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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.

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.

The Minkowski convex body theorem states that if $\Lambda \subset {\mathbb{R}}^n$ is a full lattice and $C$ is a measurable, convex, centrally symmetric set withthen $C$ contains a nonzero point of $\Lambda$.

Here $d\not\equiv 3(\mathrm{mod}4)$, so the ring of integers of a quadratic field isUnder its complex embedding it has real-lattice covolume $\sqrt{d}$. An integral ideal $I$ has index $N(I)$ in ${\mathcal{O}}_K$, so its covolume is $\sqrt{d}N(I)$. Apply Minkowski's theorem to the disk $|z|\le R$, whose area is $\pi R^2$. Taking $R^2$ down to $4\sqrt{d}N(I)/\pi$ gives a nonzero $\alpha \in I$ withSince the field norm is $N_{K/\mathbb{Q}}(\alpha )=|\alpha {|}^2$ in this imaginary quadratic field, this proves the imaginary quadratic Minkowski disk bound

Because $(\alpha )\subseteq I$, the idealis integral, represents the inverse class of $I$, and hasThus every ideal class has a bounded-norm representative. Only finitely many integral ideals have bounded norm, proving the finiteness of the ideal class group.

Now take $K=\mathbb{Q}(\sqrt{-22})$. Its ring of integers is $\mathbb{Z}[\sqrt{-22}]$, its discriminant is $-88$, andEvery class therefore has an integral representative of norm at most five. The polynomial $X^2+22$ is irreducible modulo three and modulo five, so both primes are inert and there are no ideals of norm three or five. At two it reduces to $X^2$, and the splitting of rational primes in a quadratic field givesAn ideal of norm four is ${\mathfrak{p}}_2^2=(2)$ and is principal. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm two, whereashas no integer solution. Consequently the Ideal class group of Q of square root of minus twenty-two is

Suppose finally that $y^3=x^2+22$. If $x$ were even, the right-hand side would be $2$ modulo four, which is not a cube modulo four. Hence $x$ and $y$ are odd. In ${\mathcal{O}}_K$,The two ideals on the left are coprime. Indeed, a common prime ideal would divide both $2x$ and $2\sqrt{-22}$. It cannot lie above two because their norms are odd. It would therefore lie above eleven and force $11\mid x$; but then $x^2+22$ has $11$-adic valuation exactly one, impossible for the cube $y^3$.

By unique factorization of ideals in a number field, coprimality impliesfor some ideal $\mathfrak{a}$. The class group has order two, so it has no element of order three; $[\mathfrak{a}{]}^3=1$ therefore forces $\mathfrak{a}$ to be principal. The only units are $\pm 1$, both of which can be absorbed into a cube, and hencefor integers $a,b$. Comparing the coefficient of $\sqrt{-22}$ givesThus $b=\pm 1$. For $b=1$ this requires $3a^2=23$, and for $b=-1$ it requires $3a^2=21$; neither is possible. Therefore the result on no integral solutions to y cubed equals x squared plus twenty-two is