Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 125 (Elliptic Curves)

A one-dimensional commutative formal group law over $R$ is a power series $F(X,Y)\in R[[X,Y]]$ satisfyingA homomorphism $\theta :\mathcal{F}\to \mathcal{G}$ is a series $\theta (T)\in TR[[T]]$ such that

Write $\theta (T)=uT+O(T^2)$. If $u={\theta }^{\prime }(0)$ is a unit, recursive comparison of coefficients constructs a unique compositional inverse $\psi (T)\in TR[[T]]$ with $\psi (\theta (T))=T=\theta (\psi (T))$. Apply $\psi$ to the homomorphism identity and substitute $X=\psi (U)$, $Y=\psi (V)$ to obtainThus $\psi$ is a homomorphism from $\mathcal{G}$ to $\mathcal{F}$, so $\theta$ is an isomorphism.

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A Minimal Weierstrass equation for $E/{\mathbb{Q}}_p$ is a Weierstrass equation of an elliptic curve with coefficients in ${\mathbb{Z}}_p$ whose discriminant has minimum $p$-adic valuation among all integral equations for $E$ related by admissible changes of variables.

Let $P\in E({\mathbb{Q}}_p)$ and choose projective coordinates $[X:Y:Z]$ with $X,Y,Z\in {\mathbb{Z}}_p$ and at least one coordinate a unit. Reducing the coordinates modulo $p$ givesMultiplying the primitive coordinates by a unit does not alter this point, so this defines the reduction map $E({\mathbb{Q}}_p)\to \bar{E}({\mathbb{F}}_p)$.

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For the filtration of elliptic-curve points over a local field, defineandFor $r\ge 1$, use the formal group of an elliptic curve with parameter $z=-x/y$ and putReduction induceswhile the coefficient of $p^r$ in the formal parameter gives

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For a locally compact group $G$ on which multiplication by $n$ has finite kernel and cokernel, compare a Haar measure with its pushforward under $[n]$. On the one-dimensional $p$-adic Lie group $E({\mathbb{Q}}_p)$, the derivative of $[n]$ at the identity is $n$, henceThis can also be read directly from the successive quotients in the filtration of elliptic-curve points over a local field; factors prime to $p$ act invertibly on a sufficiently small formal-group neighbourhood.

The real Lie group $E(\mathbb{R})$ has one or two circle components. On its identity component, $[n]$ has degree $n$; the component-group kernel and cokernel have the same order. ThereforeOnly primes dividing $n$ contribute to the finite-place product, and unique factorization gives

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For an endomorphism $\varphi$ of an elliptic curve, define its trace of an elliptic-curve endomorphism byPolarizing the quadratic form $\mathrm{deg}$ shows that this is the integer for whichwhere $\hat{\varphi }$ is the dual isogeny. ConsequentlyAlsoThe left side is $[\mathrm{tr}({\varphi }^2)]$, so

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Let $\pi$ be the Frobenius isogeny of an elliptic curve and putPart (a) givesLet $\alpha ,\beta$ be the two roots of $T^2-aT+p$. The points over ${\mathbb{F}}_{p^r}$ are exactly $\ker (1-{\pi }^r)$. Since the differential of $1-{\pi }^r$ is the identity, this isogeny is separable, and thereforeUsing $\mathrm{deg}\psi =(1-\psi )(1-\hat{\psi })$ in the endomorphism algebra gives the elliptic-curve point count over a finite fieldEquivalently, if $a_r={\alpha }^r+{\beta }^r$, thenand $\#E({\mathbb{F}}_{p^r})=p^r+1-a_r$.

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Suppose that $uP+v\alpha (P)=O$ with $u,v\in {\mathbb{F}}_{\ell }$. If $v=0$, then $u=0$ because $P$ has order $\ell$. If $v\ne 0$, thenApplying $\alpha$ again and using ${\alpha }^2=[-1]$ yields ${\lambda }^2=-1$ in ${\mathbb{F}}_{\ell }$. This is impossible when $\ell \equiv 3(\mathrm{mod}4)$, because $-1$ is then not a quadratic residue. Hence $P$ and $\alpha (P)$ are linearly independent in the two-dimensional vector space $E[\ell ]$.

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For $E:y^2=x^3-x$ over ${\mathbb{F}}_3$, each of $x=0,1,2$ gives the single affine point with $y=0$. Including $O$ gives $\#E({\mathbb{F}}_3)=4$, so the Frobenius trace is $a=3+1-4=0$. Therefore the Frobenius isogeny of an elliptic curve $\pi$ satisfiesOn the $11$-torsion, this reads ${\pi }^2=[-3]=[8]$. The element $8\in {\mathbb{F}}_{11}^{\times }$ has order ten and $8^5=-1$. Henceand no smaller positive power of $\pi$ is the identity on $E[11]$. A division field of an elliptic curve over a finite field has degree equal to the order of Frobenius on the torsion module, so

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For $P=[x:y]\in {\mathbb{P}}^1(\mathbb{Q})$ represented by coprime integers, its naive height on the projective line isWrite the degree-$d$ morphism as $\Phi =[F:G]$, where $F,G\in \mathbb{Z}[X,Y]$ are homogeneous of degree $d$ with no common projective zero. Bounding their coefficients gives

For the reverse inequality, the nonvanishing of the resultant of $F$ and $G$ gives homogeneous Bézout identities expressing fixed nonzero integer multiples of powers of $X$ and $Y$ as polynomial combinations of $F$ and $G$. Evaluating at $(x,y)$, removing the common divisor of $F(x,y)$ and $G(x,y)$, and taking the larger of $|x|,|y|$ givesThus $c_{1}H(P{)}^d\le H(\Phi (P))\le c_{2}H(P{)}^d$ for constants depending only on $\Phi$.

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Let $h=\log H$ and set $h_x(P)=h(x(P))$ for $P\ne O$, with $h_x(O)=0$. The given degree-four morphism and part (a) imply that a constant $C$ exists withfor every $P\in E(\mathbb{Q})$. DefineTo check the limit, put $a_r=4^{-r}{h}_x(2^{r}P)$. ThenThe geometric series converges, so $(a_r)$ is Cauchy and the limit exists. This is the canonical height of an elliptic curve; shifting the sequence by one index immediately gives $\hat{h}(2P)=4\hat{h}(P)$.

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Because the canonical height of an elliptic curve is a quadratic form, polarization makesa symmetric bilinear form on the free part of the Mordell-Weil group. If $P_i^{\prime }={\sum }_j{U}_{ij}{P}_j$ is another integral basis, then $U\in {\mathrm{GL}}_r(\mathbb{Z})$ and the Gram matrices satisfySince $\det U=\pm 1$, their determinants agree. Thus the regulator of an elliptic curve is independent of the chosen basis.

Now let $Q_1,\dots ,Q_r$ be a basis for the free part of $E^{\prime }(\mathbb{Q})$. The images $\varphi (P_i)$ span a finite-index sublattice, so modulo torsionfor an integral matrix $A$ with nonzero determinant. The height identity givesTaking determinants in the two descriptions of this Gram matrix yieldsTherefore the required formula holds with $d=(\det A{)}^{-1}\in {\mathbb{Q}}^{\times }$.

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The Kummer map of an elliptic curve gives an injectionBecause $E[n]\subseteq E(K)$, the Galois action on $E[n]$ is trivial, so a cocycle in the image is a continuous homomorphism $G_K\to E[n]$. For $P\in E(K)$ and $Q$ with $nQ=P$, its kernel fixes the Kummer extension $K(Q)/K$, which is Galois of degree at most $n^2$ and exponent dividing $n$.

The local theory of reduction of an elliptic curve shows that these extensions are unramified outside the finite set consisting of primes dividing $n$, primes of bad reduction, and archimedean places. Local fields have only finitely many extensions of any bounded degree. Together with the Hermite-Minkowski finiteness theorem, this implies that only finitely many global extensions of degree at most $n^2$ with these ramification restrictions occur. Each has only finitely many homomorphisms to the finite group $E[n]$. Hence the image of $\delta$, and therefore $E(K)/nE(K)$, is finite.

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Foruse two-isogeny descent throughThe square-class maps send a nonexceptional point to the class of its $x$-coordinate. On $E$, possible classes are $\pm 1,\pm 7$; the defining quartics and positivity exclude the negative classes, while $(1,3)$ and $(7,21)$ realize $1$ and $7$. Thus the image has order two. On $E^{\prime }$, the possible classes are $\pm 1,\pm 3$; the classes $1$ and $-3$ occur, while the quartics for $-1$ and $3$ have no primitive solution modulo $16$. Hence this image also has order two. The two-isogeny descent formulatherefore gives $r=0$.

The displayed curve has good reduction at $5$ and $11$, where direct counting givesReduction injects rational torsion of order prime to these characteristics, so its order divides $\gcd (6,18)=6$. We already have the six distinct pointsSince the rank is zero, these are all the rational points and $E(\mathbb{Q})\cong \mathbb{Z}/6\mathbb{Z}$.

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In characteristic zero, the isogeny of elliptic curves $\varphi :E\to E^{\prime }$ is finite and separable, soFor every $T\in E[\varphi ]$, translation ${\tau }_T(P)=P+T$ satisfies $\varphi {\tau }_T=\varphi$. It therefore induces a $K(E^{\prime })$-automorphism of $K(E)$. These translations are distinct, giving $\#E[\varphi ]$ automorphisms of an extension of the same degree. The extension is consequently Galois, andis an isomorphism.

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Let $Q\in E^{\prime }[\hat{\varphi }]$. The compatibility of divisor classes with pullback identifies the class ofwith $\hat{\varphi }(Q)=O$, so choose $g_Q\in K(E{)}^{\times }$ withSince $[n]Q=\varphi \hat{\varphi }(Q)=O$, choose $f_Q\in K(E^{\prime }{)}^{\times }$ withThe functions $f_Q\circ \varphi$ and $g_Q^n$ have the same divisor. Their quotient is constant, and because $K$ is algebraically closed we may rescale $g_Q$ so thatThus defineChanging either function changes $f_Q$ only by an $n$th power of a constant. The divisor relation for $Q+R$ differs from the sum of those for $Q$ and $R$ by $n$ times a principal divisor, so the map is a homomorphism. If $[f_Q]$ is trivial, then $f_Q=h^n$ for $h\in K(E^{\prime }{)}^{\times }$, whence $\mathrm{div}(h)=(Q)-(O)$; the divisor-class isomorphism forces $Q=O$. The map is therefore well-defined and injective.

For $P\in E[\varphi ]$, translation by $P$ fixes $g_Q^n=f_Q\circ \varphi$. Henceis independent of the auxiliary point $X$. This is the Weil pairing associated with $\varphi$.

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Let $D_Q=(Q)-(O)$ on $E^{\prime }$. Compatibility of the divisor-class maps with pullback gives a function $h\in K(E{)}^{\times }$ such thatIf $f_Q$ has divisor $m{D}_Q$, thenUse these functions in the divisor-evaluation formula for the Weil pairing. Pullback and pushforward satisfywhile the factor $h^m$ contributes an $m$th power and cancels from the pairing. The two evaluations are therefore identical, givingfor every $P\in E[m]$ and $Q\in E^{\prime }[m]$.

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