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Arithmetic heights measure arithmetic complexity. For a reduced rational number $x=a/b$, its multiplicative height is $H(x)=\max (|a|,|b|)$ and its logarithmic height is $h(x)=\log H(x)$. Over a number field $K$, use the absolute projective heightNormalize the absolute values on a field by the product formula. That formula makes arithmetic height independent of the chosen homogeneous representative, and the normalization makes it independent of enlarging $K$. The Northcott theorem says that algebraic points of bounded degree and bounded arithmetic height form a finite set. For rational numbers this follows immediately by bounding the numerator and denominator. In a fixed number field, the same finiteness follows by bounding the coefficients of the minimal polynomial using its conjugates and the arithmetic height.

For an elliptic curve put $h_x(P)=h([x(P):1])$ and $h_x(O)=0$. The two points above a given $x$ show that bounded $h_x$ gives finitely many $K$-rational points. More importantly, multiplication expands this arithmetic height. To see the mechanism, choose a short Weierstrass equation of an elliptic curve $y^2=x^3+Ax+B$ over $K$. Doubling givesAfter homogenization the numerator and denominator are quartics with no common zero. At infinity the numerator is nonzero; at a root $t$ of the cubic the numerator becomes $(3t^2+A{)}^2$, nonzero by nonsingularity. For a degree-$d$ morphism of the projective line, fixed homogeneous coefficients give the upper arithmetic height bound $h(f(x))\le dh(x)+C$. A nonzero resultant supplies the reverse bound: at each place it bounds the common cancellation of the two homogeneous values, with a nontrivial constant needed at only finitely many places. Applying these bounds to duplication gives a constant $C$ independent of $P$ withThis includes points mapping to $O$, by the projective formulation.

The canonical height removes the bounded errors. With the conventional factor one half, defineThe successive terms before multiplying by one half differ by at most $C{4}^{-n-1}$. The limit exists, and summing this geometric bound givesHence bounded canonical height still gives finitely many points over the fixed number field.

To obtain the pairing, the quadratic nature comes from a uniform arithmetic height parallelogram estimate, not just from doubling. If $x=x(P)$ and $z=x(Q)$, the unordered coordinates $x(P+Q),x(P-Q)$ are the roots of the binary quadratic with coefficientsThese formulas follow by adding and multiplying the two chord expressions with slopes $(y(P)-y(Q))/(x-z)$ and $(y(P)+y(Q))/(x-z)$. Their homogenizations have bidegree $(2,2)$ and no simultaneous zero. Off the diagonal $C_2\ne 0$; on the finite diagonal $x=z=t$, $C_1=4f(t)$ and $C_0=f^{\prime }(t{)}^2-8tf(t)$ for $f(t)=t^3+At+B$, so simultaneous vanishing would contradict nonsingularity. At the double point at infinity the homogeneous $C_0$ is nonzero. Local coefficient norms, or the corresponding resultant bounds, therefore giveThe arithmetic height of a binary quadratic's coefficient vector differs by $O(1)$ from the sum of the heights of its two roots. At finite places this follows from multiplicativity of the maximum coefficient norm of products; at infinite places the fixed-degree norms are comparable. ThusApply this to $2^{n}P,2^{n}Q$, divide by $2\cdot 4^n$ and take limits. The errors vanish and give the exact height parallelogram identity. Polarization consequently makesa symmetric bilinear pairing. Nonnegativity of $\hat{h}(aP+bQ)$ for integers $a,b$, and density of rational ratios, give the Cauchy-Schwarz inequality. In particular $\hat{h}(P-Q)\le 2\hat{h}(P)+2\hat{h}(Q)$. Also $\hat{h}(mP)=m^2\hat{h}(P)$ for integers $m$. Torsion points of an elliptic curve have canonical height zero because their doubling orbit is finite; conversely, if a point has canonical height zero, all its integer multiples have bounded naive height, and finiteness forces two multiples to coincide. Thus canonical height vanishes exactly on torsion.

There is an arithmetic finiteness input before height descent: $E(K)/2E(K)$ is finite, the Weak Mordell-Weil theorem. Here is its mechanism. The multiplication exact sequence gives an injective Kummer map $E(K)/2E(K)\to H^1(K,E[2])$: choose a half-point $Q$ and use the cocycle $\sigma \mapsto \sigma Q-Q$; changing $Q$ changes a coboundary, and a trivial class means a half-point can be chosen over $K$. Let $L=K(E[2])$, a Finite Galois extension. Over $L$ the torsion module is $(\mathbb{Z}/2\mathbb{Z}{)}^2$, so Kummer theory identifies its first Galois cohomology with two copies of $L^{\times }/L^{\times 2}$.

Only classes unramified outside a fixed finite set $S$ occur. Include the primes over $2$, bad-reduction primes and the ramified primes of $L/K$. At a good prime away from $2$, multiplication by two on the proper smooth elliptic model is finite étale morphism, so the torsor of halves of a point is unramified. In square-class language, valuations outside $S$ are even. Such classes form a finite group: write the principal ideal of a representative as a square times an ideal supported on $S$. The ideal class gives an element of the finite two-torsion of the localized ideal class group; after fixing that class, ambiguity is an $S$-unit modulo squares. The $S$-unit theorem makes the latter finite. The group kernel of restriction from $K$ to $L$ is contained in $H^1(\mathrm{Gal}(L/K),E[2])$, which is finite because both the group and module are finite. This proves the weak theorem, without assuming finite generation of $E(K)$.

Finally choose representatives $R_1,\dots ,R_s$ of $E(K)/2E(K)$ and put $M={\max }_i\hat{h}(R_i)$. Every $P$ has the form $2Q+R_i$, andIf $\hat{h}(P)>M+1$, this reduces canonical height by more than one half. Iterate until a point of canonical height at most $M+1$ is reached. There are only finitely many such points, by Northcott and the bounded difference of heights. Reconstructing $P=2Q+R_i$ at every stage proves that these finitely many small points and the finitely many representatives generate $E(K)$. ThereforeThe canonical pairing is positive definite on the free part. It turns rank computations into an arithmetic lattice problem, supplies independence tests for proposed generators, and bounds searches once a arithmetic height bound is available. Weak finite-quotient finiteness alone would not prove this result: the arithmetic height contraction is the step that converts it into finite generation.