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    Mordell's theorem, and $E(\QQ)$ as a finitely generated abelian group

    T-CRVMORDELL

    For an elliptic curve over $\QQ$, the rational points form $\ZZ^r$ plus a finite torsion subgroup.Suggest a slogan

    (Defined over a subfield)

    Let \(k = \bar k\), let \((E, p_0)\) be elliptic, and embed \(E \embeds \PP^2_{/k}\) by \(\abs{3p_0}\), so \(E = V(f)\) for a cubic \(f\). Say \((E, p_0)\) is defined over a subfield \(k_0 \subseteq k\) when \(f\) can be taken with coefficients in \(k_0\) and \(p_0 \in E(k_0)\). Then \(E(k_0) \leq E(k)\) is a subgroup.

    (Mordell)

    If \((E,p_0)\) is defined over \(\QQ\), then \(E(\QQ)\) is a finitely generated abelian group: \[ E(\QQ) \iso \ZZ^r \oplus E(\QQ)_{\mathrm{tors}} , \] with \(r\) the rank and the torsion subgroup finite.

    Rank \(0\), torsion \(C_3\). The Fermat cubic \(x^3 + y^3 = z^3\) has \[ E(\QQ) = \ts{ \tv{1:-1:0},\ \tv{1:0:1},\ \tv{0:1:1} } \iso C_3 , \] these being its three rational inflection points. That the list stops there is Fermat’s last theorem for \(n=3\), so this example is a restatement of a genuine theorem and not a computation.

    Rank \(1\), no torsion. For \(y^2 + y = x^3 - x\) with \(p_0 = \tv{0:1:0}\), \[ E(\QQ) = \gens{P} \iso \ZZ, \qquad P = (0,0) \text{ in affine coordinates} . \] That \(P\) has infinite order is visible after a few doublings, and the denominators are the thing to watch: \[ 2P = (1, 0), \quad 3P = (-1,-1), \quad 4P = (2,-3), \quad 5P = \qty{ \tfrac{1}{4}, -\tfrac{5}{8} }, \quad 6P = (6, 14), \quad 7P = \qty{ -\tfrac{5}{9}, \tfrac{8}{27} } . \] The denominators appear as \(d^2\) and \(d^3\) in the two coordinates, which is exactly the shape a height function measures.

    Two points require explanation: why \(E(\QQ)\) is closed under the group law, and why that subgroup is finitely generated.

    The subgroup claim is not automatic and is the reason the definition insists that \(p_0\) be rational. Addition is defined by “three collinear points sum to zero”, and a line through two rational points of a rational cubic meets it in a third point whose coordinates are rational, because substituting the line into the cubic gives a cubic in one variable with rational coefficients and two rational roots, so the third root is rational too. Inversion is reflection, also rational. If \(p_0\notin E(\QQ)\), then the identity is not rational, so \(E(\QQ)\) is not a subgroup of \(E(\bar\QQ)\) with respect to that chosen group law.

    That subgroup is where the geometry turns into arithmetic. Writing a rational point in lowest terms and clearing denominators converts \(E(\QQ)\) into the integral solutions of a homogeneous cubic in three variables, so the chord-and-tangent construction becomes an operation that manufactures new integer solutions from old ones — the classical secant method, predating any of this language. Reading it the other way, a Diophantine question about a cubic acquires a group acting on its own solution set, and Mordell’s theorem says that group is small enough to describe: finitely many generators account for every solution there is. This exchange is the entire reason elliptic curves sit in number theory rather than only in geometry.

    Finite generation is proved by combining descent with a height argument. It is a descent: the weak Mordell theorem gives \(E(\QQ)/2E(\QQ)\) finite, and a height function measuring the arithmetic size of a point turns that finiteness into generation by a bounded set. Neither half is formal, and neither is visible from the geometry over \(\bar\QQ\), where \(E\) is divisible and nothing is finitely generated.

    What the theorem does not give is the rank. Torsion is completely understood — by Mazur’s theorem, \(E(\QQ)_{\mathrm{tors}}\) is one of fifteen groups — while no algorithm is known to compute the rank \(r\) in general, and no proven uniform bound for it is known.

    What does it mean for $(E, p_0)$ to be defined over a subfield?
    Why is the set of rational points a subgroup?
    State Mordell's theorem and say what its two summands are.
    Give a curve with rank $0$ and nontrivial torsion, and one with rank $1$ and no torsion.

    Dependencies

    • PR-CRVGRPThe group law on an elliptic curve, and multiplication by $n$uses

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