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For a left $\mathbb{Z}G$-module, group cohomology iswith $\mathbb{Z}$ the trivial module. Equivalently take a projective resolution $P_{\bullet }\to \mathbb{Z}$ and the cohomology of ${\mathrm{Hom}}_{\mathbb{Z}G}(P_{\bullet },M)$. This also describes the right-derived functors of the invariant-module functor $M\mapsto M^G$.

For an explicit model, the homogeneous bar free resolution has $P_n=\mathbb{Z}[G^{n+1}]$, with diagonal $G$-action and boundary the alternating sum of coordinate deletions. It is free over $\mathbb{Z}G$: each orbit has a unique representative with first coordinate $1$. The augmented complex is exact because inserting $1$ as a first coordinate supplies a contracting homotopy of underlying abelian groups; its lack of $G$-equivariance does not affect exactness. The associated inhomogeneous group cochains are functions $f:G^n\to M$, withFor group cochains of degree zero, $(dm)(g)=gm-m$. Adjacent deletions cancel in pairs, giving $d^2=0$ and hence $H^n=\ker d/\mathrm{im}d$.

In degree zero this givesIt is also ${\mathrm{Hom}}_{\mathbb{Z}G}(\mathbb{Z},M)$, since a map is determined by the invariant image of $1$.

In degree one the cocycle condition is $f(gh)=f(g)+gf(h)$, soThese are crossed homomorphisms modulo principal crossed homomorphisms. For trivial action they reduce to ordinary group homomorphisms, so $H^1(G,M)=\mathrm{Hom}(G_{\mathrm{ab}},M)$.

There are two further useful descriptions. If $I=\ker (\mathbb{Z}G\to \mathbb{Z})$ is the augmentation ideal, a crossed homomorphism gives an equivariant map $I\to M$ by $g-1\mapsto f(g)$, because $gh-1=(g-1)+g(h-1)$. Applying $\mathrm{Hom}(-,M)$ to $0\to I\to \mathbb{Z}G\to \mathbb{Z}\to 0$ therefore givesAlso a section of $M⋊G\to G$ has the form $g\mapsto (f(g),g)$ and is a group homomorphism exactly when $f$ is a crossed homomorphism. Conjugation of sections by an element of $M$ changes $f$ by a principal crossed homomorphism. Thus $H^1$ classifies the $M$-conjugacy classes of such sections, or equivalently of complements to $M$ in the semidirect product.

In degree two, a normalized two-cocycle $c:G\times G\to M$ satisfiesDefine multiplication on $M\times G$ byThe displayed cocycle identity is precisely associativity; normalization gives the identity and inverses exist. This produces a group extension $1\to M\to E_c\to G\to 1$ with the prescribed action on $M$.

Conversely a set-theoretic section $s:G\to E$, $s(1)=1$, gives $c(g,h)=s(g)s(h)s(gh{)}^{-1}$ in the additive kernel of a group homomorphism. Associativity gives the cocycle condition. Replacing $s(g)$ by $b(g)s(g)$ changes $c$ byThe resulting extensions are equivalent through a change of coordinates in $M\times G$. Conversely any extension equivalence preserving both kernel of a group homomorphism and quotient yields such a change of section. HenceThe zero class corresponds exactly to a split extension. With trivial action these are central group extensions; addition of classes corresponds to the Baer sum. This proves the second group cohomology classifies group extensions interpretation.

For $G={\mathbb{Z}}^2$ and trivial coefficients $\mathbb{Z}$, put $R=\mathbb{Z}[x^{\pm 1},y^{\pm 1}]$. The Koszul resolution for a rank-two free abelian group iswhere $d_1(a,b)=(x-1)a+(y-1)b$ and $d_2(c)=(-(y-1)c,(x-1)c)$. To check exactness at $R^2$, reduce $d_1(a,b)=0$ modulo $x-1$. Since $y-1$ is not a zero divisor in $\mathbb{Z}[y^{\pm 1}]$, write $b=(x-1)c$, and cancellation gives $a=-(y-1)c$. Injectivity of $d_2$ follows because $R$ is a domain. The augmentation kernel is generated by $x-1,y-1$, proving the other exactness statement.

Applying ${\mathrm{Hom}}_R(-,\mathbb{Z})$ makes both differentials zero, since $x,y$ act trivially. ThusDegree-one classes assign arbitrary integers to the two generators. Geometrically these are also the cohomology groups of the torus, a classifying space for ${\mathbb{Z}}^2$; its cohomology ring is the exterior algebra on the two degree-one classes.

The integer $n$ in degree two has an explicit extension representative. On triples $(a,b,m)\in {\mathbb{Z}}^3$, setThe bilinear term is a two-cocycle. With generators $x=(1,0,0)$, $y=(0,1,0)$ and central $z=(0,0,1)$,Every central extension of ${\mathbb{Z}}^2$ by $\mathbb{Z}$ has this form: choose lifts of the two generators and the fixed generator of the kernel of a group homomorphism, then their group commutator determines $n$ and every element has a unique form $z^m{y}^b{x}^a$. Changing the lifts by central elements leaves that commutator unchanged. Thus distinct integers give inequivalent extensions, and the cocycles $nad$ add with $n$. The case $n=0$ is ${\mathbb{Z}}^3$; $n=1$ is the integer Heisenberg group. This realizes the integral central extensions of a rank-two free abelian group concretely.

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.

A crossed homomorphism is a map $\varphi :G\to M$ satisfyingFor a one-cochain, the formula in part i givesso the crossed homomorphisms are exactly the one-cocycles. A zero-cochain $m\in M$ has coboundary $g\mapsto gm-m$, the principal crossed homomorphism associated with $m$. Therefore

Write every element as $a^k{b}^l$ with $k\in \{0,1\}$. The action of this element on $M=\mathbb{Z}$ is multiplication by $(-1{)}^k$. A direct check of the four possibilities for the two exponents of $a$ showsso $\varphi$ is a crossed homomorphism and hence a one-cocycle.

It cannot be principal: if $\varphi (g)=gm-m$ for some $m\in \mathbb{Z}$, then at $g=a$ one would havewhich is impossible in the integers. Thus $[\varphi ]\ne 0$ and