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Coverings surviving all local solubility tests need not have rational points. The quotient of an isogeny Selmer group by the rational-point image is a kernel in the Tate–Shafarevich group. A nonzero class therefore passes every local test but contributes no rational point to a search. This explains why two-isogeny descent can leave a gap between a certified upper bound and found lower bound for the rank of an elliptic curve. Higher descents address finer obstructions; see Tom Fisher's paper on higher descents with rational two-torsion.

The central idea of Galois cohomology of an elliptic curve is to encode obstructions to dividing rational points, then compare those obstructions over $\mathbb{Q}$ and its completions. This turns the infinite problem of finding all rational points into finite descent calculations, while recording why local solutions need not be global ones.

Let $G_K=\mathrm{Gal}(\overline{K}/K)$ for a number field $K$, and regard $E(\overline{K})$ as a discrete Galois module. Then $H^0(K,E)=E(K)$. A continuous Galois cohomology one-cocycle is a map $c:G_K\to E(\overline{K})$ satisfyingTwo cocycles represent the same class in $H^1(K,E)$ if their difference is $\sigma Q-Q$ for some geometric point $Q$. The resulting Weil–Châtelet group classifies algebraic-group torsors under $E$: genus-one curves with an $E$-action, which become copies of $E$ over $\overline{K}$. Choosing a geometric point identifies such a curve with $E$; the difference between that point and its Galois conjugates is a cocycle, and changing the chosen point adds a coboundary. A torsor class is zero exactly when the curve has a $K$-point. Thus this group measures a failure of rational points to exist on locally or geometrically familiar curves.

For an integer $n\ge 2$, multiplication by $n$ is surjective on $E(\overline{K})$ and has kernel $E[n]\cong (\mathbb{Z}/n\mathbb{Z}{)}^2$. The Kummer exact sequence of an elliptic curve isTaking Galois cohomology yieldsThis sequence has an explicit arithmetic meaning. Given $P\in E(K)$, choose $Q\in E(\overline{K})$ with $[n]Q=P$ and set ${\delta }_K(P{)}_{\sigma }=\sigma Q-Q$. Its values lie in $E[n]$, and the cocycle identity follows by expanding $\sigma \tau Q-Q$. Choosing another $Q$ changes it by a coboundary in $E[n]$. Replacing $P$ by $P+[n]R$ with $R\in E(K)$ does not change the class. Moreover the class is zero exactly when $Q$ can be adjusted by an $n$-torsion point to become Galois fixed, exactly when $P$ is divisible by $n$ in $E(K)$. Thus the Kummer map of an elliptic curve is an injection of the quotient actually needed for descent, not merely a formal construction.

For each place $v$ of $K$, repeat this construction over the completion $K_v$. Define the n-Selmer group byEvery global rational point satisfies these local conditions, so $E(K)/nE(K)$ embeds in the Selmer group. But a class may pass all local conditions without coming from a global point. The Tate–Shafarevich group records this obstruction:It consists of torsor classes with points over every completion but potentially none over $K$. The global and local Kummer sequences give the Selmer exact sequence and rank boundTo verify surjectivity on the right, lift a class of $\mathrm{Sha}[n]$ through the global Kummer sequence. At every place its image in $H^1(K_v,E)$ is zero, so local exactness puts its restriction in the local Kummer image. The lift is therefore Selmer. The kernel on the left is the global Kummer image by global exactness. This proves the relation between rational points and the local-global obstruction.

Finiteness is what makes this framework useful. Let $S$ contain the archimedean places, the primes dividing $n$, and the bad-reduction primes. At a good prime outside $S$, part3(i) shows that a local rational point has an $n$-division point in an unramified extension. Its Kummer class is therefore unramified. Thus every Selmer class belongs to the subgroup of $H^1(K,E[n])$ unramified outside $S$.

Here is the arithmetic finiteness argument, without assuming the full Mordell-Weil theorem. Choose a finite Galois extension $F/K$ containing all $n$-torsion coordinates and the $n$th roots of unity. Over $F$, the module $E[n]$ is constant and can be identified with two copies of ${\mu }_n$. Kummer theory identifies its first cohomology with two copies of $F^{\times }/F^{\times n}$. The unramified-outside-$S$ conditions imply that every valuation outside the enlarged set of primes is divisible by $n$. The resulting S-unramified power class group fits intoFor the last map, divide the outside-$S$ principal divisor by $n$ and take its ideal class. Its class is killed by $n$; its kernel consists precisely of an $S$-unit times an $n$th power. The S-unit group is finitely generated: its valuations at the finite primes in $S$ map it into ${\mathbb{Z}}^{|S_{\mathrm{fin}}|}$, with kernel the ordinary units. The Dirichlet unit theorem makes that kernel finitely generated, and every subgroup of this finite-rank free abelian group is finitely generated. Thus the first term is finite. Finiteness of the ideal class group makes the last term finite. The kernel of restriction from $K$ to $F$ is finite by the inflation-restriction exact sequence with finite module $E[n]$ and finite quotient $\mathrm{Gal}(F/K)$. Hence the Selmer group is finite, as is $E(K)/nE(K)$. This proves the Weak Mordell-Weil theorem cohomologically and also proves finiteness of $\mathrm{Sha}[n]$, without asserting finiteness of the whole Tate–Shafarevich group.

For $K=\mathbb{Q}$, combine this finite quotient with a height descent. Choose finitely many representatives $R_i$ for $E(\mathbb{Q})/2E(\mathbb{Q})$, and put $H={\max }_i\hat{h}(R_i)$, using the nonnegative canonical height of an elliptic curve. Its height parallelogram identity gives $\hat{h}(P-R_i)\le 2\hat{h}(P)+2\hat{h}(R_i)$. Thus if $P=2Q+R_i$,Iteration eventually brings the height below $H+1$. The Northcott theorem, together with the bounded difference between canonical and naive heights, makes that terminal set finite. Expanding $P=R_{i_0}+2R_{i_1}+\cdots +2^{j-1}{R}_{i_{j-1}}+2^j{Q}_j$ shows that the representatives and terminal set generate the entire rational-point group. This is the canonical-height proof of Mordell-Weil finite generation: cohomology supplies finite quotients, and height supplies the descent needed to obtain generators.

In practical descent one computes a finite ambient cohomology group and imposes the local Kummer conditions. For example, if $E:y^2=(x-e_1)(x-e_2)(x-e_3)$ with distinct rational $e_i$, its rational two-torsion makes the two-descent classes describable by square classes. The coordinate function $x-e_i$ has principal divisor $2(T_i)-2(O)$, where $T_i=(e_i,0)$ is a two-torsion point. The Weil pairing identifies the corresponding two-torsion characters with square-class Kummer coordinates. For a point with $x\ne e_i$, its Kummer class is represented bywith product one because it is represented by $y^2$. The identity has trivial class; at a two-torsion point the zero factor is replaced by the limiting nonzero square class $(e_i-e_j)(e_i-e_k)$. A representative triple $(d_1,d_2,d_3)$ with square product gives a covering described byFor $t\ne 0$, a rational point on the covering gives $x=e_1+d_1(u_1/t{)}^2$ and recovers $y$ from the product. Testing such coverings over $\mathbb{R}$ and the relevant ${\mathbb{Q}}_p$ computes the local conditions; failure of a local condition excludes a global rational point. Passing the local tests places the class in Selmer but still leaves the possible Tate–Shafarevich obstruction. This distinction prevents local solubility from being mistaken for a proof of global solubility.

Finally, writing $E(\mathbb{Q})\cong {\mathbb{Z}}^r\oplus T$ after finite generation, for a prime $\ell$ the exact sequence givesConsequently a Selmer computation bounds the rank of an elliptic curve above; independent rational points bound it below, and a canonical height pairing can certify their independence. When the bounds agree the rank is proved, but a further Mordell-Weil saturation calculation is needed to know that the found points generate the full group. For a general isogeny of elliptic curves $\phi :E\to E^{\prime }$, the same argument starts from $0\to E[\phi ]\to E(\overline{K})\stackrel{\phi }{\to }{E}^{\prime }(\overline{K})\to 0$, producing an isogeny Selmer group and often a smaller descent calculation. Galois cohomology supplies division obstructions, finite Selmer bounds and the precise local-global gap; height descent converts the finite quotient information into finite generation.

Move the nonidentity rational 2-torsion point to $T=(0,0)$ and complete the square. After a rational change of coordinates, the elliptic curve has an integral modelThe two-isogeny formula gives a partnerand the isogeny of elliptic curves and its dual isogeny areBoth extend at the exceptional points; their kernels are the identities and the distinguished 2-torsion points, and $\hat{\varphi }\varphi =[2]$.

The two-torsion square-class homomorphism is $\alpha (O)=1$, $\alpha (T)=[b]$, and $\alpha (x,y)=[x]$ otherwise, with values in ${\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$. Its kernel is $\hat{\varphi }{E}^{\prime }(\mathbb{Q})$. Define ${\alpha }^{\prime }$ on the partner in the same way. By the prime-support bound in two-isogeny descent, representatives are signed squarefree divisors $d$ of $b$, or of $b^{\prime }$ for the partner. A class is realized precisely when its quartic covering in a two-isogeny descent has a rational solution:For $UV\ne 0$, this comes from $x=d(U/V{)}^2$ and $y=dUN/V^3$; the cases $U=0$ or $V=0$ account for the distinguished point and the identity. Test this finite list using signs and local congruences, and exhibit rational solutions for the surviving classes. The local-to-global gap in isogeny descent matters: local solubility gives an upper bound from the isogeny Selmer group, so a rank determination requires matching actual classes or further descent, not merely declaring every locally soluble cover to have a rational point.

Once the two actual images are determined, the two-isogeny rank formula isFor completeness, put $A=E(\mathbb{Q})$ and $B=E^{\prime }(\mathbb{Q})$. The Mordell-Weil theorem gives $[A:2A]=2^r|A[2]|$. The index of a subgroup formula through $\hat{\varphi }B$ givesHere $\ker \hat{\varphi }\cap \varphi A=\varphi (A[2])$, since $\hat{\varphi }\varphi =[2]$, and this image has size $|A[2]|/2$. The denominator is therefore $4/|A[2]|$, proving the displayed rank of an elliptic curve formula.

The Mordell-Weil theorem gives $E(\mathbb{Q})\cong E(\mathbb{Q}{)}_{\mathrm{tors}}\oplus {\mathbb{Z}}^r$. Move the rational point of order two to $(0,0)$ and clear denominators to obtain an integral Weierstrass equation of an elliptic curveThe two-isogeny formula gives $E^{\prime }:Y^2=X(X^2-2aX+a^2-4b)$ and dual isogenies $\varphi :E\to E^{\prime }$, $\hat{\varphi }:E^{\prime }\to E$ with $\hat{\varphi }\varphi =[2]$. Compute the torsion subgroup using reduction of torsion points on an elliptic curve at several good primes, followed by division polynomials of an elliptic curve to identify the possible points.

For the free part, use two-isogeny descent. The two-torsion square-class homomorphism isin ${\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$, with kernel $\hat{\varphi }{E}^{\prime }(\mathbb{Q})$; define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly. The prime-support bound in two-isogeny descent restricts the first image to signed square-free divisors of $b$ and the second to those of $a^2-4b$. For each candidate $d$, solve its two-isogeny descent quarticTest local solubility at the real place and the relevant p-adic fields, discard impossible classes, and search the surviving coverings for rational points. A point with $UV\ne 0$ gives $x=d(U/V{)}^2$, $y=dUN/V^3$. Include the identity and rational 2-torsion cases separately. Once both rational images are determined, the two-isogeny rank formula givesThe point searches give explicit candidates for a basis. Their canonical height pairing can certify independence; Mordell-Weil saturation and effective height bounds then certify that they generate the full free part. Rank alone does not certify generators of the whole Mordell-Weil group. If the rational images are not fully determined, the procedure instead supplies lower and upper bounds on the rank of an elliptic curve.

The rational point $P=(0,1)$ lies on $E$ and is neither the identity nor $(-1,0)$. By solution these are all the rational torsion points of an elliptic curve. Hence $P$ has infinite order and $E(\mathbb{Q})$ is infinite: its multiples $nP$ are distinct. No assertion about its being a generator or about the exact rank of an elliptic curve is needed here.

To compute the rank of an elliptic curve, translate the rational point of order two to $(0,0)$ and put the curve in the form $y^2=x^3+a{x}^2+bx$, with $a,b\in \mathbb{Z}$ after clearing denominators and $b(a^2-4b)\ne 0$. Its two-isogenous partner and the elliptic isogeny areThe maps extend over their exceptional points. The dual isogeny isThe two-torsion square-class homomorphism is $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\times }/{\mathbb{Q}}^{\times 2}$, defined by $\alpha (O)=1$, $\alpha ((0,0))=[b]$, and $\alpha ((x,y))=[x]$ otherwise. Define ${\alpha }^{\prime }$ similarly on $E^{\prime }$ with $b^{\prime }=a^2-4b$.

Here are the ingredients behind the descent procedure. For a generic chord $y=mx+c$, the cubic of intersection coordinates has root product $c^2$. As the third intersection is the negative of the group sum and negation leaves $x$ unchanged, this proves the square-class homomorphism property. For a line through $(0,0)$, the two nonzero intersection coordinates have product $b$, so the assigned class $[b]$ gives the same multiplicativity. A vertical line has two equal coordinates and contributes their square, and $O$ has identity class. These observations also cover tangencies and the exceptional sums. Its kernel is $\hat{\varphi }(E^{\prime }(\mathbb{Q}))$. Indeed, the displayed formula gives $x(\hat{\varphi }(X,Y))=(Y/(2X){)}^2$. Conversely, if a point on $E$ has $x=r^2\ne 0$, the possible preimage coordinates satisfyIts elliptic-curve discriminant is $16(r^4+a{r}^2+b)=16y^2/r^2$, a rational square. A root $X$ and $Y=2rX$, with the sign chosen appropriately, give a preimage. The identity is always in the image. The point $(0,0)$ is in the image exactly when $b$ is a square: its nonzero preimage coordinates are $X=a\pm 2\sqrt{b}$, with $Y=0$. Thus the exceptional points satisfy precisely the same kernel statement. Interchanging the curves similarly gives $\ker {\alpha }^{\prime }=\varphi (E(\mathbb{Q}))$.

A prime not dividing $b$ can occur only with even valuation in an $x$-coordinate. If its valuation is positive, the factor $x^2+ax+b$ is a unit and $2v(y)=v(x)$; if negative, $x^2$ dominates that factor and $2v(y)=3v(x)$. Thus square classes in $\mathrm{im}\alpha$ have signed square-free representatives dividing $b$, giving finitely many candidates. To test a candidate $d$, put $x=d(u/v{)}^2$ and $y=duN/v^3$. The curve equation becomesA nontrivial rational solution, including the appropriate limiting points, is exactly a realization of the class. Clear denominators and arrange $\gcd (u,v)=1$ for congruence tests. Do the same for $E^{\prime }$ with $b^{\prime }$.

The factor in the two-isogeny rank formula can also be checked directly. Let $A=E(\mathbb{Q})$, $B=E^{\prime }(\mathbb{Q})$ and $t=\#A[2]$. The index of $\hat{\varphi }(B)$ in $A$ is $\#\mathrm{im}\alpha$, while the map from $B/\varphi (A)$ onto $\hat{\varphi }(B)/2A$ has kernelThe numerator has order two. Moreover $\ker \hat{\varphi }\cap \varphi (A)=\varphi (A[2])$, since $\hat{\varphi }\varphi P=2P$, and the latter has order $t/2$. Hence this quotient has order $4/t$, andThe Mordell-Weil theorem gives $[A:2A]=2^{r}t$, where $r$ is the rank. Cancelling $t$ yieldsIn practice, local congruence and real-solubility tests remove many candidate quartics, giving an upper bound, while rational point searches realize classes and give a lower bound. These often coincide. Local solubility alone is not a proof of a rational point: an everywhere locally soluble covering can still fail globally, so a remaining gap must be reported rather than treated as an exact rank.

For the curve at hand, $a=8$, $b=-7$, $b^{\prime }=92$, andThe first image is contained in $\{1,-1,7,-7\}$ and contains $1,-7$, the latter from the point $(0,0)$. To exclude $-1$, its quartic would require coprime $u,v$ withModulo $16$, an odd fourth power is $1$ and an even fourth power is $0$. If $u$ is odd and $v$ even the right side is $15$; if $u$ is even and $v$ odd it is $7$; if both are odd it is $14$. None is among the square residues $0,1,4,9$. Thus $-1$ is impossible. The image is a subgroup, so $7$ is also impossible: its product with the known class $-7$ would be $-1$. We obtain $\mathrm{im}\alpha =\{1,-7\}$.

For $E^{\prime }$, its quadratic factor is $X^2-16X+92=(X-8{)}^2+28>0$. Every nonzero rational $X$-coordinate is therefore positive. Prime support leaves only $1,2,23,46$ as possible classes; $1$ and $23=[92]$ are supplied by $O$ and the two-torsion point. The explicit point $(18,48)$ is on $E^{\prime }$ becauseIts class is $[18]=2$. Multiplying classes supplies $46$, so $\mathrm{im}{\alpha }^{\prime }=\{1,2,23,46\}$. The two-isogeny rank formula now provesAs an additional concrete point on the original curve, the dual isogeny sends $(18,48)$ to $(16/9,116/27)$. The descent calculation proves the rank; it does not by itself certify that this point generates the full free quotient.

The two-torsion square-class homomorphism on $E$ has candidate classes $\{1,-1,17,-17\}$ by the prime-support bound in two-isogeny descent. The identity, $(0,0)$ and $(-1,4)$ realize $1,-17,-1$, so multiplicativity gives all four. The two-isogeny formula companion $E^{\prime }:y^2=x^3+68x$ has only positive candidate classes $\{1,2,17,34\}$; $(0,0)$ realizes $[68]=17$ and $(2,12)$ realizes $2$. Thus both images have order four, and the two-isogeny rank formula gives $2^r=4\cdot 4/4=4$. The rank of an elliptic curve is therefore two, without any unresolved covering class.