Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 26 / 1

Choose a Weierstrass equation of an elliptic curve, with origin $O$ at infinity. For $P,Q$, draw their chord, or the tangent if they coincide, and call its third intersection $R$, counted with intersection multiplicity. The line through $R$ and $O$ gives a third point $S$; define $P+Q=S$. This is the chord-and-tangent group law. The inverse of an affine point on $y^2+a_{1}xy+a_{3}y=x^3+a_2{x}^2+a_{4}x+a_6$ is $(x,-y-a_{1}x-a_3)$, obtained from its vertical line through $O$. These conventions give identity $O$, and cover tangent and flex cases by multiplicity.

Here is a proof of associativity, without assuming that the geometric construction already defines a group. Work first over an algebraic closure and use divisors on an algebraic curve. The Riemann-Roch theorem says $\ell (D)-\ell (K_E-D)=\mathrm{deg}D+1-g$. For a smooth genus-one curve the canonical divisor has degree zero, so every degree-one divisor has $\ell (D)=1$. Thus every divisor class of degree one has a unique effective representative, necessarily a single point. Existence follows from a nonzero section; uniqueness follows because two distinct effective representatives would give linearly independent sections in the same one-dimensional space. Consequentlyis a bijection: add $(O)$ to any degree-zero divisor class, and use its unique degree-one representative.

A line through $P,Q,R$ cuts out a divisor linearly equivalent to $3(O)$, since the line at infinity has triple intersection at the Weierstrass flex $O$. Equivalently, the ratio of the line equation to that of the line at infinity has principal divisor $(P)+(Q)+(R)-3(O)$. Thus the three corresponding degree-zero classes sum to zero. Applying the same fact to $O,R,S$ shows that the point produced by the chord-and-tangent process satisfiesAddition of divisors modulo principal divisors is associative. The two parenthesizations of $P+Q+T$ therefore have identical classes; injectivity of the displayed bijection makes them the same point. Hence the chord-and-tangent law is associative. This is associativity of the chord-and-tangent law via divisor classes. Its construction is defined over the original field, so the same conclusion holds for its rational points.

The nonzero elliptic-curve discriminant forces $b\ne 0$. At $P$, the derivative with respect to $y$ is $b$ and that with respect to $x$ is zero, so the tangent is $y=0$. Substitution gives $x^2(x+b)=0$: its third point is $(-b,0)$. Applying the negation formula givesThe line through $P$ and $2P$ has slope $1-a$, because their x-coordinates are distinct. Substitution of $y=(1-a)x$ givesIts third point is $(1-a,(1-a{)}^2)$, interpreted with multiplicity if it coincides with a previously used point. Negating it yieldsBy associativity, $5P=O$ exactly when $3P=-2P$. Equality of their x-coordinates forces $1-a=-b$, and when $a=b+1$ their y-coordinates also agree and are zero. ThereforeSince $P\ne O$, its order in this case is exactly five. These are the torsion multiples in a two-parameter Weierstrass family formulas.

Use $P=(0,0)$ in the family from part (b). For order four, take $a=b=1$, givingIts discriminant is $-15$, and $2P=(-1,0)$ is a nonzero point equal to its own inverse. Thus $4P=O$ and $2P\ne O$.

For order five, take $a=2,b=1$:The discriminant is $-11$, and $a=b+1$ gives exact order five by part (b).

For order six, take $a=0,b=-2$:Its discriminant is $80$. Here $2P=(2,2)$ and $3P=(1,1)$; negation is $(x,y)\mapsto (x,2-y)$, so $3P$ is nonzero 2-torsion. Neither $2P$ nor $3P$ is $O$, excluding orders two and three, while $6P=O$. All coefficients and points are rational, and the nonzero discriminants certify genuine elliptic curves.