute values so that the product formula is ${\prod }_v|a{|}_v^{n_v}=1$ for $a\ne 0$, where $n_v=[K_v:{\mathbb{Q}}_v]$; at a complex place use the ordinary modulus with weight two. For $P=[a_0:\cdots :a_N]$, defineOnly finitely many factors differ from one. Rescaling all coordinates multiplies the local maxima by $|a{|}_v$, whose weighted product is one, so the projective height is well-defined. It is at least one by comparison with any nonzero coordinate. On enlarging $K$ to $L$, the identity ${\sum }_{w\mid v}[L_w:{\mathbb{Q}}_v]=[L:K][K_v:{\mathbb{Q}}_v]$ makes the normalized product unchanged. Thus this is an absolute height on algebraic projective points. For coprime rational integer coordinates it is simply their maximum absolute value.

For a projective variety $V/K$ and an embedding $i:V\hookrightarrow {\mathbb{P}}^N$, take $h_i=h\circ i$. Changing the basis of the same projective linear system changes the height by a bounded amount: the coordinate transformation and its inverse have fixed local coefficient bounds, equal to one outside finitely many places. If $L$ is ample and $L^m$ is very ample, set $h_L=m^{-1}{h}_i$ for an embedding from $L^m$, up to a bounded function. A general line bundle can be written $(L\otimes A)\otimes A^{-1}$ with both $A$ and $L\otimes A$ very ample, defining its height as the difference of their embedding heights.

This is the Weil height machine, withThe first relation follows from the Segre embedding, whose local maximum is the product of the two maxima; tensor powers similarly use the Veronese embedding. The second follows by pulling back the defining sections. These constructions make the choices equivalent modulo bounded functions. For a fixed homogeneous morphism with no base point, upper local height bounds come from its coefficients, and lower bounds from the Projective Nullstellensatz expressing coordinate powers in the ideal of its defining forms. The constants are one outside a finite set of places. This gives the uniform bounded errors and shows why heights belong to line-bundle classes rather than being independent of every projective embedding.

The Northcott lemma states that for fixed $B$ and $D$, the algebraic points of ${\mathbb{P}}^N$ with $H(P)\le B$ and $[\mathbb{Q}(P):\mathbb{Q}]\le D$ form a finite set. In particular the $K$-rational points of bounded projective height are finite. Here is a proof, including the degree bound.

First let $\alpha$ have degree $e\le D$ and $H(\alpha )=H([1:\alpha ])\le B$. Write its primitive minimal polynomial as $a_e{\prod }_{j=1}^e(T-{\alpha }_j)$, with positive integer $a_e$. The height-Mahler measure formula isIts finite-place part follows from multiplicativity of the nonarchimedean polynomial coefficient norm: a primitive integer polynomial has norm one at each prime, so factoring it bounds the product of the roots' local maxima by the inverse leading-coefficient norm. The archimedean part is the displayed product over conjugates. Combining them with the product formula gives the identity.

Expanding the polynomial now bounds each integer coefficient byThere are finitely many degrees and finitely many such integer coefficient lists, each with finitely many roots. Hence there are only finitely many possible $\alpha$. This proves the minimal-polynomial coefficient proof of Northcott finiteness for algebraic numbers.

On the projective chart $a_j\ne 0$, each ratio ${\alpha }_i=a_i/a_j$ has degree at most the point's field degree andThe equality uses the product formula. Each coordinate ratio therefore belongs to a finite set, and there are finitely many charts. This proves the projective Northcott theorem. Restricting to $V$ gives the result for a chosen embedding. For an ample line-bundle height, the comparison $m{h}_L=h_i+O(1)$ reduces its bounded-height sets to this case.

Both qualifications matter: a degree bound cannot be dropped, since infinitely many roots of unity have height one; and the finiteness assertion need not hold for a nonample line bundle, such as the trivial bundle's bounded height on a variety with infinitely many rational points.

Fix an elliptic curve $E/K$ with origin $O$ and a Weierstrass equation of an elliptic curve. Define the naive logarithmic $x$-height by $h_x(P)=h([x(P):1])$ for $P\ne O$ and $h_x(O)=0$. The $x$-map has degree two and pulls the projective hyperplane bundle back to ${\mathcal{O}}_E(2[O])$. Thusis a Weil height machine representative for the symmetric ample line bundle $L={\mathcal{O}}_E([O])$. This factor of one half fixes the convention for the normalized height on an elliptic curve used below.

The Theorem of the Cube gives the quadratic geometric identities. Pull its alternating line-bundle identity back along $(P,Q)\mapsto (P,Q,-Q)$ and use $[-1{]}^{\ast }L\cong L$. For the addition and subtraction maps $m,d:E^2\to E$, this givesOn heights, therefore,uniformly on $E(K{)}^2$. In particular, restriction to the diagonal gives $[2{]}^{\ast }L\cong L^{\otimes 4}$ and a uniform constant $C$ with $|h_E(2P)-4h_E(P)|\le C$. The general multiplication formula is $[n{]}^{\ast }L\cong L^{\otimes n^2}$: the cube identity applied to $[n],[1],[-1]$ gives a second-difference recurrence, whose initial classes are $[0{]}^{\ast }L=0$ and $[1{]}^{\ast }L=L$ in the Picard group.

Define the canonical height of an elliptic curve byThe existence and bounded comparison are elementary: consecutive terms differ by at most $C{4}^{-n-1}$, so their telescoping series converges andThe bound is uniform in $P$, and the limit is nonnegative. Shifting the index gives $\hat{h}(2P)=4\hat{h}(P)$. This is the unique function within bounded distance of $h_E$ with that doubling rule: the difference of two such functions at $P$ is $4^{-n}$ times their bounded difference at $2^{n}P$, and hence zero. A change of Weierstrass model changes the naive height only by a bounded function, so the normalized limit is independent of that model.

Apply the approximate parallelogram identity to $2^{n}P,2^{n}Q$, divide by $4^n$, and pass to the limit. The bounded error disappears, yielding the exact height parallelogram identityEvenness follows already from $x(-P)=x(P)$. The second-difference recurrence along $nP$, with initial values at $0,P$, consequently gives $\hat{h}(nP)=n^2\hat{h}(P)$ for every integer $n$. The arithmetic cube identity from 2(i), scaled in the same way, also becomes an exact third-difference identity for $\hat{h}$.

The canonical height pairingis symmetric and biadditive: the difference $⟨P+Q,R⟩-⟨P,R⟩-⟨Q,R⟩$ is half the vanishing third difference. Its diagonal is $\hat{h}(P)$. Positivity gives the Cauchy-Schwarz inequality for this pairing without first assuming finite generation. Indeed $\hat{h}(mP+nQ)\ge 0$ for integers $m,n$; dividing by $n^2$ and using density of rational $m/n$ shows that the associated quadratic polynomial is nonnegative on the real line. Its discriminant therefore gives $⟨P,Q{⟩}^2\le \hat{h}(P)\hat{h}(Q)$.

The Northcott theorem and the bounded comparison make $\{P\in E(K):\hat{h}(P)\le B\}$ finite. The map $x:E\to {\mathbb{P}}^1$ has finite fibres, so bounded $x$-height indeed controls the number of points, including $O$. This proves the important criterionFor a torsion point the integer quadratic rule gives zero. Conversely, height zero makes all $2^{n}P$ belong to one finite bounded-height set; two coincide, forcing $(2^j-2^i)P=O$ for some $j>i$. This argument also proves torsion finiteness over the fixed number field.

The normalized height is functorial for an isogeny of elliptic curves $\varphi :E\to E^{\prime }$:The pullback of ${\mathcal{O}}_{E^{\prime }}([O^{\prime }])$ has degree $\mathrm{deg}\varphi$ and is symmetric. Its difference from that many copies of $[O]$ is a symmetric degree-zero line bundle, hence a two-torsion Picard class. Its height is bounded, so the ordinary height relation has a bounded error. Commuting the isogeny with $[2^n]$ and taking the normalized limit removes that error. This explains the natural quadratic scaling under multiplication and the usefulness of normalized heights in comparing isogenous curves.

Here is their role in the Mordell-Weil theorem. The separate arithmetic input is the Weak Mordell-Weil theorem, asserting finiteness of $E(K)/2E(K)$. One route uses the Kummer exact sequence of an elliptic curve to inject this quotient into $H^1(K,E[2])$. The division torsors are unramified outside finitely many places containing those over $2$ and those of bad reduction. After a finite Galois extension splitting $E[2]$, their classes are pairs of square classes with even valuations outside this finite set. These square classes form a finite group: the obstruction to being represented by an S-unit lies in the finite ideal class group's two-torsion, and the S-unit group is finitely generated, hence finite modulo squares. The inflation-restriction kernel is a cohomology group of a finite group with finite coefficients, also finite. This supplies the finite quotient without using the full theorem in advance.

Choose representatives $R_1,\dots ,R_s$ for that finite quotient and put $H={\max }_i\hat{h}(R_i)$. Every point has the form $P=2Q+R_i$. Nonnegativity and the exact parallelogram identity giveIf $\hat{h}(P)>H+1$, this decreases the height by more than $1/2$. Repeating must reach the finite set $T=\{P:\hat{h}(P)\le H+1\}$. Unwinding the halving expressions writes the initial point as an integer linear combination of the $R_i$ and a point of $T$. Thus $E(K)$ is finitely generated, the canonical-height proof of Mordell-Weil finite generation. The normalized height supplies the descent and Northcott supplies its finite endpoint; neither by itself supplies the weak theorem's finite quotient.

After finite generation, $E(K)$ is a finite torsion subgroup plus a free abelian group of finite rank. The canonical height pairing is positive-definite on the associated real vector space. To justify the latter rather than assuming it, if its nonnegative quadratic form had a real null direction, rounding longer and longer multiples of that direction to lattice points would produce infinitely many points of bounded height: the null-direction cross terms vanish, leaving only uniformly bounded rounding errors. This contradicts Northcott finiteness. Hence the free part is a Euclidean height lattice. Heights can consequently bound point searches, certify independence through positive Gram determinants, and organize the arithmetic complexity of rational points.