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Inside an algebraic closure, choose compatible roots $t^{1/n}$. The fields $\mathbb{C}((t^{1/n}))$ exhaust all finite extensions by finite extensions of the complex Laurent series field. Restriction from a degree-$m$ field to a degree-$n$ field for $n\mid m$ sends an exponent modulo $m$ to its reduction modulo $n$. The absolute Galois group is their inverse limit, namely the profinite completion of $\mathbb{Z}$. This is a procyclic group, rather than the direct-limit torsion group of all complex roots of unity.

The value ${\mathrm{cd}}_p(G)$ is the least $n$ such that all cohomology in degrees greater than $n$ vanishes for every discrete $p$-primary torsion module. Vanishing in degree $n+1$ for all such modules suffices by dimension shifting. Dimension does not increase on closed subgroups. For a field, use its absolute Galois group.

For a primitive $m$th root of unity ${\zeta }_m$, every Galois automorphism sends ${\zeta }_m$ to a unique ${\zeta }_m^a$ with $a$ invertible modulo $m$. The assignment $\sigma \mapsto a$ is a group homomorphism, since composition multiplies the exponents, and is injective because ${\zeta }_m$ generates the cyclotomic field. The irreducibility of cyclotomic polynomials gives degree $\phi (m)$, equal to the order of the target, so the map is an isomorphism. The cyclotomic character on an absolute Galois group factors through this finite quotient and generally has a nontrivial kernel.

A Galois character is a continuous one-dimensional Galois representation: the absolute Galois group acts by multiplication by a scalar in a coefficient field $F$. A quadratic character over a coefficient field of characteristic different from two has values in $\{1,-1\}$; after identifying this group with $\mathbb{Z}/2\mathbb{Z}$, its kernel fixes a quadratic extension, or the base field for the trivial character. Its restriction to an inertia group is trivial exactly when that extension is unramified at the corresponding place.

For an elliptic curve $E/K$, regard $E(K^{\mathrm{sep}})$ as a discrete Galois module for the absolute Galois group $G_K$. Its continuous Galois cohomology has $H^0(K,E)=E(K)$; $H^1(K,E)$ classifies torsors under $E$. For an integer $n$ invertible in $K$, the Kummer exact sequence of an elliptic curve gives $0\to E(K)/nE(K)\to H^1(K,E[n])\to H^1(K,E)[n]\to 0$. If $[n]Q=P$, the first map sends $P$ to $\sigma \mapsto \sigma Q-Q$. Changing $Q$ changes the cocycle by a coboundary; this class vanishes exactly when $P$ has a division point over $K$.

A discrete Galois module over $K$ is an abelian group with a continuous action of the absolute Galois group $G_K$. Continuity means that each element has an open stabilizer. Examples are the algebraic points of an elliptic curve, its finite torsion module $E[m]$, and the group of roots of unity ${\mu }_m$. Galois cohomology takes continuous cocycles with values in such a module.

A Galois representation is a continuous group representation of an absolute Galois group, usually on a vector space or lattice over a local field. Examples include the cyclotomic character and the representations attached to modular forms. The coefficient topology is part of the definition.

Suppose an elliptic curve $E/K$ has all four geometric 2-torsion points rational. For $P\in E(K)$ choose $Q\in E(\overline{K})$ with $2Q=P$. The function $c_P(\sigma )=\sigma (Q)-Q$ takes values in $E[2]$ and is a group homomorphism from the absolute Galois group to $E[2]$: its usual cocycle law becomes additivity because the action on $E[2]$ is trivial. Replacing $Q$ by another half, or $P$ by $P+2R$ with $R\in E(K)$, leaves it unchanged. Adding chosen halves proves additivity in $P$. Its kernel in $E(K)/2E(K)$ is zero, since $c_P=0$ precisely when the chosen half is rational. This constructs the injective map directly, without assuming any theorem of two-descent on an elliptic curve or finite generation of the Mordell-Weil group.

For $K=\mathbb{Q}(\zeta )$, every Galois automorphism sends the primitive root $\zeta$ to another primitive $m$th root of unity, uniquely ${\zeta }^a$ for $a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }$. The Cyclotomic character of a finite cyclotomic extension isIf ${\chi }_m(\sigma )=a$ and ${\chi }_m(\tau )=b$, then $(\sigma \tau )(\zeta )=\sigma ({\zeta }^b)={\zeta }^{ab}$, proving the group homomorphism property. Its kernel is trivial: an automorphism fixing $\zeta$ fixes the field it generates. Thus the character is injective. Here its domain is the finite cyclotomic Galois group; the corresponding map from an absolute Galois group need not be injective.

Let $K$ be a field, $K_s$ a separable closure and $G_K$ its absolute Galois group. A discrete Galois module $A$ is an abelian group with a continuous $G_K$-action, so each element has an open stabilizer. Galois cohomology uses continuous cochains. Its zeroth group is $H^0(K,A)=A^{G_K}$, and its first group iswhereThus cocycles measure a failure of descent, and coboundaries account for changing a chosen lift. A short exact sequence of Galois modules gives a long exact sequence in group cohomology. Its connecting map sends an invariant element of a quotient to the cocycle $\sigma b-b$ of a chosen lift $b$; changing the lift changes this by a coboundary. These constructions provide the bridge between arithmetic divisibility and cohomology.

For Kummer theory, assume the characteristic does not divide $n$. The multiplicative exact sequence isThe final arrow is surjective because a root of $T^n-a$ is separable for $a\ne 0$. Hilbert's theorem 90 says $H^1(K,K_s^{\times })=0$. Here is its usual finite-extension proof. For a multiplicative cocycle $c_{\sigma }$ on a finite Galois extension $L/K$, choose $a\in L$ so thatSuch an $a$ exists by the Artin independence theorem. The cocycle identity gives $\tau A=c_{\tau }^{-1}A$. With $b=A^{-1}$ this becomes $c_{\tau }=\tau b/b$, a coboundary. Every continuous cocycle with values in $K_s^{\times }$ is defined over a sufficiently large finite Galois extension, so the same result holds for the absolute group.

The long exact sequence in group cohomology now givesThe class of $a$ maps to $c_{\sigma }=\sigma (\alpha )/\alpha$, where ${\alpha }^n=a$. This cohomological statement does not require ${\mu }_n\subset K$; it retains the actual Galois action on ${\mu }_n$. If all $n$th roots of unity do lie in $K$, the action is trivial, and after choosing a primitive root of unity the cocycles are characters into $\mathbb{Z}/n\mathbb{Z}$. Thus adjoining an $n$th root of $a$ describes a cyclic extension of degree dividing $n$, and Kummer classes classify the corresponding characters. For $n=2$, this gives the familiar quadratic extensions and the square-class group of a field.

Now let $K$ be a number field and $E/K$ an elliptic curve. The multiplication-by-n morphism is a surjective isogeny on $E(\overline{K})$, with kernel $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$. The Kummer exact sequence of an elliptic curveyieldsExplicitly, choose $Q$ with $nQ=P$ and set $\delta (P{)}_{\sigma }=\sigma Q-Q$. This lies in $E[n]$, satisfies the cocycle identity and is unchanged as a class if $Q$ is replaced by another division point. If it is a coboundary $\sigma T-T$ for $T\in E[n]$, then $Q-T$ is rational over $K$ and $P=n(Q-T)$. Conversely such a rational division point makes the cocycle zero. This proves the injectivity of the Kummer map of an elliptic curve on $E(K)/nE(K)$.

To prove the Weak Mordell-Weil theorem, it remains to prove that this particular cohomological image is finite. The full group $H^1(K,E[n])$ need not be finite, so the injection alone is not enough. Choose a finite set $S$ of finite places containing the bad-reduction places and those dividing $n$. At any other place, the construction in Question 3 works over the local field: reduce $P$, find an $n$-division point over a finite residue extension, lift it by the Hensel lemma, and correct its error by the inverse formal multiplication. Thus $P$ has an $n$-division point over the maximal unramified extension. Choosing that lift locally makes the cocycle zero on inertia. For an arbitrary lift its restriction is a coboundary, so the cohomology class is unramified. Therefore the image lies in first Galois cohomology unramified outside a finite set:We prove the required finiteness of this group. Take a finite Galois extension $L/K$ containing all of $E[n]$ and ${\mu }_n$, and enlarge $S$ to include any remaining ramification of $L/K$. Let $S_L$ be the places above it. The inflation-restriction exact sequence bounds the kernel of restriction to $L$ bya finite group, since both its group of arguments and its group of values are finite. Over $L$ the torsion module is constant and can be identified with ${\mu }_n^2$, so Kummer theory givesRestrictions of our unramified classes remain unramified outside $S_L$. If a Kummer class $[a]$ is unramified at such a place $v$, its $n$th root can be taken in an unramified local extension. The extended valuation still has integer values, and $nv(\sqrt[n]{a})=v(a)$. Hence $v(a)$ is divisible by $n$. Each coordinate therefore lies in the S-unramified power class groupThis group is finite. Indeed, the ideal of $a$ away from $S_L$ is ${\mathfrak{a}}^n$. Sending $[a]$ to $[\mathfrak{a}]$ yields the exact sequenceFor its kernel, a principal $\mathfrak{a}$ lets us divide $a$ by an $n$th power to obtain an $S_L$-unit. For surjectivity, an ideal class killed by $n$ has an $n$th power which is principal away from $S_L$, giving such an $a$. The S-unit group is finitely generated and the ideal class group is finite; hence both end groups are finite, proving finiteness of S-unramified Kummer classes. Restriction has finite image in $L(S_L,n{)}^2$ and finite kernel, so $H_S^1(K,E[n])$ is finite. We concludewhich is the Weak Mordell-Weil theorem and its Kummer-theoretic proof of the weak Mordell-Weil theorem.

This framework also defines the Selmer group of an elliptic curve by imposing that a class in $H^1(K,E[n])$ lie in the local Kummer image at every completion. Its classes are unramified outside a fixed finite set, so the same finiteness proof applies. Its relation to rational points isThe Tate–Shafarevich group here measures the difference between locally soluble torsors and globally soluble ones. Finally, the weak theorem gives finite divisibility quotients; adding the height descent lemma and bounded-height finiteness proves the stronger Mordell-Weil theorem. Those height inputs are additional to the cohomological finiteness just established.

Galois cohomology and finite quotients. Let $k$ be a number field and $G_k=\mathrm{Gal}(\overline{k}/k)$ its absolute Galois group. For a discrete abelian $G_k$-module $M$, $H^0(k,M)=M^{G_k}$. A continuous 1-cocycle is a map $c:G_k\to M$ with $c_{\sigma \tau }=c_{\sigma }+\sigma c_{\tau }$; a coboundary has the form $c_{\sigma }=\sigma m-m$. Their quotient is $H^1(k,M)$. Short exact sequences of modules give the long exact sequence in group cohomology; the connecting map measures the obstruction to lifting a fixed element to a fixed preimage. This interpretation gives a concrete descent map for an elliptic curve.

For an integer $m\ge 2$, multiplication on the geometric points gives the exact sequenceHere multiplication is a separable isogeny in characteristic zero, with $m^2$ geometric kernel points. For $P\in E(k)$ choose $Q$ with $mQ=P$. Then $c_{\sigma }=\sigma Q-Q$ lies in $E[m]$, satisfies the cocycle identity, and changing $Q$ changes it by a coboundary. Its class $\delta (P)$ vanishes exactly when $P\in mE(k)$: if $c_{\sigma }=\sigma T-T$ for $T\in E[m]$, the point $Q-T$ is rational and divides $P$. Thus the Kummer exact sequence of an elliptic curve includesThe construction also proves that $\delta$ is a homomorphism. The final group can be interpreted as classes of elliptic-curve torsors killed by $m$; imposing soluble completions leads to the Selmer group of an elliptic curve. For weak Mordell-Weil, the important task is to bound the arithmetic image of $\delta$, not to assume that the entire first cohomology group is finite.

Choose a finite set $S$ containing the bad primes, primes dividing $m$, and archimedean places. At a good finite place $v\notin S$, the class $\delta (P)$ is unramified. Here is the local reason. Over a finite extension of the residue field choose an $m$-division point of the reduction of $P$; multiplication is surjective on the points over its algebraic closure. Pass to the corresponding unramified extension of $k_v$ and lift this nonsingular point by the Hensel lemma. The discrepancy between $m$ times its lift and $P$ lies in the kernel of reduction. The formal group of an elliptic curve describes that kernel, and multiplication by $m$ is bijective there because its leading coefficient is a unit. Correcting the lift therefore gives an actual division point in the unramified closure. Inertia fixes that point, so the Kummer cocycle vanishes on inertia. The same good-reduction argument shows that $E[m]$ itself is unramified outside $S$.

We now prove the required finiteness of unramified cohomology classes. Choose a finite Galois extension $L/k$ over which $E[m]$ and the $m$th roots of unity are all rational, and enlarge $S$ to include its ramified primes. The inflation-restriction exact sequence shows that restriction to $G_L$ has finite kernel: that kernel comes from $H^1(\mathrm{Gal}(L/k),E[m])$, a finite group since both its group and coefficient module are finite. Over $L$, a choice of basis identifies $E[m]$ with ${\mu }_m^2$. The multiplicative Kummer sequence, together with Hilbert 90, givesFor clarity, Hilbert 90 says that a multiplicative cocycle for a finite Galois extension is a coboundary. If its values are $c_{\sigma }$, choose $a$ so that $b={\sum }_{\sigma }{c}_{\sigma }\sigma (a)\ne 0$. Then $\tau (b)=c_{\tau }^{-1}b$, so $c_{\tau }=\tau (b^{-1})/b^{-1}$. Such an $a$ exists because distinct field automorphisms are linearly independent: a relation with the fewest nonzero terms, evaluated on $at$ and compared with its value on $t$ multiplied by one automorphism's value on $a$, would produce a shorter nonzero relation. Continuity reduces the absolute-Galois version to a finite extension. This proves the cohomological vanishing used in the multiplicative Kummer sequence.

An unramified Kummer class represented by $a\in L^{\times }$ has $v_w(a)\equiv 0(\mathrm{mod}m)$ at every finite $w\notin S_L$. Indeed an $m$th root of $a$ fixed by inertia belongs to an unramified local extension, whose value group is still the integers. DefineFactoring the fractional ideal away from $S_L$ as an $m$th power gives the exact sequenceTo check it, send $[a]$ to the ideal class of its $m$th root ideal. Changing $a$ by an $m$th power changes that ideal by a principal ideal. The kernel consists precisely of S-units modulo $m$th powers, and every $m$-torsion ideal class has a representative whose $m$th power is principal, proving surjectivity. The S-unit group is finitely generated by the Dirichlet unit theorem and the finite collection of allowed prime valuations. The localized ideal class group is a quotient of the finite ordinary ideal class group. Both end groups, and hence $L(S_L,m)$, are finite. This is the finiteness of S-unramified Kummer classes argument.

The restrictions of the elliptic Kummer image lie in two copies of this finite group, and the restriction kernel is finite. The injection $\delta$ therefore provesThis is the Weak Mordell-Weil theorem. The unramified restriction is essential: even $H^1(k,{\mu }_2)=k^{\times }/k^{\times 2}$ is generally infinite. Cohomology turns divisibility into a cocycle; good reduction restricts its ramification; units and ideal classes supply the arithmetic finiteness.

Height functions and finite generation. Write $d=[k:\mathbb{Q}]$ and use absolute values extending the usual rational ones, with local degrees $n_v=[k_v:{\mathbb{Q}}_v]$. The logarithmic height of a projective pair isThe product formula makes it independent of scaling. For $t\in k$, $h(t)=h(t:1)$ is nonnegative. On a short Weierstrass equation $E:y^2=x^3+Ax+B$, put $h_x(P)=h(x(P))$ and $h_x(O)=0$.

Bounded height gives only finitely many rational points. To see the scalar finiteness behind the Northcott theorem, an algebraic number $t$ of degree $e$ has primitive minimal polynomial $a_0{\prod }_{i=1}^e(T-t_i)$, and the product formula gives $e^{eh(t)}=|a_0|{\prod }_i\max (1,|t_i|)$. If $e\le d$ and $h(t)\le H$, every coefficient has absolute value at most $2^d{e}^{dH}$. There are only finitely many such integer polynomials and hence finitely many roots. Each x-coordinate lifts to at most two points of $E$, with $O$ handled separately. Thus $\{P\in E(k):h_x(P)\le H\}$ is finite.

The second ingredient is height expansion under doubling:Indeed the duplication formula is the degree-four morphism of the projective x-line represented by the coprime quartic formsThey are coprime because the elliptic curve is nonsingular. At each place their coefficient bounds give $\max (|F{|}_v,|G{|}_v)\le C_v\max (|X{|}_v,|Z{|}_v{)}^4$. Bezout identities expressing powers of each coordinate as combinations of $F,G$ give the reverse inequality up to another local constant. One obtains these identities by the Euclidean algorithm on the two affine charts and clearing denominators. Only finitely many non-Archimedean places have a constant different from one. Taking weighted logarithms and summing proves the uniform bounded error. This is height growth under a morphism of the projective line, applied with degree four.

For a fixed $R=(r,s)\in E(k)$, the addition formula also givesFor $x(P)\ne r$, its x-coordinate isThe curve equation bounds $|y{|}_v$ by a local constant times $\max (1,|x{|}_v{)}^{3/2}$. Both numerator and denominator are consequently bounded by a local constant times $\max (1,|x{|}_v{)}^2$. The projective-height definition gives the claimed inequality; the finitely many exceptional points with $x=r$ are absorbed into the constant. For $R=O$ the estimate is immediate.

By weak Mordell-Weil, choose finitely many representatives $R_1,\dots ,R_t$ for $E(k)/2E(k)$. Every $P$ can be written $P=2Q+R_i$. Combining the last two inequalities giveswith one constant $C_0\ge 0$ for all the representatives. This is naive height contraction in elliptic halving descent. If $h_x(P)>2C_0+1$, the new point $Q$ has height at least one half less than that of $P$. Repetition must therefore reach the finite set of points of height at most $2C_0+1$. Reversing the identities $P_j=2P_{j+1}+R_{i_j}$ shows that every point is generated by this finite terminal set together with the finite list of representatives. ThusThe structure theorem then gives $E(k)\cong E(k{)}_{\mathrm{tors}}\times {\mathbb{Z}}^r$ with finite torsion subgroup: this is the Mordell-Weil theorem. Finiteness modulo two alone does not imply finite generation; the contracting height argument is what bridges that gap.

The bounded duplication error also constructs the canonical height of an elliptic curve:The successive differences are bounded by a summable geometric sequence, so the limit exists, $\hat{h}(P)=\frac{1}{2}{h}_x(P)+O_E(1)$, and $\hat{h}(2P)=4\hat{h}(P)$. It is nonnegative. Torsion points have canonical height zero because their multiples form a finite set. Conversely, if the canonical height is zero, all $2^{j}P$ have bounded naive height; Northcott finiteness makes two of them equal and hence makes $P$ torsion. The canonical height removes the bounded error from the arithmetic growth law and identifies exactly the torsion directions, while the elementary naive-height descent already proves finite generation.

Fix a separable closure $k^s$. The Tate module and rational Tate module arewhere the transition maps are multiplication by $\ell$. Since $\ell$ is different from the characteristic, $E[{\ell }^n](k^s)\simeq (\mathbb{Z}/{\ell }^n{)}^2$ and these transition maps are surjective. Choosing compatible bases gives $T_{\ell }(E)\simeq {\mathbb{Z}}_{\ell }^2$, and $T_{\ell }(E)/{\ell }^n{T}_{\ell }(E)\simeq E[{\ell }^n]$. Both spaces carry the natural continuous action of the absolute Galois group of $k$ and of the base-defined endomorphism ring of an elliptic curve.

For (i), choose a nonrational $a\in R$. Such an element exists because the intersection of the given two-dimensional rational subalgebra with the integral endomorphism ring contains a full lattice. The characteristic equation of an elliptic endomorphism isSince $a\in K$ is nonrational, its minimal polynomial over $\mathbb{Q}$ has degree two. Comparing the two monic equations gives $\mathrm{tr}(a)=a+\bar{a}$ and $\mathrm{deg}(a)=a\bar{a}$. The same is the characteristic polynomial on the rational Tate module: the Weil pairing gives determinant $\mathrm{deg}(a)$, and applying it to $1-a$ gives trace $1+\mathrm{deg}(a)-\mathrm{deg}(1-a)$. Thus on $V_{\ell }(E)$ the characteristic polynomial isIf $K_{\ell }$ is a quadratic field over ${\mathbb{Q}}_{\ell }$, a unital $K_{\ell }$-action on this two-dimensional ${\mathbb{Q}}_{\ell }$-space makes it a one-dimensional $K_{\ell }$-space. If $K_{\ell }\simeq {\mathbb{Q}}_{\ell }\times {\mathbb{Q}}_{\ell }$, the two roots above are distinct in ${\mathbb{Q}}_{\ell }$ and their eigenspaces each have dimension one. The two idempotent factors therefore have equal rank one, so the module is again free of rank one. This also proves faithfulness in the split case. Hence

For (ii), first prove an integral divisibility statement. If $a\in R$ maps $T_{\ell }(E)$ into ${\ell }^n{T}_{\ell }(E)$, then its induced action on $E[{\ell }^n]$ is zero. The kernel of an isogeny of $[{\ell }^n]$ is therefore contained in the kernel of $a$. The quotient property of this isogeny of elliptic curves gives a unique endomorphism $b$ withThis factorization is defined over $k$: applying any descent automorphism preserves it, and uniqueness fixes $b$. Inside the rational endomorphism ring, $b=a/{\ell }^n$ belongs to $K$, so $b\in R$. ConsequentlyHere the prime-to-characteristic torsion is étale, so vanishing on its geometric points is vanishing on the entire kernel scheme.

Every element of $R_{\ell }$ preserves the Tate module by continuity. Conversely suppose $\varphi \in K_{\ell }$ preserves it. Choose $n\ge 0$ with ${\ell }^n\varphi \in R_{\ell }$, and approximate it by $a\in R$ modulo ${\ell }^n{R}_{\ell }$. Write ${\ell }^n\varphi =a+{\ell }^{n}c$ with $c\in R_{\ell }$. Then $a{T}_{\ell }(E)\subseteq {\ell }^n{T}_{\ell }(E)$, so the divisibility statement gives $a/{\ell }^n\in R$. It follows that $\varphi =a/{\ell }^n+c\in R_{\ell }$. Thus the CM Tate module stabilizer isThis does not presume freeness of the integral Tate module over a nonmaximal order.

For (iii), every element of $R$ is defined over $k$, so the Galois representation commutes with $R$ and therefore with $K_{\ell }$. By (i), its action on $V_{\ell }(E)$ is multiplication by an element of $K_{\ell }^{\times }$. The action and its inverse preserve $T_{\ell }(E)$; by (ii), this scalar and its inverse belong to $R_{\ell }$. Hence

For a complete discretely valued field with perfect residue field $k$, the Witt residue sequence is the split exact sequenceThe first map identifies the unramified Brauer classes. The residue map is canonical; a splitting is chosen by a uniformizer $\pi$ and sends an unramified character $\chi$ to the cyclic Brauer class $(\chi ,\pi )$. Changing $\pi$ may change this splitting by an unramified class. This is the version of Witt's theorem used here.

Now suppose $k={\mathbb{F}}_q$. Fix arithmetic Frobenius $F_k:a\mapsto a^q$. Define the local Brauer invariant byHere is an explicit cohomological construction. Set $E=K^{\mathrm{nr}}$ and $G=\mathrm{Gal}(E/K)\cong \hat{\mathbb{Z}}$. The permitted vanishing $\mathrm{Br}(E)=0$, together with Hilbert 90 and inflation, identifies $\mathrm{Br}(K)$ with $H^2(G,E^{\times })$. Normalize the valuation $v(\pi )=1$. Its induced map sends a multiplicative two-cocycle $c$ to the integer-valued two-cocycle $v(c)$, givingThe middle isomorphism is the connecting map for $0\to \mathbb{Z}\to \mathbb{Q}\to \mathbb{Q}/\mathbb{Z}\to 0$, since positive-degree continuous cohomology of the trivial module $\mathbb{Q}$ vanishes.

To see that the valuation map is an isomorphism, its module sequence splits using $\pi$, and the unit part has zero second cohomology. For each finite unramified cyclic extension $E_m/K$, cyclic cohomology identifies that unit contribution with ${\mathcal{O}}_K^{\times }/N_{E_m/K}{\mathcal{O}}_{E_m}^{\times }$. The norm on residue-field units is surjective. On the principal-unit quotient of level $r\ge 1$, the norm is the residue trace, through $N(1+{\pi }^{r}b)\equiv 1+{\pi }^r\mathrm{Tr}(\overline{b})(\mathrm{mod}{\pi }^{r+1})$. The trace of a finite separable field extension is surjective; successive corrections and completeness therefore make the norm surjective on all units. Passing to the direct limit gives the asserted vanishing. Finally, a continuous character of $\hat{\mathbb{Z}}$ is uniquely specified by its value on arithmetic Frobenius, and that value can be any element of $\mathbb{Q}/\mathbb{Z}$. The invariant map is an isomorphism.

In concrete terms, if $E_m/K$ is the unramified extension of degree $m$ and $\sigma$ lifts arithmetic Frobenius, the cyclic algebra with $ub=\sigma (b)u$ and $u^m=a$ hasIndeed its standard two-cocycle has value $a$ when the powers of $\sigma$ wrap past $m$ and value one otherwise. Valuation turns this into the connecting cocycle for the character with Frobenius value $v_K(a)/m$. This also fixes the sign of the normalization.

For restriction and corestriction of local Brauer invariants, let $l$ be the residue field of $L$, $e$ the ramification index and $f=[l:k]$. A finite extension of local fields has $[L:K]=ef$, $v_L{|}_K=e{v}_K$ and $v_K(N_{L/K}b)=f{v}_L(b)$. Every class of $\mathrm{Br}(K)$ is $(\chi ,{\pi }_K)$, with $\chi$ unramified. After restriction to $L$, its character has Frobenius value $f\chi (F_k)$ and its parameter has valuation $e$. ThusFor corestriction, write a class of $\mathrm{Br}(L)$ as $({\chi }_l,{\pi }_L)$. Choose a character ${\chi }_k$ of the finite-field absolute group such that ${\mathrm{Res}}_{l/k}{\chi }_k={\chi }_l$. This is possible because restriction is multiplication by $f$ on $\mathbb{Q}/\mathbb{Z}$, which is divisible. Inflate both characters to the local fields. The norm projection formula for cyclic Brauer pairings givesThis projection identity follows from the cohomological transfer and cup-product identity; on multiplicative degree-zero coefficients transfer is the field norm, and the identity in higher degree follows by the cochain definition or dimension shifting. The norm parameter has $K$-valuation $f$, so its invariant is $f{\chi }_k(F_k)={\chi }_l(F_l)$. ThereforeThis argument uses the character-and-norm description directly and does not cancel multiplication by $[L:K]$ in $\mathbb{Q}/\mathbb{Z}$, where such cancellation would be invalid. For purely inseparable steps in equal characteristic, absolute Galois groups identify, restriction uses inclusion of multiplicative coefficient modules, and Brauer corestriction uses the field norm on them. The same projection calculation applies. Factoring a general finite extension into separable and purely inseparable steps covers all finite extensions in the question.

Lastly, a class with invariant $a/b$ in lowest terms has Brauer period $b$. The unramified extension of degree $b$ splits it, so its division-algebra degree divides $b$. Conversely a maximal subfield of the division algebra splits it and has degree equal to that division-algebra degree; the restriction formula forces $b$ to divide that degree. These standard splitting-field properties are among the permitted inputs. Thus the index of a central simple algebra equals its period over a local field. The central division algebras of degree $n$ are exactly those with invariants $a/n$ and $\gcd (a,n)=1$. The Brauer class determines its division representative uniquely, so there areisomorphism classes, where $\phi$ is the Euler totient function. This includes the single degree-one algebra $K$.

We first give the common reduction for the three conditions, then prove the cycle from (i) to (ii), from (ii) to (iii), and from (iii) back to (i). Assume here that $\mathrm{char}K\ne p$. The Galois module of roots of unity ${\mu }_p$ fits into the Kummer sequencewhich is surjective on the right because $p$th-root polynomials are separable. Restrict it to the absolute Galois group of any algebraic extension $L/K$. The long exact sequence gives the Kummer cohomology divisibility criterionContinuous cohomology of a profinite group in positive degree with discrete coefficients is torsion: a continuous normalized cocycle becomes zero on a sufficiently small open subgroup, and restriction followed by corestriction multiplies its class by the finite index. For an abelian group, p-primary torsion consists of elements killed by powers of $p$, while p-divisibility means that multiplication by $p$ is surjective. For a torsion group $A$, $A\{p\}=0$ if and only if $A[p]=0$, because a nonzero element of $p$-power order has a multiple of exact order $p$. Consequently the two conditions on the multiplicative cohomology groups are jointly equivalent toThe coefficient module $K_s^{\times }$ is used with its restricted Galois action as in the question. Purely inseparable algebraic extensions give equivalent absolute Galois groups; the ${\mu }_p$ reduction therefore also covers those extensions.

The cohomological dimension at a prime condition ${\mathrm{cd}}_p(K)\le n$ means that $H^q(G_K,M)=0$ for every discrete $p$-primary torsion Galois module $M$ and every $q>n$. Cohomological dimension does not increase on a closed subgroup. Since absolute Galois groups of algebraic extensions identify with closed subgroups, up to the purely inseparable identification just noted, condition (i) gives $H^{n+1}(L,{\mu }_p)=0$ for every algebraic $L/K$. The Kummer reduction proves (i) implies (ii).

All geometric 2-torsion is rational: its points are $O,(\alpha ,0),(\beta ,0),(\gamma ,0)$. We will construct the required finite quotient directly, without assuming any theorem about finite generation or two-descent on an elliptic curve.

Choose $Q\in E(\overline{K})$ with $2Q=P$ and define the halving cocycle with rational two-torsionSuch a half exists because the nonconstant multiplication morphism $[2]$ on a smooth projective elliptic curve is surjective over an algebraic closure. Since $E[2]$ is rational, $c_P(\sigma \tau )=c_P(\sigma )+c_P(\tau )$. A different half is $Q+T$ with rational $T\in E[2]$ and gives the same cocycle. Replacing $P$ by $P+2R$, with $R\in E(K)$, allows the half $Q+R$ and again changes nothing. Adding chosen halves shows additivity in $P$. If the cocycle is zero, the half is fixed by the absolute Galois group and lies in $E(K)$, so $P\in 2E(K)$. This proves, rather than assumes, an injectionThe fixed field of $\ker c_P$ is exactly $K(Q)$. Its Galois group is the image of $c_P$, so its degree is at most four. This also follows from the permitted biquadratic description.

Let $S$ be the finite set of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At a place $v\notin S$ the displayed integral cubic has unit elliptic-curve discriminant, hence good reduction of an elliptic curve, and the residue characteristic is odd. Choose a half $\overline{Q}$ of the reduction of $P$ over the algebraic closure of the residue field. Its coordinates belong to a finite residue extension; let $L_v/K_v$ be the corresponding finite unramified extension. Smoothness and the Hensel lemma lift $\overline{Q}$ to some $Q_0\in E(L_v)$. The difference $P-2Q_0$ reduces to zero and belongs to the formal kernel of a minimal Weierstrass equation. In its integral formal group law the doubling series is $2t+O(t^2)$, with unit linear coefficient. The prime-to-residue-characteristic multiplication on a formal group gives a unique $R$ in this kernel such that $2R=P-2Q_0$. Thus $Q_0+R$ is a half of $P$ defined over an unramified local extension. Every other half differs by rational 2-torsion, so it too is unramified. We have proved the unramified halving fields for a split cubic assertion at every $v\notin S$.

There are only finitely many bounded-degree extensions with restricted ramification of $K$ of degree at most four. This is the permitted number-field finiteness consequence of the Hermite–Minkowski theorem: bounded local degrees bound the discriminant exponents at the finitely many allowed ramified primes, so absolute discriminants are bounded. Each possible halving field $L$ is Galois, and there are only finitely many homomorphisms $\mathrm{Gal}(L/K)\to E[2]$. The injection constructed above therefore has finite image. Consequently $E(K)/2E(K)$ is finite. No elliptic-curve finiteness or descent theorem was used in this argument.

For $K=\mathbb{Q}$, write $s=\#S$. Each coordinate of $c_P$ in a basis of $E[2]\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ is a quadratic Galois character unramified outside $S$. Its quadratic extension has a signed square-free integer representative $d=\pm {\prod }_{p\in S}{p}^{e_p}$, $e_p\in \{0,1\}$: an odd valuation outside $S$ would ramify there. There are at most $2^{s+1}$ choices, including the trivial character. Hence the pair of characters gives at most $2^{2s+2}$ cocycles. Each prime in $S$ is at most $2m$, because the three root differences have absolute values at most $2m$ and $2\le 2m$. Among the integers up to $2m$ there are at most $m$ primes: two, together with at most $m-1$ odd candidates. The rational halving quotient bound from root size is therefore

Galois cohomology records the obstruction to choosing Galois-invariant division points. Let $G_K=\mathrm{Gal}(\overline{K}/K)$ be the absolute Galois group and give its algebraic-point modules the discrete topology. A continuous $1$-cocycle with values in a Galois module $M$ is a function $c:G_K\to M$ satisfyingA group coboundary has the form $c_{\sigma }=\sigma a-a$. Quotienting cocycles by coboundaries defines $H^1(K,M)$. For the finite module $E[m]$ these cocycles have finite image and factor through finite data; continuity is essential because $G_K$ is a profinite group.

In characteristic zero, multiplication by $m\ge 2$ on the algebraic points of an elliptic curve is surjective, with kernel $E[m]\cong (\mathbb{Z}/m\mathbb{Z}{)}^2$. The short exact sequence of Galois modulesgives the Kummer exact sequence of an elliptic curveExplicitly, choose $Q$ with $mQ=P$ and set $\delta (P{)}_{\sigma }=\sigma Q-Q$. Changing $Q$ by an $m$-torsion point changes the cocycle by a coboundary. The class vanishes exactly when a suitable choice of $Q$ is Galois-fixed, that is, when $P\in mE(K)$. This proves the injection directly and identifies its arithmetic meaning.

The entire group $H^1(K,E[m])$ need not be finite. For example, Kummer theory gives $H^1(K,{\mu }_m)=K^{\ast }/K^{\ast m}$, which has classes supported on arbitrarily many different primes. The finite part needed for the Weak Mordell-Weil theorem comes from a ramification restriction on the image of $\delta$.

Choose a finite set $S$ of places containing the archimedean places, the primes dividing $m$ and all primes of bad reduction of an elliptic curve. At a finite place $v\notin S$, the curve has good reduction and $m$ is invertible in its valuation ring. The elliptic curve extends to a smooth proper group scheme, and $[m]$ on that model is finite etale. A point $P\in E(K_v)$ extends to an integral section by properness. Its division-point fibre is consequently finite etale over the valuation ring. Over the maximal unramified extension it has a point, so the Kummer cocycle restricts trivially to the inertia group. The module $E[m]$ itself is unramified there for the same reason. Thuswhere $H_S^1$ denotes classes unramified outside $S$.

Here is a proof that this restricted group is finite. Take a finite Galois extension $L/K$ containing all coordinates of $E[m]$ and the $m$th roots of unity, and enlarge $S$ by its ramified primes. Over $L$ the module is trivial and, after choosing a basis, is isomorphic to ${\mu }_m^2$. Hilbert theorem 90 and the multiplicative Kummer sequence therefore giveFor unramified Kummer classes with bounded prime support, a class unramified outside $S_L$ has valuations divisible by $m$ at every prime outside $S_L$: the valuation of an $m$th root in an unramified extension is integral. Consequently both coordinates lie in

To show this set finite, write the outside-$S_L$ divisor of $a$ as $mD$. The ideal class of $D$ belongs to the $m$-torsion of the ideal class group of the ring of $S_L$-integers. This gives the exact sequenceThe last map is onto: if $mD$ is principal in the $S_L$-ideal group, a generator represents a class with outside valuations divisible by $m$. The kernel consists exactly of S-units modulo $m$th powers. The Dirichlet unit theorem, with the finitely many inverted primes adjoined, makes the S-unit group finitely generated; the ideal class group is finite, and localization only quotients it. Both ends of the sequence are therefore finite.

Finally, the inflation-restriction exact sequence bounds the kernel of restriction from $H^1(K,E[m])$ to $H^1(L,E[m])$ by the finite group $H^1(\mathrm{Gal}(L/K),E[m])$. The unramified subgroup has finite image, contained in $L(S_L,m{)}^2$, and finite kernel, so it is finite. The Kummer injection now provesThis is the Weak Mordell-Weil theorem, proved without first assuming finite generation of $E(K)$.

For computation one refines the unramified group by local solvability. The Selmer group of an elliptic curve consists of classes whose restriction at every completion lies in the corresponding local Kummer image. It is finite and sits inwhere the Tate–Shafarevich group $\mathrm{Sha}$ measures classes in $H^1(K,E)$ that become trivial at every completion. Thus locally soluble descent equations can give an upper bound without every class coming from a rational point. No finiteness assumption on the whole Tate-Shafarevich group is needed for the weak theorem. Combining the finite quotient with the height descent lemma from the height essay yields the full Mordell-Weil theorem. Replacing multiplication by $m$ with a smaller isogeny gives the same cohomological framework for two-isogeny descent.

For an abelian variety $A$ and a prime number $\ell$ different from the characteristic, the inverse limit of its geometric ${\ell }^n$-torsion, with transition multiplication by $\ell$, is its Tate module. For an elliptic curve, it is free of rank two over the p-adic integers and satisfies $T_{\ell }(E)/{\ell }^n{T}_{\ell }(E)\simeq E[{\ell }^n]$. The absolute Galois group acts continuously and commutes with all base-defined endomorphisms.