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Kummer theory studies extracting roots through Galois actions, and on an elliptic curve the corresponding operation is division of points. The bridge is a cocycle: different Galois conjugates of a chosen division point differ by torsion. Controlling the ramification of these cocycles is what proves the Weak Mordell-Weil theorem.

Classically, for a characteristic-zero field $K$, the sequenceis exact. Hilbert's theorem 90 gives $H^1(K,{\overline{K}}^{\times })=0$, so Galois cohomology identifiesThe class of $a$ is represented by $\sigma \mapsto \sigma (\sqrt[m]{a})/\sqrt[m]{a}$. If ${\mu }_m\subset K$, the module is constant and these are continuous characters into ${\mu }_m$; the associated fields are cyclic Kummer extensions of degree dividing $m$. Several such classes give abelian extensions of exponent dividing $m$. Without the roots-of-unity hypothesis the cohomological description still holds, but the action on ${\mu }_m$ must not be discarded.

For an elliptic curve over a number field $K$, multiplication by $m\ge 2$ gives the exact sequence of Galois modulesThe map is onto because it is a nonconstant isogeny over an algebraically closed field. For $P\in E(K)$ choose $Q\in E(\overline{K})$ with $mQ=P$, and setIt satisfies $\delta (P{)}_{\sigma \tau }=\delta (P{)}_{\sigma }+\sigma \delta (P{)}_{\tau }$, the Galois 1-cocycle identity. Changing $Q$ by a torsion point of an elliptic curve changes the cocycle by a coboundary. If the class is zero, a corresponding change of $Q$ makes it Galois-fixed, so $P\in mE(K)$. Conversely a rational division point gives the zero class. Thus the Kummer map of an elliptic curve is injective on the quotient, and the cohomology sequence givesThis is the Kummer exact sequence of an elliptic curve. The last group parametrizes the appropriate torsor classes. It is important that the entire middle group need not be finite; an unrestricted Kummer class can ramify at arbitrarily many primes.

The Weak Mordell-Weil theorem states $E(K)/mE(K)$ is finite for every number field $K$ and every $m\ge 1$. The case $m=1$ is trivial. To prove the remaining cases, choose a finite set $S$ containing the primes of bad reduction and the primes dividing $m$. At a prime outside $S$, the elliptic curve extends to a smooth proper group scheme over the valuation ring, and $[m]$ is finite étale. Properness extends a local rational point to a section; its division fibre is a finite étale torsor over that ring. Therefore its Kummer class is unramified. This is unramified division torsors at good primes, and shows that ${\delta }_m(E(K))$ lies in first Galois cohomology unramified outside a finite set.

Let $L=K(E[m])$, enlarged if necessary to contain ${\mu }_m$, and include in $S$ the finitely many primes ramified in $L/K$. This is a finite Galois extension. Over $L$ the torsion module is constant, isomorphic to $(\mathbb{Z}/m\mathbb{Z}{)}^2$, and so to two copies of ${\mu }_m$. Restricting a cocycle to $G_L$ therefore reduces it to two classical Kummer classes unramified outside the primes above $S$. Such a class is represented by $a\in L^{\times }$ satisfyingWrite this power-class group as $L(S_L,m)$. The implication follows locally because, away from residue characteristic dividing $m$, the valuation of an unramified $m$th-root extension must be divisible by $m$.

Its finiteness comes from two standard arithmetic finiteness results, rather than from an assertion that all of $L^{\times }/(L^{\times }{)}^m$ is finite. The valuations outside $S_L$ associate to $a$ an ideal whose $m$th power is principal. There is an exact sequenceThe left group is finite by the finite generation of the S-unit group, and the right group is finite by finiteness of the ideal class group. This proves finiteness of S-unramified Kummer classes. Hence there are only finitely many restricted cocycles over $L$. The kernel of restriction from $K$ is contained in $H^1(\mathrm{Gal}(L/K),E[m])$, which is finite because both group and module are finite. The restricted cohomology group over $K$ is therefore finite, and its injected subgroup $E(K)/mE(K)$ is finite too. This completes a Kummer-theoretic proof of the weak Mordell-Weil theorem.

For arithmetic computation one adds local conditions. The Selmer group of an elliptic curve consists of classes in $H^1(K,E[m])$ whose restriction at every completion $K_v$ lies in the image of the corresponding local Kummer map. They are unramified outside the same finite set, so this group is finite. The failure of a everywhere locally soluble torsor to have a global point is measured by the Tate–Shafarevich group, defined as the kernel of the local restriction map on $H^1(K,E)$. The resulting exact sequence isThus the Selmer group of an elliptic curve gives a computable upper bound for the quotient; actual rational points supply lower bounds. Equality need not follow just from local solubility. Finiteness of this fixed torsion subgroup does not prove finiteness of the whole Tate–Shafarevich group.

Finally, weak finiteness alone is not finite generation: the additive group $\mathbb{Q}$ has trivial quotients by multiplication and is not finitely generated. The additional ingredient for the Mordell-Weil theorem is height descent. Choose representatives $P_i$ of $E(K)/mE(K)$ and write $P=P_i+mQ$. The quadratic growth of the canonical height of an elliptic curve, together with bounds for translation by the finitely many $P_i$, makes $Q$ have smaller height whenever $P$ is sufficiently large. Northcott theorem gives only finitely many points of bounded height, and repeated descent yields a finite generating set. This explains the roles of Kummer theory, local information and heights without conflating the weak and full theorems.

A height function measures arithmetic size, which is what makes an otherwise infinite descent terminate. For a number field $K$, normalize absolute values to extend the standard real and $p$-adic ones, and write $n_v=[K_v:{\mathbb{Q}}_v]$. The logarithmic projective height isThe product formula makes it independent of the chosen homogeneous coordinates, and the local-degree normalization makes it independent of the number field containing them. Over $\mathbb{Q}$, $h([a:b])=\log \max (|a|,|b|)$ for coprime integer coordinates. Set $h_x(P)=h(x(P))$ on an elliptic curve, with $h_x(O)=0$.

Two features are essential. First, the Northcott theorem says that points of bounded projective height and bounded field degree form a finite set. For points on a fixed elliptic curve over $K$, each $x$-coordinate has at most two preimages, so bounded $h_x$ gives finitely many points. Second, a degree-$d$ morphism of the projective line satisfies $h(f(z))=dh(z)+O(1)$, uniformly in $z$. The upper bound comes from evaluating its homogeneous polynomials; for the lower bound, their lack of a common zero gives a resultant identity bounding the input coordinates by the output coordinates at each place. Summing the local bounds gives the asserted uniform constant.

The duplication map on the $x$-line has degree four. Nonsingularity ensures that its numerator and denominator have no common projective zero. ThereforeTelescoping defines the canonical height of an elliptic curveThe error in successive terms is bounded by a geometric series, proving convergence and the uniform bounded difference. It also gives $\hat{h}(2P)=4\hat{h}(P)$ and nonnegativity. The usual addition formula gives the approximate height parallelogram identity for $h_x$; equivalently, the unordered pair of sum and difference on the $x$-line has bidegree $(2,2)$. Applying that identity to $2^{j}P,2^{j}Q$ and passing to the limit givesIn particular $\hat{h}(nP)=n^2\hat{h}(P)$. Its polarization is the canonical height pairing, a positive semidefinite bilinear form even before finite generation has been proved. The Cauchy-Schwarz inequality for this pairing givesAlso $\hat{h}(P)=0$ precisely for torsion points of an elliptic curve: one direction follows from periodic multiples, and in the other direction all multiples have bounded $h_x$, so the Northcott theorem makes two multiples equal. Bounded canonical height of an elliptic curve likewise gives a finite set of $K$-rational points.

Now the Weak Mordell-Weil theorem gives finitely many representatives $R_1,\dots ,R_s$ for $E(K)/2E(K)$. Put $H={\max }_i\hat{h}(R_i)$. Write any point as $P=2Q+R_i$. ThenRepeatedly applying this height descent lemma eventually reaches height at most $H+1$: after $j$ steps the height is at most $H+2^{-j}(\hat{h}(P)-H)$. The set of points with height at most $H+1$ is finite. Reading the relations $P=2Q+R_i$ backwards shows that this finite set together with the $R_i$ generates $E(K)$. ConsequentlyHeights turn weak Mordell-Weil finiteness into the Mordell-Weil theorem. The canonical height pairing subsequently equips the free part with a positive definite quadratic form, useful for bounding searches and measuring independent generators; this interpretation is a consequence of the proof, not an assumption used in the descent.

The Hasse-Minkowski theorem says that a nondegenerate quadratic form over a number field has a nonzero isotropic vector over that field if and only if it has one over every completion. In particular, over $\mathbb{Q}$ one tests $\mathbb{R}$ and every ${\mathbb{Q}}_{\ell }$. For the present integral form, a rational isotropic vector can be multiplied by a common denominator to give an integer one. This is the Hasse-Minkowski principle.

Write $P=p$, $R=p+12$ and $S=p+24$. All three primes are odd and exceed three: $p=2$ or $3$ would make $p+12$ composite. The second supplementary law for quadratic reciprocity gives

Necessity at $P$. Scale any nonzero vector over ${\mathbb{Q}}_P$ so that its coordinates lie in ${\mathbb{Z}}_P$ and at least one is a unit. Reduction of its equation modulo $P$ gives $y^2=2z^2$, since twelve is invertible modulo $P$. If the Legendre symbol $(2/P)$ is $-1$, this forces $P\mid y,z$. The original equation then forces $P\mid x$, because its other two terms are divisible by $P^2$. This contradicts the normalization. Thus

Sufficiency at the coefficient primes. Suppose $(2/P)=1$. A nonzero solution modulo $P$ is $(x,y,z)=(0,\sqrt{2},1)$. Its $y$-derivative is nonzero modulo $P$, so fixing the other two coordinates and applying the Hensel lemma lifts it to ${\mathbb{Q}}_P$.

Since $S\equiv P(\mathrm{mod}8)$, the same Legendre symbol criterion gives $(2/S)=1$. Modulo $S$, the equation reduces to $y^2=2x^2$, and the vector $(1,\sqrt{2},0)$ is a simple zero in the $y$ coordinate. It lifts to ${\mathbb{Q}}_S$. Modulo $R$, the coefficients of $x^2$ and $z^2$ are respectively $-12$ and $12$, so $(1,0,1)$ is a zero with nonzero $x$-derivative. It lifts to ${\mathbb{Q}}_R$.

The remaining odd primes. If $\ell$ is odd and divides none of $PRS$, the reduced ternary quadratic form is nondegenerate. An elementary finite-field argument finds a zero with $z=1$: if its coefficients are $A,B,C$, the setsboth have $(\ell +1)/2$ elements, so intersect. The resulting vector has nonzero $z$-derivative $2C$, and the Hensel lemma lifts it. This isotropy of nondegenerate ternary quadratic forms over finite fields verifies every remaining odd completion without an extra congruence condition.

The dyadic and real places. Setting $y=1,z=2$ reduces the dyadic problem toThe right side is a 2-adic unit congruent to $-3-4=1$ modulo eight. The unit square classes of the p-adic integers computed above show that it is a square in ${\mathbb{Q}}_2$. This also explains the printed hint: its integer can be taken as $m=-(p+21)/2$. Over $\mathbb{R}$, the positive and negative coefficients make the form indefinite; for instance $x=0,z=1,y=\sqrt{S/R}$ gives a nonzero zero.

All completions now have a nonzero isotropic vector. The Hasse-Minkowski theorem, followed by clearing denominators, proves

A number-field completion is finite. Let $F$ be the given number field and $L=F_{\mathfrak{p}}$ its completion. Normalize its absolute value on a field so that the restriction to $\mathbb{Q}$ is $|\cdot {|}_p$: if $v_{\mathfrak{p}}(p)=e$, use $|x|=p^{-v_{\mathfrak{p}}(x)/e}$. Changing the original absolute value by this positive power does not change its completion. The closure of $\mathbb{Q}$ in $L$ is an isometric copy of ${\mathbb{Q}}_p$.

Take a $\mathbb{Q}$-basis of $F$ and let $V$ be its ${\mathbb{Q}}_p$-span in $L$. Choose a maximal linearly independent subset of that finite spanning set. The compact-sphere proof of finite-dimensional non-Archimedean norm equivalence applies to this basis too, so $V$ is complete and therefore closed in $L$. Since $F\subset V$ is dense in $L$, it follows that $V=L$. Hence

Every finite extension is obtained this way. Let $E/{\mathbb{Q}}_p$ be finite. Use its extending Non-Archimedean absolute value, whose completeness and uniqueness were proved above. In characteristic zero the primitive element theorem gives $E={\mathbb{Q}}_p(\alpha )$; scaling $\alpha$ by a power of $p$, we may arrange that it is integral. Its monic minimal polynomial $f$ belongs to ${\mathbb{Z}}_p[T]$. If the degree is one, $E={\mathbb{Q}}_p$ is already the completion of $\mathbb{Q}$.

Otherwise choosewhere ${\alpha }^{\prime }$ runs over the conjugates in an algebraic closure. Approximate the coefficients of $f$ by integers, obtaining a monic $g\in \mathbb{Z}[T]$ of the same degree. Since $f^{\prime }(\alpha )\ne 0$, sufficiently accurate approximation gives $|g(\alpha )|<|g^{\prime }(\alpha ){|}^2$ and makes $|g(\alpha )/g^{\prime }(\alpha )|<\delta$. The strong form of Hensel lemma in $E$ produces a root $\beta \in E$ of $g$ with $|\beta -\alpha |<\delta$.

For completeness, the required Krasner's lemma follows from the unique extension of an absolute value to a finite extension. In a finite Galois extension containing $\alpha ,\beta$, every automorphism $\sigma$ fixing $\beta$ is an isometry, andIf $\sigma \alpha \ne \alpha$, this contradicts the choice of $\delta$. Thus every automorphism fixing $\beta$ fixes $\alpha$, and Galois correspondence gives ${\mathbb{Q}}_p(\alpha )\subseteq {\mathbb{Q}}_p(\beta )$. Since $\beta \in E$, equality follows. In particular $g$ is irreducible over ${\mathbb{Q}}_p$, and therefore over $\mathbb{Q}$.

Put $F=\mathbb{Q}(\beta )\subset E$. It is a number field, and the closure of $F$ in $E$ contains ${\mathbb{Q}}_p(\beta )=E$, so $F$ is dense. LetEvery element of the ring of integers of a number field has absolute value at most one: otherwise its highest-degree term would strictly dominate all other terms of a monic integer equation. Thus reduction into the residue field of $E$ makes $\mathfrak{p}$ a prime ideal, and $\mathfrak{p}\cap \mathbb{Z}=p\mathbb{Z}$. The restriction of the valuation to $F$ is the valuation of this prime, up to normalization. Indeed, the localized discrete valuation ring $({\mathcal{O}}_F{)}_{\mathfrak{p}}$ maps into ${\mathcal{O}}_E$, its units have absolute value one, and its uniformizer has absolute value less than one. Density and completeness giveThis proves the local realization as a completion of a number field.

For a finite extension $E/K$ of number fields, the inverse different is the fractional idealThe different ideal is its inverse. The trace pairing is nondegenerate because extensions of number fields are separable, so this definition gives a full fractional ideal. The inclusion ${\mathcal{O}}_E\subseteq {\mathfrak{D}}_{E/K}^{-1}$ shows that ${\mathfrak{D}}_{E/K}$ is integral.

Write its prime ideal factorization as ${\mathfrak{D}}_{E/K}={\prod }_{\mathfrak{P}}{\mathfrak{P}}^{d_{\mathfrak{P}}}$. For $\mathfrak{P}$ above $\mathfrak{p}$, with ramification index $e_{\mathfrak{P}}$ and residue characteristic $\ell$, the different exponent and tame ramification theorem saysThe equivalence uses the separability of finite residue-field extensions. In particular, $d_{\mathfrak{P}}=0$ precisely at unramified prime ideals; in the wild case $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}$. This theorem does not require a Galois extension. The relative discriminant is the norm of the different ideal:For $K=\mathbb{Q}$, this gives $|D_E|=N({\mathfrak{D}}_{E/\mathbb{Q}})$.

Now take ${\alpha }^3=m$ and $K=\mathbb{Q}(\alpha )$. For any prime number $q\mid m$, the square-free integer hypothesis makes $X^3-m$ an Eisenstein polynomial at $q$. It is therefore irreducible, and $K$ is a pure cubic number field of degree $3$. Since $\alpha$ is an algebraic integer, $A=\mathbb{Z}[\alpha ]$ is an order in a number field. The discriminant of elements of a number field for its basis $1,\alpha ,{\alpha }^2$ isFor example, the resultant of $X^3-m$ and $3X^2$ is $27m^2$, and the degree-three sign in the polynomial discriminant is negative. If $I=[{\mathcal{O}}_K:A]$, the discriminant-index formula for an integral lattice givesOnly prime numbers dividing $3m$ can therefore divide $I$.

For $q\mid m$, the Eisenstein polynomial gives a totally ramified extension of ${\mathbb{Q}}_q$ of degree $3$. Thus $K$ has a unique prime ideal $\mathfrak{P}$ above $q$, with ramification index $3$ and residue-field degree $1$. Since $q\ne 3$, this is tame ramification, and the different exponent and tame ramification theorem gives $d_{\mathfrak{P}}=2$. Hence $v_q(D_K)=2$. Comparing with $v_q(-27m^2)=2$ in the discriminant-index formula for an integral lattice yields $v_q(I)=0$.

At $3$, use the shifted Eisenstein polynomial of $\beta =\alpha -m$:Because $3\nmid m$, the constant term is divisible by $3$. Moreover,when $3\nmid m$: the factors $m-1$ and $m+1$ cannot both be divisible by $3$. The hypotheses therefore give $v_3(m^3-m)=1$, so the translated polynomial is an Eisenstein polynomial at $3$. The resulting completion is a totally ramified extension of degree $3$, with residue-field degree $1$, and its ramification index is divisible by the residue characteristic. Thus its different exponent is at least $3$.

It follows that $v_3(D_K)\ge 3$. But $v_3(-27m^2)=3$, soforces $v_3(I)=0$ and $v_3(D_K)=3$. No prime number divides $I$. We have proved the integral basis of a nonexceptional pure cubic field:Using local Eisenstein polynomials here only establishes the local ramification indices; it does not assume that $\mathbb{Z}[\alpha ]$ is already the full ring of integers of a number field.

For a number field $K$, let ${\mathcal{O}}_K$ be its ring of integers of a number field, $r_1$ the number of real embeddings and $r_2$ the number of conjugate pairs of nonreal complex embeddings, so $[K:\mathbb{Q}]=r_1+2r_2$. The Dirichlet unit theorem sayswhere $\mu (K)$ is the finite group of roots of unity in $K$. Thus an imaginary quadratic number field has only roots of unity as units: usually $\{\pm 1\}$, with four for $d=-1$ and six for $d=-3$. A real quadratic number field has units $\pm {ϵ}^n$ for a fundamental unit $ϵ>1$ and integers $n$.

For $K=\mathbb{Q}(\sqrt{26})$, ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{26}]$ and the field norm is $x^2-26y^2$. The element $ϵ=5+\sqrt{26}$ has norm $-1$ and is a unit. It is the smallest positive unit greater than one. Indeed for $1<u<ϵ$, its conjugate is $\pm 1/u$, so the coefficient $y=(u\mp 1/u)/(2\sqrt{26})$ is a positive integer less than two. Therefore $y=1$, and $x^2=26\pm 1$ forces $x=5$, giving $u=ϵ$, a contradiction. The Dirichlet unit theorem now gives $\boxed{{\mathcal{O}}_K^{\times }=\{\pm (5+\sqrt{26}{)}^n:n\in \mathbb{Z}\}}$. Its least positive norm-one generator is ${ϵ}^2=51+10\sqrt{26}$.

For the generalized Pell equation, write $\alpha =x+y\sqrt{26}$ with norm $N=\pm 10$. Change the sign if necessary and multiply by a power of $ϵ$ to place it in $\sqrt{10}\le \alpha <ϵ\sqrt{10}$. These operations preserve integrality and the absolute norm, though the latter operation can change the norm's sign. Since ${\alpha }^{\prime }=N/\alpha$, we have $x=(\alpha +N/\alpha )/2\ge 0$ and $y=(\alpha -N/\alpha )/(2\sqrt{26})\ge 0$. The upper bound gives $y<4$. Checking $y=0,1,2,3$ in $x^2=26y^2\pm 10$ leaves only $(x,y)=(4,1),(6,1)$. Both reduced elements lie in the chosen interval. Therefore all integral solutions of either sign areNegative powers are integral because ${ϵ}^{-1}=\sqrt{26}-5$. In the first family the norm is $-10(-1{)}^n$ and in the second it is $10(-1{)}^n$, so this also separates the two equations. The reduction proves exhaustion rather than merely generating examples.

Heights measure arithmetic size, and their finiteness and growth properties turn finite divisibility information into finite generation. For an elliptic curve $E$ over a number field $K$, the resulting Mordell-Weil theorem isThe distinction between this statement and the Weak Mordell-Weil theorem, which only asserts that $E(K)/mE(K)$ is finite, is exactly where heights enter.

For a rational point of the projective line, choose coprime integers $a,b$ and define the logarithmic projective heightFor a fixed number field, its absolute logarithmic version iswith the usual normalized absolute values. The product formula makes this independent of scaling $(a,b)$, and the local degree formula makes it independent of enlarging the number field. Heights are nonnegative and are useful precisely because they control numerator and denominator together, unlike an ordinary real absolute value.

The Northcott theorem states that algebraic points of bounded degree and bounded height form a finite set. Over $\mathbb{Q}$ this is immediate from the coprime numerator-denominator description. In general, bounded degree and height bound the coefficients of the primitive minimal polynomial: their elementary symmetric functions are bounded by its Mahler measure, which is the appropriate power of the multiplicative height. There are only finitely many such integer polynomials.

On a Weierstrass equation of an elliptic curve, define the naive logarithmic height $h_x(P)=h(x(P))$ and $h_x(O)=0$. Since an $x$-coordinate has at most two points above it, Northcott theorem implies that only finitely many points of $E(K)$ have bounded $h_x$.

Duplication descends to a rational map of the projective $x$-line of degree $4$. More generally, if a morphism $\varphi :{\mathbb{P}}^1\to {\mathbb{P}}^1$ has degree $d$, thenuniformly in $z$. Write $\varphi$ as two homogeneous degree-$d$ polynomials without common zeros. Bounding their coefficients gives the upper estimate at each place. A nonzero resultant, or homogeneous Bezout identities, prevents simultaneous cancellation and gives the lower estimate; only finitely many places contribute a nonzero constant. Applied to duplication, this gives the height growth under a morphism of the projective line estimate

Define the canonical height of an elliptic curve byThe factor $1/2$ is a convention making it the height for the degree-one symmetric divisor $O$. If $a_n=4^{-n}{h}_x(2^{n}P)$, then $|a_{n+1}-a_n|\le C_E{4}^{-(n+1)}$, so the limit exists by a geometric-series estimate. Summing that estimate also provesThus bounded canonical height still gives only finitely many $K$-points, and $\hat{h}(2P)=4\hat{h}(P)$ exactly.

The addition formula on the quotient by negation gives the uniform approximate parallelogram identityThis height parallelogram identity follows as follows. One way to obtain it is to express the unordered pair of their $x$-coordinates as a morphism ${\mathbb{P}}^1\times {\mathbb{P}}^1\to {\mathrm{Sym}}^2{\mathbb{P}}^1$ of bidegree $(2,2)$; the height of that unordered pair is the sum of the two heights, up to a bounded constant. It is also the height identity for the symmetric divisor $O$. Apply this identity to $2^{n}P,2^{n}Q$, divide by $2\cdot 4^n$, and take limits. The errors disappear, givingTogether with evenness and $\hat{h}(O)=0$, the recurrence implies $\hat{h}(mP)=m^2\hat{h}(P)$ for every integer $m$. The associated canonical height pairingis bilinear. Nonnegativity of the height implies its Cauchy-Schwarz inequality, by applying it to integer combinations and then rational approximations. In particular,The height vanishes precisely at torsion points: torsion gives a finite set of multiples, whereas if $\hat{h}(P)=0$, every multiple of $P$ has bounded naive height and therefore the set of multiples is finite by Northcott theorem.

Now apply the Weak Mordell-Weil theorem with $m=2$, proved cohomologically in the other essay. Choose a finite set $R$ of representatives for $E(K)/2E(K)$, and put $H={\max }_{R_i\in R}\hat{h}(R_i)$. Write an arbitrary point as $P=2Q+R_i$. ThenIf $\hat{h}(P)>H+1$, this decreases the height by more than $1/2$. Repeat until the current point has height at most $H+1$. There are only finitely many such points. Unwinding the equations $P=2Q+R_i$ shows that $E(K)$ is generated by this finite bounded-height set together with $R$. This is the height descent lemma, and proves the Mordell-Weil theorem.

Once finite generation is established, the structure theorem for finitely generated abelian groups gives the displayed decomposition. The canonical height pairing is positive definite on the real vector space of the free part: a null direction would, by approximation with integer combinations of generators, give infinitely many lattice points of bounded height, contradicting Northcott theorem. Its determinant is the regulator of an elliptic curve, and its geometry gives practical tools for searching for generators and comparing independence. Thus heights supply both the finiteness mechanism in the proof and the quadratic size function used in arithmetic computations.

Galois cohomology and the finite division quotient. Let $K$ be a number field, $G_K=\mathrm{Gal}(\overline{K}/K)$ with its profinite topology, and $M$ a discrete continuous $G_K$-module. Its Galois cohomology is continuous group cohomology. In degrees zero and one,Thus a crossed homomorphism measures the failure of a choice to be Galois-invariant, and changing the choice changes the cocycle by a coboundary. A short exact sequence of such modules gives a long exact sequence in group cohomology. The connecting map sends an invariant element to the cocycle obtained by lifting it and comparing its Galois translates.

For an elliptic curve $E/K$ and $n\ge 2$, multiplication by $n$ is surjective on $E(\overline{K})$ and has kernel $E[n]$. The Kummer exact sequence of an elliptic curve therefore givesExplicitly, choose $nQ=P$ and set $\kappa (P)(\sigma )=\sigma Q-Q$. Replacing $Q$ changes this by a coboundary; replacing $P$ by $P+nR$, $R\in E(K)$, does not change it. If its class is zero, some torsion translate of $Q$ is Galois-invariant, so $P\in nE(K)$. This proves injectivity of the Kummer map of an elliptic curve.

The whole group $H^1(K,E[n])$ need not be finite. The crucial restriction is ramification. Let $S$ contain the finite primes of bad reduction of an elliptic curve and those dividing $n$. At $v\notin S$, good reduction and prime-to-residue-characteristic multiplication imply that every division point of a local point is in an unramified extension, by the formal-group lifting argument. Thus every global Kummer class restricts trivially to inertia there.

Choose a finite Galois extension $L/K$ containing $E[n]$ and the $n$th roots of unity, and enlarge $S$ to include its ramified primes. Over $L$, the torsion module is constant and can be identified with ${\mu }_n^2$. By Kummer theory,For classes unramified outside the primes $S_L$ above $S$, valuations outside $S_L$ must be divisible by $n$. Their restriction images therefore lie in $L(S_L,n{)}^2$, the square of an S-unramified power class group. Its finiteness can be seen without assuming the Mordell-Weil theorem: the finiteness of S-unramified Kummer classes follows fromThe S-unit group is finitely generated by the Dirichlet unit theorem and its valuations at $S_L$. The localized ideal class group is a quotient of the ordinary ideal class group of $L$, which is finite by the finiteness of the ideal class group. The Minkowski bound for ideal classes supplies an integral ideal of bounded norm in each class, and only finitely many integral ideals have bounded norm.

The kernel of restriction from $H^1(K,E[n])$ to $H^1(L,E[n])$ is also finite: a cocycle trivial after restriction can, after subtracting a coboundary, be made to factor through the finite group $\mathrm{Gal}(L/K)$. There are only finitely many maps from that group to the finite module $E[n]$. This is the finite-extension kernel of a Kummer map mechanism. Consequently the possible Kummer classes form a finite set, provingThis is the Weak Mordell-Weil theorem. Imposing membership in the local Kummer image at every place refines the finite unramified collection to the n-Selmer group. Its relation with the obstruction to a globally rational point iswhere the last group is the $n$-torsion of the Tate–Shafarevich group. Thus local solubility gives a finite computable upper bound but can leave a genuine global obstruction.

Heights and finite generation. For the other essay, normalize the absolute values on a field $K$ so that the product formula holds. The Absolute logarithmic Weil height of $[x_0:x_1]\in {\mathbb{P}}^1(K)$ isAt the infinite places the local degrees are one or two and the usual real or complex modulus is used. The product formula makes this independent of scaling; it is also unchanged by extending $K$. Over $\mathbb{Q}$, using coprime integral coordinates gives $h([u:v])=\log \max (|u|,|v|)$, the naive height on the projective line. The Northcott theorem says that points of bounded height and bounded field degree form a finite set; in particular this holds over the fixed field $K$. One way to see the finiteness is the height-Mahler measure formula: for an algebraic number of degree $d$, the coefficients of its primitive minimal integer polynomial are bounded by binomial factors times $\exp (dh)$. Bounds on $d$ and $h$ therefore leave only finitely many integer polynomials and hence only finitely many algebraic numbers.

Set $h_E(P)=h(x(P))$ for $P\ne O$, and $h_E(O)=0$. Since $x:E\to {\mathbb{P}}^1$ has degree two, bounded $h_E$ gives only finitely many points of $E(K)$. The height growth under a morphism of the projective line states that a degree-$d$ morphism gives $h(f(t))=dh(t)+O(1)$. The upper bound follows from its homogeneous coordinate polynomials; the lower bound follows from their having no common zero, using a resultant identity at each place. Applied to the degree-four duplication map, it gives a uniform constant $C$ withConsequently $4^{-j}{h}_E(2^{j}P)$ is Cauchy: the absolute difference of consecutive terms is at most $C{4}^{-(j+1)}$. Define the canonical height of an elliptic curve byIt is nonnegative and satisfies $\hat{h}(2P)=4\hat{h}(P)$. The factor one-half is the conventional normalization for the divisor $(O)$, since the first-coordinate height belongs to $2(O)$.

The addition-divisor identity also gives the bounded-error height identityuniformly in $P,Q$. This is the height form of the identity for the sum and difference pullbacks of the line bundle associated to $2(O)$; it includes points where the affine addition formulas have a zero denominator. Apply it to $2^{j}P,2^{j}Q$, divide by $2\cdot 4^j$, and take the limit. Thus $\hat{h}$ satisfies the parallelogram law, and its polarization is a bilinear form, the height pairing. Nonnegativity implies the Cauchy-Schwarz inequality for that pairing, by evaluating its nonnegative quadratic polynomial on $mP+nQ$ for all integers $m,n$ and approximating real ratios by rational numbers. In particularThese properties do not presuppose finite generation. Bounded canonical height gives a finite set by the bounded difference and the Northcott theorem. Torsion points have canonical height zero. Conversely, if $\hat{h}(P)=0$, all $2^{j}P$ have bounded naive height, so two coincide; their difference makes $P$ torsion. Thus the height detects precisely the free part, once finite generation is proved.

Use the Weak Mordell-Weil theorem with $n=2$ and choose finitely many coset representatives $R_i$. Write $P=2Q+R_i$, and put $H={\max }_i\hat{h}(R_i)$. The height inequality yieldsIteration enters the finite set $\{P:\hat{h}(P)\le H+1\}$, since after $j$ steps the height is at most $H+2^{-j}\max (\hat{h}(P)-H,0)$. Unwinding $P=2Q+R_i$ expresses every point using that finite set and the $R_i$. This proves the canonical-height proof of Mordell-Weil finite generation and henceThe Fundamental theorem of finitely generated abelian groups supplies this decomposition. The Mordell-Weil theorem combines a finite quotient from arithmetic Galois cohomology with a contracting height descent; either ingredient alone would not establish finite generation.

For a number field with $r$ real embeddings ${\sigma }_i$ and $s$ conjugate pairs of complex embeddings ${\tau }_j$, use the logarithmic embedding of number field unitsIt is a group homomorphism from multiplication to addition. A unit has field norm $\pm 1$, so the coordinates sum to zero. Deleting one coordinate identifies this hyperplane with ${\mathbb{R}}^{r+s-1}$ and gives the desired map.

If its image coordinates remain in a bounded set, the omitted coordinate is bounded too, and every conjugate of $u$ has bounded absolute value. The coefficients of its monic minimal polynomial are integers and elementary symmetric expressions in at most $[K:\mathbb{Q}]$ bounded conjugates. They are therefore bounded integers. Only finitely many such polynomials, and finitely many roots of them, exist. Thus the inverse image of every bounded set is finite. In particular the kernel is finite and the image is discrete. This argument includes the zero-dimensional target case; it does not yet assert the full-rank part of Dirichlet unit theorem.

TakeWe justify that this is a Dedekind domain. It is an integral domain inside the number field $K=\mathbb{Q}(\sqrt{-5})$. As an additive group it is free of rank two over $\mathbb{Z}$. The subgroup theorem for finitely generated abelian groups says that every subgroup of such a group is finitely generated. Thus every ideal has finitely many additive generators, and those same elements generate it as a $D$-ideal. This proves that $D$ is Noetherian.

We use the standard formula for the ring of integers of a quadratic field: for squarefree $d$, it is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Since $-5\equiv 3(\mathrm{mod}4)$, $D$ is the ring of integers of a number field $K$, meaning all elements of $K$ integral over $\mathbb{Z}$. We also use transitivity of integral extensions. If an element of $K$ is integral over $D$, it is integral over $\mathbb{Z}$ because $D$ is integral over $\mathbb{Z}$, and therefore belongs to $D$. Thus $D$ is an integrally closed domain.

Every nonzero prime ideal $P$ is maximal. Indeed, choose $0\ne \alpha =a+b\sqrt{-5}\in P$. Multiplying by its algebraic conjugate puts the nonzero integer $N(\alpha )=a^2+5b^2$ in $P$. Hence the integral domain $D/P$ is a quotient of the finite ring $D/N(\alpha )D$, and is finite. Every finite integral domain is a field, since multiplication by a nonzero element is injective and therefore surjective. The element $2$ has no inverse in $D$, since $1/2\notin \mathbb{Z}[\sqrt{-5}]$, so $D$ is not a field. Consequently $(0)$ is not maximal, and a maximal ideal above it is nonzero. Together with the preceding result, this gives Krull dimension exactly one.

Finally, consider the idealReduction modulo $2$ with $\sqrt{-5}\mapsto 1$ identifies $D/I$ with ${\mathbb{F}}_2$, so the additive index $[D:I]$ is $2$. If $I$ were the principal ideal $(a+b\sqrt{-5})$, multiplication by its nonzero generator would have integer matrixin the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, soThere are no integer solutions: $b\ne 0$ makes the right side at least $5$, while $b=0$ would require $a^2=2$. Thus $I$ is a nonprincipal ideal in the integers adjoined a square root of minus five, and $D$ is a Dedekind domain that is not a principal ideal domain.

The Hilbert class field $\tilde{K}$ is the maximal everywhere unramified abelian extension of the number field $K$, with complete splitting at real places if infinite places are included. Global class field theory gives the canonical Artin reciprocity mapso $[\tilde{K}:K]$ is the class number of $K$.

A finite extension $L/K$ of non-Archimedean local fields is an unramified extension when its ramification index is one and its residue-field degree equals $[L:K]$. Equivalently, its maximal ideal is generated by a uniformizer of $K$ and its residue-field extension is separable.

A finite extension of number fields is an everywhere unramified extension of number fields when every nonzero prime ideal of $K$ is unramified in $L$. Under the convention that includes infinite places, one also requires every real embedding of $K$ to remain real.

The Dirichlet unit theorem states that for a number field with $r$ real embeddings and $s$ conjugate pairs of complex embeddings,where $\mu (K)$ is its finite group of roots of unity. Here $K$ is real quadratic, so $(r,s)=(2,0)$ and $\mu (K)=\{\pm 1\}$. Thus its unit group has rank one.

The elementis a unit because its field norm isIt is the smallest unit greater than one in the chosen real embedding. Indeed, write such a unit as $u=a+b\sqrt{2}$ and let $u^{\prime }=a-b\sqrt{2}=N(u)/u$ be its algebraic conjugate. Since $|u^{\prime }|<1<u$, both $a=(u+u^{\prime })/2$ and $b=(u-u^{\prime })/(2\sqrt{2})$ are positive integers. Hence $a,b\ge 1$ and $u\ge 1+\sqrt{2}$. The rank-one logarithmic lattice in Dirichlet's theorem is therefore generated by $\log \epsilon$. Consequently the units of Q of square root two are

Let ${\sigma }_1,\dots ,{\sigma }_n:K\hookrightarrow \mathbb{C}$ be the embeddings of the number field $K$. The discriminant of elements of a number field isIf the ${\alpha }_j$ are $\mathbb{Q}$-linearly dependent, the embedding matrix has dependent columns and the discriminant vanishes. Conversely, if they are independent, they form a $\mathbb{Q}$-basis of the $n$-dimensional vector space $K$. The trace pairing of a characteristic-zero number field is nondegenerate, so its Gram matrix in this basis is nonsingular. Hence

For each place of a number field $v$, let $K_v$ be the corresponding completion, and for finite $v$ let ${\mathcal{O}}_v$ be its valuation ring. The adele ring is the restricted productIts restricted product topology has basic open sets ${\prod }_v{U}_v$, where every $U_v\subseteq K_v$ is open and $U_v={\mathcal{O}}_v$ at all but finitely many finite places.