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We bound the torsion subgroups by reduction of torsion points on an elliptic curve. At a prime $p$ of good reduction, reduction is injective on the prime-to-$p$ torsion. Indeed, a point in the kernel is represented by the formal group of an elliptic curve on the maximal ideal. Multiplication by an integer prime to $p$ has a unit linear coefficient and is invertible there, so the kernel has no nonzero torsion of such order.

Both curves have good reduction at $2$ and $5$. For $E$ this follows from $\Delta =-91^3$, which is a unit at both primes. For $E^{\prime }$ at $2$, its reduction is $x^3+y^3+z^3=0$ and its three partial derivatives cannot vanish at a projective point. At $5$, the singularity calculation in part (i) would require $351-8=343$ to vanish modulo $5$, which it does not; $3$ and $13$ are also units there.

Direct counting gives:

For $E$ over ${\mathbb{F}}_5$, the numbers of ordinate solutions at $x=0,1,2,3,4$ are $2,2,2,2,0$. For $E^{\prime }$ in the chart $z=1$, they are $1,2,0,3,2$, with the single infinity point $(1:-1:0)$. These give the displayed totals without identifying the finite-field groups through the given isogeny of elliptic curves.

For either rational torsion subgroup, every odd-primary part injects at $2$, so the whole odd torsion has order dividing three. The two-primary part injects at $5$ and must be trivial, since the reduced group has odd order nine. Thus either rational torsion subgroup has order at most three. For $E$, the point $P$ already has order three, soFor the plane cubic $E^{\prime }$, the chosen identity $O^{\prime }$ is a plane cubic flex. The fact that three-torsion points are flexes of a plane cubic therefore identifies $E^{\prime }[3]$ with its nine geometric flexes. Part (i) found only the identity among the rational flexes. No nonidentity rational three-torsion exists, and the preceding bound excludes every other order. HenceThis also illustrates that isogenous elliptic curves can have different rational point groups; the given degree-three isogeny does not imply equality of rational torsion.

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.

Choose the real embeddings ${\sigma }_i$ and one from each complex pair ${\tau }_j$. The logarithmic embedding of number field units isIt is a group homomorphism from multiplication to addition. A unit has absolute norm one, so the coordinate sum is zero and its image lies in a Euclidean hyperplane of dimension $r+s-1$.

Here is the required discreteness argument. There are only finitely many algebraic integers of $K$ whose conjugate absolute values are bounded by a fixed constant: their monic minimal polynomials have degree at most $n$, and each integer coefficient is an elementary symmetric function of bounded roots, hence belongs to a fixed finite range. There are finitely many such polynomials and finitely many roots of each. A bounded set of logarithms bounds all conjugate absolute values, so every compact region in the logarithmic hyperplane meets the unit image in finitely many points. Thus the image is a discrete additive subgroup.

Use the allowed Euclidean-group theorem: a discrete subgroup of a finite-dimensional real vector space is free abelian on finitely many vectors, with rank at most the dimension. Therefore $\lambda ({\mathcal{O}}_K^{\times })$ has rank at most $r+s-1$. Its kernel is finite by the same bounded-conjugate argument, now with every absolute value one. Each element of a finite multiplicative group is a root of unity, and conversely every root of unity has all logarithms zero. Thus the kernel is exactly the finite group of roots of unity, and exactly the torsion subgroup of units.

Lift generators of the free logarithmic image to units; together with the finite kernel they generate every unit. HenceThis proves the requested upper bound without invoking the existence half of the Dirichlet unit theorem. The field $\mathbb{Q}(\sqrt{11})$ is a subfield of $\mathbb{R}$, and its only real roots of unity are $\boxed{1,-1}$.

The torus map acts on $H_1(T^2;\mathbb{Z})$ byIts determinant is one, so it is an orientation-preserving homeomorphism and acts as the identity on both $H_0(T^2)$ and $H_2(T^2)$.

To calculate the homology of its mapping torus, cover the time circle by two arcs. Each inverse image has the homotopy type of $T^2$, while their intersection has two components of that homotopy type. The map in the Mayer–Vietoris sequence can be writtenAdding the output coordinates and using the input coordinate $a+b$ separates an identity map from $f_{\ast }-I$. Canceling the identity summands gives the Wang sequence, and henceIn degree one,Its image and kernel are both generated by $(1,-1)$; its cokernel is $\mathbb{Z}$ since this image is primitive. Its Smith normal form is $\mathrm{diag}(1,0)$, so its torsion subgroup is zero. On the other two nonzero homology groups of the torus, $f_{\ast }-I=0$. The displayed short exact sequences split because their right-hand groups are free. ThereforeIn particular, the mapping torus contributes an additional degree-one class from the time circle, as well as the invariant degree-one class in the fiber.

The elliptic-curve discriminant of $y^2=x^3-12x^2+20x$ is $16\cdot 20^2\cdot (144-80)=409600=2^{14}{5}^2$. Thus three and eleven are primes of good reduction. Counting the reduced points gives four points over ${\mathbb{F}}_3$ and sixteen over ${\mathbb{F}}_{11}$. For example, the numbers of affine points above successive $x=0,1,2$ modulo three are $1,1,1$. Modulo eleven, for $x=0,\dots ,10$, they are $1,2,1,2,0,0,2,2,2,2,1$; add the point at infinity in each case.

The reduction of torsion points on an elliptic curve is injective on torsion prime to the residue characteristic. The three-primary rational torsion injects into the group of order sixteen at eleven, so is trivial; likewise the eleven-primary torsion injects into the group of order four at three. Every other primary component injects at both primes. Therefore the rational torsion subgroup has order dividing four. All four rational 2-torsion points are visible, so

The rational point $P=(1,3)$ lies on the curve. The tangent slope in the elliptic-curve addition formula is $(3-24+20)/6=-1/6$, giving $2P=(361/36,-323/216)$. The integral substitution $X=x-4$ changes the equation to $y^2=X^3-28X-48$; the doubled point has $X=217/36$. The Nagell–Lutz theorem says that a rational torsion point on this integral short equation has integral coordinates, so $2P$, and hence $P$, is nontorsion. An infinite-order point is $(1,3)$.

We prove the rank bound by two-isogeny descent. For $a=-12,b=20$, the two-isogeny formula gives the partnerand maps $\varphi :E\to E^{\prime }$, $\psi :E^{\prime }\to E$ with $\psi \varphi =[2]$. The two-torsion square-class homomorphism $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\ast }/({\mathbb{Q}}^{\ast }{)}^2$ is $\alpha (x,y)=[x]$ away from $O,(0,0)$, with values $1$ and $[20]=[5]$ at those exceptional points. Its kernel is $\psi E^{\prime }(\mathbb{Q})$. The corresponding map ${\alpha }^{\prime }$ on $E^{\prime }$ has kernel $\varphi E(\mathbb{Q})$.

The prime-support bound in two-isogeny descent can be checked directly here. For a prime $\ell \nmid 20$, if $v_{\ell }(x)>0$ then $x^2-12x+20$ is a unit, so $2v_{\ell }(y)=v_{\ell }(x)$ is even; if $v_{\ell }(x)<0$, the $x^3$ term has strictly least valuation and $2v_{\ell }(y)=3v_{\ell }(x)$ is even. Thus the square class of $x$ is supported only on two and five. Also $x(x-2)(x-10)\ge 0$ forces $x\in [0,2]\cup [10,\infty )$, so nonzero $x$ is positive. The possible classes are $1,2,5,10$, and the four torsion points provide all of them:Consequently $\#\alpha (E(\mathbb{Q}))=4$.

The same valuation argument on $E^{\prime }$ gives possible signed classes $1,-1,2,-2$, since its coefficient $b^{\prime }=64$ has only the prime two. The identity gives class $1$, and $(-4,8)\in E^{\prime }(\mathbb{Q})$ gives class $-1$. For $d=\pm 2$, the two-isogeny descent quartic iswith integral coprime $U,V$ after scaling a rational solution. If $U$ is odd, its right side is $2$ or $10$ modulo sixteen for $d=2$, and $14$ or $6$ for $d=-2$, according to the parity of $V$. None is a square. If $U=2u$ is even, $V$ is odd and the right side is $32(\pm u^4+3u^2{V}^2\pm V^4)$. The bracket is odd for either parity of $u$, so its 2-adic valuation is five, again impossible for a square. The signed divisor-two obstruction for an isogeny covering excludes both classes. Hence ${\alpha }^{\prime }(E^{\prime }(\mathbb{Q}))=\{1,-1\}$.

For completeness, the index correction in the two-isogeny index formula over a number field is one: $\ker \psi =\{O,(0,0)\}$ and $\varphi ((2,0))=(0,0)$, so the whole kernel lies in $\varphi E(\mathbb{Q})$. ThusThe Mordell-Weil theorem and the four rational 2-torsion points identify this index as $2^{r+2}$. Therefore $2^{r+2}=8$ andThis is the rank-one elliptic curve with roots zero two and ten. We have exhibited a point of infinite order but have not needed to prove that it generates the free part.

Call this simultaneous antipodal quotient of two spheres $M$. The simultaneous antipodal map acts freely on $S^2\times S^2$, so $M$ is a connected closed manifold of dimension four. Each antipodal factor has mapping degree $-1$; their product preserves orientation. Hence $M$ is orientable.

The product $S^2\times S^2$ is simply connected, so this double covering space is the universal cover. Consequentlyusing the abelianization of the fundamental group. The Euler characteristic under a finite covering gives $\chi (M)=4/2=2$. Its rational first Betti number is zero, and Poincare duality gives $b_3=b_1=0$ and $b_0=b_4=1$. Thus $b_2=0$ as well.

The universal coefficient theorem for cohomology now gives $H^1(M;\mathbb{Z})=0$ andHere $\mathrm{Hom}(H_2(M;\mathbb{Z}),\mathbb{Z})=0$ because $b_2=0$. Integral Poincare duality further gives $H^3(M;\mathbb{Z})\cong H_1(M;\mathbb{Z})=\mathbb{Z}/2$, and $H^4(M;\mathbb{Z})=\mathbb{Z}$.

Let $a,b,c$ generate these groups in degrees $2,3,4$ respectively. The only potentially nonzero positive-degree cup product is $a^2$. But $2a^2=(2a)a=0$, while $H^4(M;\mathbb{Z})$ has no nonzero torsion subgroup, so $a^2=0$. Every other positive product vanishes by dimension. Thuswith every product of positive-degree elements equal to zero. The degree-three torsion is essential: it would be lost by computing only rational cohomology.

We use two precise facts about the formal group of an elliptic curve. At a prime $p$ of good reduction, its kernel of reduction of an elliptic curve is identified by the uniformizer $t=-x/y$ with the group $\hat{E}(p{\mathbb{Z}}_p)$. For odd $p$ this group is torsion-free. To see the second fact, the integral invariant differential of a formal group law has the form $(1+{\sum }_{j\ge 1}{c}_j{T}^j)dT$, with $c_j\in {\mathbb{Z}}_p$. Integrating constructs the formal logarithmIt is a group homomorphism to the additive group. For $0\ne t\in p{\mathbb{Z}}_p$ and $j\ge 2$,when $p$ is odd. Hence the series converges and $v_p({\log }_F(t))=v_p(t)$, so it is injective. The target has no nonzero torsion. Therefore reduction is injective on the entire rational torsion subgroup at an odd prime of good reduction, including its $p$-primary part.

For the present elliptic curve, $\Delta =64D^6$, so every odd $p\nmid D$ is a prime of good reduction. If $p\equiv 3(\mathrm{mod}4)$, the Legendre symbol of $-1$ is $-1$. The values $x$ and $-x$ cancel in the sum of the Legendre symbols of $x^3-D^{2}x$, yieldingThe rational torsion subgroup injects into each of these groups, so it is finite and its order $M$ divides every such $p+1$.

For any odd prime $\ell$, the Dirichlet theorem on primes in arithmetic progressions supplies infinitely many $p$ with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$. Discard the finitely many dividing $D$. Since $\ell \nmid p+1$, it cannot divide $M$. Similarly choose $p\equiv 3(\mathrm{mod}8)$, again avoiding $D$; then $v_2(p+1)=2$, so $M\mid 4$.

There are already four rational 2-torsion points,which are distinct because a squarefree integer $D$ is nonzero. ThusIts order is four. In fact the argument works for every nonzero integer $D$; squarefreeness is not needed for this torsion conclusion.

For $p\nmid 2k$, the displayed integral equation has nonzero elliptic-curve discriminant modulo $p$, so it has good reduction. Let $\chi$ be the Legendre symbol, with $\chi (0)=0$. Counting the two, one, or zero possible ordinates over each $x$ givesIf $p\equiv 3(\mathrm{mod}4)$, then $\chi (-1)=-1$ and the terms for $x$ and $-x$ cancel. The term at zero vanishes. Thus the count is $p+1$.

For the converse, suppose $p\equiv 1(\mathrm{mod}4)$, and put $h=(p-1)/2$. The Euler criterion and the sum of powers over a finite field show that the Trace of Frobenius satisfiesIndeed, among the positive exponents in this polynomial, of degree $3(p-1)/2<2(p-1)$, only $p-1$ has a nonzero sum over ${\mathbb{F}}_p$, and that sum is $-1$. Since $(x^3+kx{)}^h=x^h(x^2+k{)}^h$, the coefficient formula for trace of Frobenius modulo p yieldsNeither $k$ nor the binomial coefficient vanishes modulo $p$. Hence $a_p\ne 0$, proving the point-count criterion for y squared equals x cubed plus k x:

We use the following formal logarithm fact, also useful for the next part. For a formal group law over ${\mathbb{Z}}_p$, its logarithm has coefficients $c_j/j$ with $c_j\in {\mathbb{Z}}_p$, obtained by integrating its integral invariant differential. If $p$ is odd and $t\in p{\mathbb{Z}}_p\setminus \{0\}$, every term of degree $j\ge 2$ has valuation strictly greater than $v_p(t)$, since $j{v}_p(t)-v_p(j)>v_p(t)$. Thus the logarithm converges and is nonzero at $t$. Its homomorphism into the additive characteristic-zero group proves torsion-freeness of the formal group over Qp for odd p. Consequently reduction injects the entire rational torsion subgroup at any odd prime of good reduction, including its $p$-primary torsion. In particular the torsion subgroup is finite, and its order divides $p+1$ at every good $p\equiv 3(\mathrm{mod}4)$.

By the Dirichlet theorem on primes in arithmetic progressions, choose a good prime $p\equiv 3(\mathrm{mod}8)$. Then $v_2(p+1)=2$, bounding the two-primary part by four. For any odd prime $\ell$ dividing the torsion order, the Chinese remainder theorem and the Dirichlet theorem on primes in arithmetic progressions give a good prime $p$ with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$, avoiding the finitely many primes dividing $2k$. But then $\ell \nmid p+1$, a contradiction. Therefore

The order four does occur. For $k=4$, the elliptic-curve addition formula gives $2(2,4)=(0,0)$, so $(2,4)$ has order four. More precisely, the only nonzero rational 2-torsion point when $k>0$ is $(0,0)$, so order four means a cyclic group with a rational half of this point. The duplication formula isSuch a half must have $x(P{)}^2=k$. Writing $k=s^2$, positivity of $y(P{)}^2=2kx(P)$ forces $x(P)=s>0$. Then $y(P{)}^2=2s^3$, so $2s$ is a rational square, equivalently $s=2t^2$ for an integer $t\ge 1$. Conversely $(2t^2,4t^3)$ is a half of $(0,0)$ when $k=4t^4$. The rational torsion on y squared equals x cubed plus positive k x is therefore cyclic of order four exactly for these $k$, and otherwise cyclic of order two.

Use the standard complex K-theory Thom class convention. For a rank-$r$ complex vector bundle, its K-theory Euler class is the pullback of the K-theory Thom class along the zero section, namelyHere ${\Lambda }^j$ denotes an exterior power, and a Hermitian metric identifies the dual bundle with the conjugate vector bundle. Choosing ${\lambda }_{-1}(E)$ instead changes the K-theory Euler class by the unit $(-1{)}^r[\det E]$, so its kernel and cokernel computations are unchanged.

For the disk bundle $D(E)$ and sphere bundle $S(E)$, the K-theory Thom class identifies $K^i(D(E),S(E))$ with $K^i(X)$. The exact sequence of this pair becomes the K-theory Gysin sequence of a sphere bundle:The groups are two-periodic by Bott periodicity.

View Real projective space as the circle bundle of ${\gamma }^{\otimes 2}$ over Complex projective space, where $\gamma \to {\mathbb{CP}}^n$ is the tautological bundle. Indeed, $S^{2n+1}/\{\pm 1\}={\mathbb{RP}}^{2n+1}$ maps to $S({\gamma }^{\otimes 2})$ by $[v]\mapsto v\otimes v$; in each fiber squaring identifies exactly $v$ and $-v$.

Use Complex K-theory of complex projective space in the formThe K-theory Euler class of ${\gamma }^{\otimes 2}$ is $1-[\gamma {]}^{-2}$. Multiplying it by the unit $-[\gamma {]}^2$ gives $1-[\gamma {]}^2=2t-t^2$. The K-theory Gysin sequence of a sphere bundle therefore givesUnder the projection $q:{\mathbb{RP}}^{2n+1}\to {\mathbb{CP}}^n$, the pullback vector bundle $q^{\ast }\gamma$ is the complexification of a real vector bundle $L$ of the tautological bundle of Real projective space: the fiber map sends $[v,z]$ to $zv$. Thus $t$ pulls back to $1-[L]$. Put $\mu =[L]-1$. The relation $L^{\otimes 2}\cong 1$ gives ${\mu }^2=-2\mu$, and $t^{n+1}=0$ gives $2^n\mu =0$. HenceFor completeness, these relations give precisely the abelian group $\mathbb{Z}\oplus \mathbb{Z}/2^n$. In the basis $t,\dots ,t^n$, the image of multiplication by $2t-t^2$ is generated by $2t-t^2,2t^2-t^3,\dots ,2t^n$; its relation matrix is triangular with determinant $2^n$, and the relations express every $t^j$ as $2^{j-1}t$. Thus there is exactly one cyclic torsion subgroup of order $2^n$.

If $n\ge 1$ and $a=a_0+a_{1}t+\cdots +a_n{t}^n$ is in the kernel, successive coefficients of $(2t-t^2)a$ give $2a_0=0$, $2a_1-a_0=0$, and so on, up to $2a_{n-1}-a_{n-2}=0$. Hence $a_0=\cdots =a_{n-1}=0$, with $a_n$ arbitrary. For $n=0$ the multiplier is zero on $R=\mathbb{Z}$. In all cases,

Finally let $E=L^{\oplus k}$ with $k\ge 1$, over $B={\mathbb{RP}}^{2n+1}$. Since $L^{\ast }\cong L$, multiplicativity of the K-theory Euler class givesOn $K^0(B)=\mathbb{Z}\oplus (\mathbb{Z}/2^n)\mu$, the image of multiplication by this K-theory Euler class is the subgroup generated by $2^{k-1}\mu$: multiplication by an integer produces every such multiple, while multiplication by $\mu$ only adds multiples of $2^k\mu$. ThereforeMultiplication by $\mu$ is zero on $K^{-1}(B)\cong \mathbb{Z}$, since $2^n\mu =0$ and this abelian group has no nonzero torsion elements. Thus the K-theory Gysin sequence of a sphere bundle gives a short exact sequenceIt splits as an exact sequence of abelian groups because its quotient is a free abelian group. ConsequentlyThe convention $\mathbb{Z}/1=0$ covers $n=0$ and $k=1$. For example, when $k=1$ the sphere bundle is $(S^{2n+1}\times S^1)/\{(v,z)\sim (-v,-z)\}$, a mapping torus of the antipodal map; its Topological K-theory has no nonzero torsion elements, agreeing with the formula. If $k=0$ is allowed, the rank-zero vector bundle has an empty unit sphere bundle, so $K^0(S(E))=0$ instead.

For an elliptic curve over a number field $K$, the Mordell-Weil theorem writes $E(K)$ as a finite torsion subgroup plus a free abelian group of finite rank of an abelian group $r$. This integer is the rank of the elliptic curve over $K$. It can increase under field extension. A two-isogeny descent supplies rank bounds from rational square-class images; the canonical height pairing can certify independence of rational points.

This elliptic curve has torsion subgroup $\{O,(0,0),(2,0),(10,0)\}\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ and the infinite-order point $(1,3)$. Good-reduction point counts at three and eleven are four and sixteen, which bound the torsion order by four using reduction of torsion points on an elliptic curve. Doubling $(1,3)$ gives $(361/36,-323/216)$; after the shift $X=x-4$ to $y^2=X^3-28X-48$, its nonintegral coordinate proves nontorsion by the Nagell–Lutz theorem. The two-isogeny formula gives the partner $Y^2=X^3+24X^2+64X$. The first two-torsion square-class homomorphism has image $\{1,2,5,10\}$, while the second has image $\{1,-1\}$ by the signed divisor-two obstruction for an isogeny covering. The two-isogeny rank formula therefore gives $2^r=4\cdot 2/4=2$, so the rank is one. The argument does not determine the index of the subgroup generated by $(1,3)$.

For nonzero integral $D$, the elliptic curve $E_D:y^2=x^3-D^{2}x$ has rational torsion precisely $O,(0,0),(D,0),(-D,0)$. At every good prime $p\equiv 3(\mathrm{mod}4)$, cancellation of the Legendre symbols at $x$ and $-x$ gives $\#E_D({\mathbb{F}}_p)=p+1$. The torsion-freeness of the formal group over Qp for odd p makes the rational torsion subgroup inject into this group. The Dirichlet theorem on primes in arithmetic progressions excludes odd factors of its order and bounds its two-primary part by four. The four displayed rational 2-torsion points then give equality.

For $d\in {\mathbb{Q}}^{\times }$ the rational torsion subgroup of $X^3+Y^3+d{Z}^3=0$ is cyclic of order $3$ if $d$ is a cube, cyclic of order $2$ if $d/2$ is a cube, and trivial otherwise.

To bound torsion, use the Weierstrass equation of an elliptic curve $y^2=x^3-432d^2$. At every good prime $\ell \equiv 2(\mathrm{mod}3)$, cubing is a bijection on ${\mathbb{F}}_{\ell }$, so the number of points is $\ell +1$. Reduction injects the prime-to-$\ell$ torsion, since multiplication by an integer prime to $\ell$ is invertible in the formal group of an elliptic curve. For any prime $q>3$, the Dirichlet theorem on primes in arithmetic progressions supplies a good $\ell \equiv 2(\mathrm{mod}3)$, $\ell \equiv 1(\mathrm{mod}q)$, excluding $q$-torsion. Good primes $\ell \equiv 5(\mathrm{mod}12)$ exclude order $4$, and good primes $\ell \equiv 11(\mathrm{mod}18)$ exclude order $9$. Thus every rational torsion point is killed by $6$.

The flexes of a diagonal cubic give rational three-torsion exactly when $d$ is a cube, and then it has order $3$. Rational two-torsion requires $x^3=432d^2$, whose sole possible rational root exists exactly when $d/2$ is a cube. This equivalence follows by comparing prime valuations modulo $3$, including the valuation at $2$. The two conditions cannot hold together because $2$ is not a rational cube. The asserted classification follows.

For positive integral $k$, the rational torsion subgroup of $y^2=x^3+kx$ has order dividing four. Indeed, torsion-freeness of the formal group over Qp for odd p, reduction of torsion points on an elliptic curve, and the point-count criterion for y squared equals x cubed plus k x bound its order by $p+1$ at all good primes $p\equiv 3(\mathrm{mod}4)$. The Dirichlet theorem on primes in arithmetic progressions excludes every odd torsion prime and bounds the two-primary part by four, by choosing $p\equiv 3(\mathrm{mod}8)$. There is exactly one nonzero rational 2-torsion point, namely $(0,0)$. A rational half of it must have $x^2=k$ by the elliptic-curve addition formula. Writing $k=s^2$ with $s>0$, its equation is $y^2=2s^3$, so $s=2t^2$ and $k=4t^4$. Conversely $(2t^2,4t^3)$ has order four. Thus the cyclic group of order four occurs exactly for $k=4t^4$, $t\ge 1$. The order bound itself holds for every nonzero integral $k$ of either sign: positivity was used only to classify the rational points of orders two and four.

On an integral Weierstrass equation of an elliptic curve with good reduction at a prime $p$, a nonidentity rational point reducing to the identity cannot have both coordinates integral at $p$. Indeed, an affine integral pair reduces to an affine point, distinct from the point at infinity. This can prove nonintegrality of a large linear combination without computing its coordinates. The nonidentity condition is essential and can often be checked using the known torsion subgroup.

For an elliptic curve over $\mathbb{Q}$, reduction at a good prime $p$ is injective on the $\ell$-primary torsion subgroup when $\ell \ne p$: its kernel is a formal group, where multiplication by $\ell$ is bijective. Hence a rational point of order $\ell$ forces $\ell \mid \#\tilde{E}({\mathbb{F}}_p)$. If for every possible prime $\ell$ some good $p\ne \ell$ has a point count not divisible by $\ell$, rational torsion is trivial. In particular two distinct good primes $p,r$ with coprime counts, $p\nmid \#\tilde{E}({\mathbb{F}}_r)$ and $r\nmid \#\tilde{E}({\mathbb{F}}_p)$ suffice: the coprime counts exclude other primes, and the two cross conditions exclude $p,r$. This argument does not incorrectly apply prime-to-$p$ injectivity to $p$-power torsion.

For odd $p$, a one-dimensional formal group law over ${\mathbb{Z}}_p$ has a formal logarithm ${\log }_F(T)=T+{\sum }_{j\ge 2}{c}_j{T}^j/j$ with $c_j\in {\mathbb{Z}}_p$. This follows by integrating its integral invariant differential. For $t\in p{\mathbb{Z}}_p\setminus \{0\}$, $v_p(c_j{t}^j/j)>v_p(t)$ for $j\ge 2$. Hence the series converges and $v_p({\log }_F(t))=v_p(t)$, so its kernel is zero. As it is a homomorphism into a characteristic-zero additive group, there is no nonzero torsion. In particular, at an odd prime of good reduction over $\mathbb{Q}$, reduction injects the whole rational torsion subgroup, including its $p$-primary part.