Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 125 (Elliptic Curves)

For points $P,Q$ on a smooth plane cubic, draw the line through them, using the tangent at $P$ when $P=Q$, and let its third intersection with the cubic be $R$. The chord-and-tangent group law defines $P+Q$ as the reflection of $R$ in the $x$-axis; the point at infinity $O$ is the identity and $-(x,y)=(x,-y)$.

If $P,Q\in E(\mathbb{Q})$, their chord or tangent has rational coefficients. Substitution into the cubic gives a polynomial with rational coefficients for which two intersection roots are rational, so the third is rational as well. The identity and inverse of every rational point are rational, and closure under addition follows. Assuming the elliptic-curve law is a group law, $E(\mathbb{Q})$ is therefore a subgroup of $E(\overline{\mathbb{Q}})$.

For $E:y^2=x^3-9x+9$, the addition formulas usewith $m=(3x_1^2-9)/(2y_1)$ for doubling. They giveand

Let $(x,y)\ne O$ be rational. If the 3-adic valuation satisfies $v_3(y)\ge 2$, first $v_3(x)$ cannot be negative: otherwise $x^3$ is the unique term of least valuation in $x^3-9x+9$, giving $2v_3(y)=3v_3(x)<0$. If $v_3(x)=0$, the right side has valuation zero. If $v_3(x)\ge 1$, its three terms have valuations at least $3$, at least $3$, and exactly $2$, so $2v_3(y)=2$. In every case $v_3(y)\le 1$.

The Lutz–Nagell theorem says that ifhas nonzero discriminant and $P=(x,y)\in E(\mathbb{Q})$ is a torsion point, then $x,y\in \mathbb{Z}$ and either $y=0$ or

For integrality, fix a prime $p$. If a rational point has nonintegral coordinates, its primitive projective coordinates reduce to $O$, so it belongs to the kernel of reduction. The parameter $t=-x/y$ identifies this kernel with the formal group of an elliptic curve over $p{\mathbb{Z}}_p$. The formal logarithm, with the standard separate first-step argument at $p=2$, shows that this group has no nonzero rational torsion. A rational torsion point therefore has nonnegative $p$-adic valuations in both coordinates for every prime $p$, hence integral coordinates.

Suppose now that $y\ne 0$. The point $2P$ is again a nonzero torsion point and hence integral. Its $x$-coordinate is $m^2-2x$, whereThus $m^2$ is an integer. A rational number whose square is integral is integral, so $2y\mid 3x^2+a$ and in particular $y^2\mid (3x^2+a{)}^2$. The curve equation and the identitythen prove $y^2\mid 4a^3+27b^2$. This is the divisibility proof in the Nagell–Lutz theorem.

HereBy the Lutz–Nagell theorem, a nonzero rational torsion point has integral coordinates and either $y=0$ or $y^2\mid 3^6$. Part b gives $v_3(y)\le 1$, so $y\in \{0,\pm 1,\pm 3\}$. The cubic $x^3-9x+9$ has no integral zero. Substitution of $y=\pm 1,\pm 3$ gives exactlyThis proves the required inclusion.

Since $2Q=-Q$, the point $Q$ has order three. The point $P$ is not torsion because $2P=(9/4,3/8)$ is nonintegral, contradicting Nagell–Lutz. If $P+Q$ or $P-Q$ were torsion, adding the torsion point $\mp Q$ would make $P$ torsion; their negatives are excluded in the same way. Hence

The degree of an isogeny $\varphi :E\to E^{\prime }$ is the degree of the induced finite extension of function fields. Saying that degree is a quadratic form on $\mathrm{End}(E)$ meansand thatis bilinear, equivalentlyThe trace of an elliptic-curve endomorphism isThe relation ${\varphi }^2-[\mathrm{tr}\varphi ]\varphi +[\mathrm{deg}\varphi ]=0$ impliesthe trace of the square of an elliptic-curve endomorphism.

The Hasse theorem for elliptic curves states that for $E/{\mathbb{F}}_q$,Let $\pi$ be the Frobenius isogeny of an elliptic curve, put $a=\mathrm{tr}(\pi )$, and note that $\mathrm{deg}\pi =q$ andFor integers $m,n$, quadraticity givesThis binary quadratic form cannot have positive discriminant, since rational numbers $m/n$ are dense, so $a^2-4q\le 0$. Substitution proves the bound. This is the degree-form proof of the Hasse bound.

Both endpoints occur. The curve $E:y^2=x^3-x$ over ${\mathbb{F}}_3$ is supersingular with trace zero. Over ${\mathbb{F}}_9$, its Frobenius is $[-3]$, soIts nontrivial quadratic twist over ${\mathbb{F}}_9$ has the opposite trace and therefore haspoints.

For a short Weierstrass equation $E:y^2=x^3+ax+b$ in characteristic different from two, the invariant differential on an elliptic curveis nonzero and regular, including at $O$.

Write $X=x\circ \varphi$. Since $X(-P)=X(P)$, it is an even rational function and therefore belongs to $k(x)$; write $X=p(x)/q(x)$ with coprime polynomials $p,q$. The pullback of a nonzero invariant differential is invariant, so for some $c\in k^{\times }$,Sincewe obtainThus

Take $(x,y)\in E_r(K)\setminus \{O\}$. Because $v(x)<0$ and $p,q$ have degrees $d,d-1$, their leading terms dominate, givingThe polynomial $p^{\prime }q-p{q}^{\prime }$ has degree $2d-2$ with nonzero leading coefficient, the same degree as $q^2$. Their quotient consequently has valuation zero at $x$, soThe isogeny sends $O$ to $O$, hence $\varphi (E_r(K))\subseteq E_r^{\prime }(K)$.

Use the alternative affine coordinatesso that $x=t/w$ and $y=-1/w$. The equation becomesRecursive coefficient comparison gives a unique series $w(t)\in t^{3}k[[t]]$. Since $v(w(t))=3v(t)$ and hence $v(x)=-2v(t)$, $v(y)=-3v(t)$, this gives the formal coordinates on a short Weierstrass curve identification

A one-dimensional commutative formal group law over a ring $R$ is a power series $F(X,Y)\in R[[X,Y]]$ satisfying $F(X,0)=X$, commutativity, and associativity. A morphism $f:F\to G$ is a series $f(T)\in TR[[T]]$ satisfying

Express the isogeny of part b in the formal coordinates of part d and setPart c shows that points approaching $O$ map to points approaching $O^{\prime }$, so $f(t)\in tk[[t]]$. Since $\varphi$ is a group homomorphism, applying the parameter $t^{\prime }$ to $\varphi (P+Q)=\varphi (P)+\varphi (Q)$ givesThus $f$ is the formal-group morphism induced by an isogeny $\hat{E}\to \widehat{E^{\prime }}$.

Let $G=\mathrm{Gal}(L/K)$. If $a\in K^{\times }$ becomes an $n$th power in $L$, choose $\alpha \in L$ with ${\alpha }^n=a$. Thenis a cocycle with values in the finite group ${\mu }_n(L)$. Changing $\alpha$ changes it by a coboundary, and its class is trivial exactly when $a$ was already an $n$th power in $K$. Thus the kernel injects into the finite group $H^1(G,{\mu }_n(L))$ and is finite. This is the multiplicative case of the finite-extension kernel of a Kummer map.

Assume ${\mu }_n\subset K$ and put $\alpha =\sqrt[n]{a}$. All roots of $X^n-a$ are $\zeta \alpha$ with $\zeta \in {\mu }_n\subset K$, so $K(\alpha )$ is its splitting field and is a Finite Galois extension. The mapis an injective group homomorphism.

For a finite Galois extension $L/K$, the kernel ofis finite. Indeed, if $P=nQ$ for $Q\in E(L)$, then $\sigma \mapsto \sigma Q-Q$ is a cocycle in the finite $G$-module $E[n](L)$, and the resulting map from the kernel to $H^1(G,E[n](L))$ is injective.

The analogue of part b assumes $E[n]\subset E(K)$. Given $P\in E(K)$ and $Q\in E(\overline{K})$ with $nQ=P$, every conjugate of $Q$ is $Q+T$ for some $T\in E[n]$. Hence $K(Q)/K$ is Galois andis an injective homomorphism. These are the two elliptic forms of the Kummer pairing.

Pass to the finite extension $L=K(E[n])$. Part c shows that the kernel of $E(K)/nE(K)\to E(L)/nE(L)$ is finite, so it is enough to control the image after $n$-torsion becomes rational. The Kummer map of an elliptic curveassociates to $P$ the finite extension generated by one $n$-division point of $P$.

Let $S$ contain the places over $n$, all archimedean places, and all places of bad reduction. The local theory of elliptic curves shows that these Kummer classes are unramified outside $S$. Because $E[n]$ is finite and constant over $L$, such classes are controlled by finitely many $S$-unramified power classes. Finiteness of the class group and finite generation of the unit group make that power-class group finite. The Kummer image is therefore finite, proving the Weak Mordell-Weil theorem that $E(K)/nE(K)$ is finite; this is the Kummer-theoretic proof of the weak Mordell-Weil theorem.

Let a rational right triangle have legs $A,B$, hypotenuse $C$, and area $AB/2=D$. After interchanging the legs if necessary, setUsing $C^2=A^2+B^2$ and $AB=2D$ givesso $P=(x,y)$ lies on the congruent number elliptic curve $E_D:D{y}^2=x^3-x$. The triangle is nondegenerate, so $y\ne 0$ and $2P\ne O$.

Conversely the three displayed lengths in the question satisfyand their area isThus every such triangle is ${\Delta }_P$.

The change of variables $X=5x$, $Y=25y$ identifies $E_5$ withA full two-isogeny descent, equivalently the standard full 2-descent for a cubic with three rational roots, givesThe local conditions at $2$, $5$, and infinity leave exactly these eight square-class combinations; primes outside $\{2,5\}$ have even valuations and contribute none. Hence the quotient has dimension three. Since the rational 2-torsion has dimension two, the rank is one.

The Lutz–Nagell theorem on the integral model, or reduction at two good primes, excludes odd torsion and torsion of order greater than two. Thereforeand $(-4/5,6/25)$ generates the free part. This is the Mordell-Weil group of the congruent number curve for five.

For $x(P)=A/B^2$ in lowest terms, take the logarithmic naive height $h(P)=\log \max \{|A|,B^2\}$. Its required properties areandwith constants depending only on the curve. Define the canonical height of an elliptic curve byThe first bounded-error relation makes this a convergent telescoping correction to $h(P)$. Apply the second relation to $2^{r}P,2^{r}Q$, divide by $4^r$, and let $r\to \infty$ to obtainAlso $\hat{h}(2P)=4\hat{h}(P)$, and the parallelogram identity then gives $\hat{h}(nP)=n^2\hat{h}(P)$ for every integer $n$. Thus $\hat{h}$ is a quadratic form.

By part b, writeThe canonical height of an elliptic curve vanishes on torsion and satisfies $\hat{h}(mG+T)=m^2\hat{h}(G)$. If $\hat{h}(P)=\hat{h}(Q)\ne 0$, their nonzero integer coefficients therefore have equal squares, sofor some $T\in E_5[2]$.

Negation leaves the $x$-coordinate unchanged. Addition by the four 2-torsion points changes $x$ amongSubstitution in the three side formulas for ${\Delta }_P$, using $5y^2=x^3-x$, only changes signs and permutes the three values. Hence ${\Delta }_Q$ and ${\Delta }_P$ are the same right triangle up to reordering their sides.