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Fix a prime number $p$. Take the full subcategory of finitely generated abelian groups with no element of order $p^2$. It is closed under finite direct sums and subgroups. If $t_p(G)$ is the $p$-primary torsion subgroup, the map $G\to R(G)=G/(p{t}_p(G))$ is a reflection into this subcategory: every map to an allowed group kills $p{t}_p(G)$. Therefore ordinary kernels in a category and reflected cokernels in a category supply all finite categorical limits and colimits. No nonzero finitely generated group has zero reflection, so the epimorphisms are precisely surjections and are all conormal epimorphisms. Multiplication by $p^2$ on $\mathbb{Z}$ is a monomorphism but not a normal monomorphism: its cokernel here is $\mathbb{Z}/p\mathbb{Z}$, whose kernel is $p\mathbb{Z}$, not $p^2\mathbb{Z}$. Multiplication by $p$ is normal; composing it with itself gives this nonnormal monomorphism.

Let $X$ be a finite CW complex, with $c_i$ topological cells of dimension $i$ and none above $d$. Its Euler characteristic is the displayed alternating sum. For a finite simplicial complex, $c_i$ counts its $i$-dimensional simplexes, each counted once rather than once for each containing simplex. In dimension two this becomes $V-E+F$. A topological cell's interior is counted; identified parts of its boundary are counted in their own lower-dimensional topological cells.^[ref]

The cellular homology theorem identifies cellular homology with singular homology, so the count equals the alternating rational Betti numbers. These are the free ranks of integral homology; torsion subgroups do not contribute to them. Over any field, the alternating homology dimensions still give the same Euler characteristic, although individual Betti numbers can depend on the field.

Integral homology is homology with coefficients in $\mathbb{Z}$. It retains both the free and torsion parts of each homology group.

For a rational homology three-sphere $Y$, the linking form is the nonsingular pairingReversing the orientation of $Y$ negates this form.

Here the paired elements lie in the torsion subgroup of first homology.

The saturation of a subgroup $H$ of the Mordell-Weil group consists of all points with a positive multiple in $H$. Saturation includes the full torsion subgroup. For a full-rank $H$ containing that torsion, it equals $E(K)$, but the index of $H$ may exceed one. Checking that $\ell P\in H$ forces $P\in H$ for every prime divisor of a certified index bound proves that $H$ is already the whole group. A canonical height pairing proves independence of found points; saturation answers the separate question of whether a full-rank subgroup is the full group.

The elliptic-curve discriminant is $37$, so the elliptic curve has good reduction at both $2$ and $3$. At a prime $\ell$ of good reduction, the reduction of torsion points on an elliptic curve is injective on torsion of order prime to $\ell$. Here is the formal reason: the kernel of reduction of an elliptic curve is described by an integral formal group law on $\ell {\mathbb{Z}}_{\ell }$, and its multiplication series is $[n](t)=nt+O(t^2)$. If $\ell \nmid n$, the first term has strictly smaller valuation than every higher term for $t\ne 0$. Thus this kernel has no nonidentity point killed by $n$.

Apply this to each prime-primary component of the rational torsion subgroup. A component at $2$ injects into ${\tilde{E}}_3({\mathbb{F}}_3)$, of order $7$, and is therefore trivial. A component at $3$ injects into ${\tilde{E}}_2({\mathbb{F}}_2)$, of order $5$, and is trivial. Every other prime-primary component injects into both groups, so its order divides $\gcd (5,7)=1$. This proves the two-prime bound on rational elliptic torsion in the present case:

At $2$ there is good reduction, and the surjectivity of good reduction over a local field gives an exact sequenceThe quotient has order $5$ and is consequently a cyclic group. For every odd integer $n$, multiplication by $n$ is an automorphism of $E_1({\mathbb{Q}}_2)$: its formal group law has multiplication series $nt+O(t^2)$ with unit linear coefficient, whose inverse formal power series has integral coefficients and converges on $2{\mathbb{Z}}_2$. This is the prime-to-residue-characteristic multiplication on a formal group.

Thus the odd-primary torsion subgroup injects into the order-five quotient. It also realizes that quotient: choose a lift $P$ of any reduced point, so $[5]P\in E_1$. There is a unique $Q\in E_1$ with $[5]Q=[5]P$. Then $P-Q$ is killed by five and has the prescribed reduction. This proves the prime-to-p torsion lifts uniquely at good reduction directly here. Therefore, for an odd prime $p$,The five-primary subgroup consists of the unique five-torsion lifts of the five reduced points. These lifts need not be rational over $\mathbb{Q}$; the preceding answer about rational torsion points of an elliptic curve does not contradict their existence over the p-adic field.

Assume good reduction, and write $N_p=\#\tilde{E}({\mathbb{F}}_p)$. The surjectivity of good reduction over a local field and the formal group law determine the torsion as follows. If $p\nmid n$, multiplication by $n$ is an automorphism of $E_1$, because its multiplication series has an integral compositional inverse with unit linear coefficient. Therefore reduction gives an isomorphismTo construct the inverse, lift a reduced point to $P$, solve $[n]Q=[n]P$ uniquely in $E_1$, and use $P-Q$. This proves the prime-to-p torsion lifts uniquely at good reduction, including uniqueness, rather than merely comparing orders.

For residue-characteristic torsion, use a deep logarithm subgroup of a formal group. Take $m=1$ for odd $p$, or $m=2$ at two. The formal logarithm identifies $E_m$ with $p^m{\mathbb{Z}}_p$, so $E_m$ has no torsion. The finite quotient grouptherefore receives the torsion subgroup injectively. There is a complete test for which cosets actually contain torsion. For a coset $g$ of order $n$, choose a lift $P$. Then $[n]P\in E_m$. Its coset has a torsion representative exactly whenWhen this condition holds, let $Q\in E_m$ be the unique point with $L(Q)=L([n]P)/n$. Then $T=P-Q$ is killed by $n$ and represents $g$. Conversely, any torsion representative has order precisely $n$, since an element killed in the quotient would belong to the torsion-free group $E_m$. It must therefore arise from this construction. Changing $P$ by $R\in E_m$ changes $L([n]P)$ by $nL(R)$, so the test is independent of the chosen lift. This proves the torsion test using a deep formal logarithm subgroup and gives the full torsion by a finite list of cosets and a logarithmic divisibility test.

For odd $p$, one has $E_m=E_1$, so all torsion points of an elliptic curve inject into $\tilde{E}({\mathbb{F}}_p)$. Nevertheless a reduced $p$-primary point need not lift to a torsion point: if its order is $p^r$, the test is $L([p^r]P)\in p^{r+1}{\mathbb{Z}}_p$. For $p=2$, the subgroup $E_2$ is torsion-free but $E_1/E_2$ has order two. Thus $E_1$ can have a two-torsion point, and one must use $G=E/E_2$ of order $2N_2$ rather than assert that all torsion injects under reduction. The prime-two torsion bound for good reduction follows: the reduction kernel contributes at most a subgroup of order two. These distinctions are essential to a correct determination of local torsion subgroups.

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.

Use the two-prime bound on rational elliptic torsion with the counts six at three and eight at five from solution. For a prime $\ell \ne 3,5$, the $\ell$-primary torsion subgroup injects into both reduced groups by reduction of torsion points on an elliptic curve. Its order therefore divides the corresponding primary part of $\gcd (6,8)=2$. The three-primary subgroup injects into the group of order eight at five and is trivial; the five-primary subgroup injects into the group of order six at three and is trivial. Hence the entire rational torsion subgroup has order at most two.

The factorization $x^3+x^2+x+1=(x+1)(x^2+1)$ supplies $T=(-1,0)$, a nonidentity point of order two under the elliptic curve group law. ThereforeThe primary-component argument avoids assuming injectivity on residue-characteristic torsion.

Fix an elliptic curve $E/K$ with origin $O$ and a Weierstrass equation of an elliptic curve. Define the naive logarithmic $x$-height by $h_x(P)=h([x(P):1])$ for $P\ne O$ and $h_x(O)=0$. The $x$-map has degree two and pulls the projective hyperplane bundle back to ${\mathcal{O}}_E(2[O])$. Thusis a Weil height machine representative for the symmetric ample line bundle $L={\mathcal{O}}_E([O])$. This factor of one half fixes the convention for the normalized height on an elliptic curve used below.

The Theorem of the Cube gives the quadratic geometric identities. Pull its alternating line-bundle identity back along $(P,Q)\mapsto (P,Q,-Q)$ and use $[-1{]}^{\ast }L\cong L$. For the addition and subtraction maps $m,d:E^2\to E$, this givesOn heights, therefore,uniformly on $E(K{)}^2$. In particular, restriction to the diagonal gives $[2{]}^{\ast }L\cong L^{\otimes 4}$ and a uniform constant $C$ with $|h_E(2P)-4h_E(P)|\le C$. The general multiplication formula is $[n{]}^{\ast }L\cong L^{\otimes n^2}$: the cube identity applied to $[n],[1],[-1]$ gives a second-difference recurrence, whose initial classes are $[0{]}^{\ast }L=0$ and $[1{]}^{\ast }L=L$ in the Picard group.

Define the canonical height of an elliptic curve byThe existence and bounded comparison are elementary: consecutive terms differ by at most $C{4}^{-n-1}$, so their telescoping series converges andThe bound is uniform in $P$, and the limit is nonnegative. Shifting the index gives $\hat{h}(2P)=4\hat{h}(P)$. This is the unique function within bounded distance of $h_E$ with that doubling rule: the difference of two such functions at $P$ is $4^{-n}$ times their bounded difference at $2^{n}P$, and hence zero. A change of Weierstrass model changes the naive height only by a bounded function, so the normalized limit is independent of that model.

Apply the approximate parallelogram identity to $2^{n}P,2^{n}Q$, divide by $4^n$, and pass to the limit. The bounded error disappears, yielding the exact height parallelogram identityEvenness follows already from $x(-P)=x(P)$. The second-difference recurrence along $nP$, with initial values at $0,P$, consequently gives $\hat{h}(nP)=n^2\hat{h}(P)$ for every integer $n$. The arithmetic cube identity from 2(i), scaled in the same way, also becomes an exact third-difference identity for $\hat{h}$.

The canonical height pairingis symmetric and biadditive: the difference $⟨P+Q,R⟩-⟨P,R⟩-⟨Q,R⟩$ is half the vanishing third difference. Its diagonal is $\hat{h}(P)$. Positivity gives the Cauchy-Schwarz inequality for this pairing without first assuming finite generation. Indeed $\hat{h}(mP+nQ)\ge 0$ for integers $m,n$; dividing by $n^2$ and using density of rational $m/n$ shows that the associated quadratic polynomial is nonnegative on the real line. Its discriminant therefore gives $⟨P,Q{⟩}^2\le \hat{h}(P)\hat{h}(Q)$.

The Northcott theorem and the bounded comparison make $\{P\in E(K):\hat{h}(P)\le B\}$ finite. The map $x:E\to {\mathbb{P}}^1$ has finite fibres, so bounded $x$-height indeed controls the number of points, including $O$. This proves the important criterionFor a torsion point the integer quadratic rule gives zero. Conversely, height zero makes all $2^{n}P$ belong to one finite bounded-height set; two coincide, forcing $(2^j-2^i)P=O$ for some $j>i$. This argument also proves torsion finiteness over the fixed number field.

The normalized height is functorial for an isogeny of elliptic curves $\varphi :E\to E^{\prime }$:The pullback of ${\mathcal{O}}_{E^{\prime }}([O^{\prime }])$ has degree $\mathrm{deg}\varphi$ and is symmetric. Its difference from that many copies of $[O]$ is a symmetric degree-zero line bundle, hence a two-torsion Picard class. Its height is bounded, so the ordinary height relation has a bounded error. Commuting the isogeny with $[2^n]$ and taking the normalized limit removes that error. This explains the natural quadratic scaling under multiplication and the usefulness of normalized heights in comparing isogenous curves.

Here is their role in the Mordell-Weil theorem. The separate arithmetic input is the Weak Mordell-Weil theorem, asserting finiteness of $E(K)/2E(K)$. One route uses the Kummer exact sequence of an elliptic curve to inject this quotient into $H^1(K,E[2])$. The division torsors are unramified outside finitely many places containing those over $2$ and those of bad reduction. After a finite Galois extension splitting $E[2]$, their classes are pairs of square classes with even valuations outside this finite set. These square classes form a finite group: the obstruction to being represented by an S-unit lies in the finite ideal class group's two-torsion, and the S-unit group is finitely generated, hence finite modulo squares. The inflation-restriction kernel is a cohomology group of a finite group with finite coefficients, also finite. This supplies the finite quotient without using the full theorem in advance.

Choose representatives $R_1,\dots ,R_s$ for that finite quotient and put $H={\max }_i\hat{h}(R_i)$. Every point has the form $P=2Q+R_i$. Nonnegativity and the exact parallelogram identity giveIf $\hat{h}(P)>H+1$, this decreases the height by more than $1/2$. Repeating must reach the finite set $T=\{P:\hat{h}(P)\le H+1\}$. Unwinding the halving expressions writes the initial point as an integer linear combination of the $R_i$ and a point of $T$. Thus $E(K)$ is finitely generated, the canonical-height proof of Mordell-Weil finite generation. The normalized height supplies the descent and Northcott supplies its finite endpoint; neither by itself supplies the weak theorem's finite quotient.

After finite generation, $E(K)$ is a finite torsion subgroup plus a free abelian group of finite rank. The canonical height pairing is positive-definite on the associated real vector space. To justify the latter rather than assuming it, if its nonnegative quadratic form had a real null direction, rounding longer and longer multiples of that direction to lattice points would produce infinitely many points of bounded height: the null-direction cross terms vanish, leaving only uniformly bounded rounding errors. This contradicts Northcott finiteness. Hence the free part is a Euclidean height lattice. Heights can consequently bound point searches, certify independence through positive Gram determinants, and organize the arithmetic complexity of rational points.

A preadditive category has an abelian group of morphisms between each pair of objects, and composition is additive in both variables. An additive category is a preadditive category with a zero object and finite products in a category; equivalently one may require finite coproducts. The equivalence, and the stronger assertion that these constructions agree canonically, follow from the argument below. An abelian category is an additive category with kernels in a category and cokernels in a category, in which every monomorphism is a categorical kernel and every epimorphism is a categorical cokernel. A monomorphism with the first property is a normal monomorphism; an epimorphism with the second is a conormal epimorphism.

Let $P={\prod }_{i=1}^n{A}_i$, with projections $p_i$. There is a unique $i_j:A_j\to P$ with $p_i{i}_j={\delta }_{ij}{1}_{A_j}$, interpreting the other components as zero. Since every $p_j$ sends ${\sum }_i{i}_i{p}_i$ to $p_j$, the product's universal property givesGiven arrows $f_i:A_i\to B$, put $f={\sum }_i{f}_i{p}_i$. Then $f{i}_j=f_j$. Conversely any arrow with these restrictions equals $f{1}_P={\sum }_{i}f{i}_i{p}_i$, so it is uniquely this sum. Thus $(P,i_i)$ is a coproduct in a category. The empty product is the zero object and also the empty coproduct. The dual argument starts from coproducts and constructs projections; it proves the equivalent definition. Hence finite products and coproducts are the same biproducts, with the displayed identities giving the canonical identification.

For the image factorization in an abelian category, let $h:A\to B$, let $c:B\to Q$ be its cokernel, and let $i:I\to B$ be the kernel of $c$. Since $ch=0$, write $h=ie$. We prove that $e:A\to I$ is epic. Suppose $t:I\to Z$ satisfies $te=0$ and let $l:L\to I$ be its categorical kernel. Then $e=lr$ for some $r:A\to L$. The composite $il$ is monic and hence is a normal monomorphism, say $il=\ker z$ for $z:B\to W$. Since $zh=zilr=0$, the categorical cokernel property makes $z$ factor through $c$, and therefore $zi=0$. The categorical kernel property of $il$ gives $s:I\to L$ with $ils=i$. Cancellation of $i$ yields $ls=1_I$; since $l$ is monic this also gives $sl=1_L$. Thus $l$ is invertible and $t=0$. In a preadditive category, $ue=ve$ now implies $(u-v)e=0$, hence $u=v$. This proves $e$ is an epimorphism and supplies the required factorization.

For uniqueness, an epimorphism that is a categorical cokernel is the categorical cokernel of its own categorical kernel. Indeed, if $e=\mathrm{coker}r$ and $k=\ker e$, then $r$ factors through $k$; an arrow killing $k$ therefore kills $r$ and factors uniquely through $e$. Dually, a normal monomorphism is the categorical kernel of its own categorical cokernel. If $h=mp$ is any epi–mono factorization, monicity of $m$ gives $\ker p=\ker h$. Both $p$ and $e$ are consequently categorical cokernels of this same categorical kernel. Their universal properties give a unique isomorphism $\theta$ between the intermediate objects with $\theta p=e$, and cancellation of $p$ gives $i\theta =m$. Therefore the epi–mono factorization is unique up to the unique compatible isomorphism.

Here is an additive category with truncated primary torsion, with the prime fixed as $2$. Its objects are finitely generated abelian groups having no element of order $4$, and its morphisms are all group homomorphisms. Equivalently their $2$-primary torsion subgroup has exponent at most $2$. It is a full subcategory of abelian groups, closed under finite direct sums and subgroups. Its hom groups and bilinear composition are inherited, and the zero group and finite direct sums belong to it, so it is additive. Categorical kernels are ordinary subgroup categorical kernels, which are still finitely generated and have no element of order $4$.

To construct categorical cokernels, for an arbitrary finitely generated abelian group $G$ let $t_2(G)$ be its $2$-primary torsion subgroup, and defineThe Fundamental theorem of finitely generated abelian groups shows that this quotient lies in the subcategory: it leaves the free and odd-primary parts unchanged and replaces each cyclic $2$-primary summand by either $0$ or $\mathbb{Z}/2\mathbb{Z}$. Every homomorphism $G\to H$ to an allowed group kills $2t_2(G)$, because it sends $2$-primary torsion into a subgroup killed by $2$. Thus $G\to R(G)$ has the universal property of a reflection. For $f:A\to B$, its categorical cokernel in the subcategory isThis has exactly the categorical cokernel universal property by the preceding factorization. Finite categorical limits exist because equalizers are categorical kernels of differences and finite products exist. Finite colimits exist dually because coequalizers are categorical cokernels of differences and finite coproducts exist.

The epimorphisms in this example are precisely the surjective homomorphisms. A surjection is epic already in abelian groups. Conversely, if $f$ is epic in the subcategory, its categorical cokernel is zero, so $R(B/f(A))=0$. A nonzero finitely generated abelian group has nonzero reflection: a nonzero free or odd-primary summand survives, and a nonzero finite $2$-primary summand has a nonzero quotient modulo twice itself. Hence $B/f(A)=0$ and $f$ is surjective. Its ordinary categorical kernel belongs to the subcategory, and its quotient by that categorical kernel is its codomain, which also belongs to the subcategory. It is therefore the categorical cokernel of its categorical kernel. This proves that every epimorphism is conormal.

The monomorphismis not normal. Its categorical cokernel in this category is reduction modulo $2$, since $R(\mathbb{Z}/4\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$. The categorical kernel of that categorical cokernel is the inclusion $2\mathbb{Z}\hookrightarrow \mathbb{Z}$, whereas the image of $m$ is $4\mathbb{Z}$. A normal monomorphism is the categorical kernel of its own categorical cokernel, as proved above, so $m$ cannot be normal. In fact multiplication by $2$ is itself the categorical kernel of reduction modulo $2$, up to the identification $\mathbb{Z}\cong 2\mathbb{Z}$. Its composite with itself is multiplication by $4$. Thus this example already exhibits two normal monomorphisms whose composite is not normal.

For the general assertion, let $m:A\to B$ be a nonnormal monomorphism in any additive category of the specified kind. Put $c=\mathrm{coker}m$ and $k:K\to B$ be its categorical kernel, and factor $m=kg$. Both $m$ and $k$ are monic, so $g$ is monic. Set $d=\mathrm{coker}g$ and $l:L\to K$ be its categorical kernel. Then $k,l$ are normal, and $g=lr$ for some $r$.

Suppose their composite were normal, say $kl=\ker t$. Since $tm=tklr=0$, $t$ factors through $c$, whence $tk=0$. The categorical kernel property of $kl$ supplies $s:K\to L$ with $kls=k$; cancellation shows $ls=1_K$, and monicity makes $l$ an isomorphism. Because $dl=0$, this forces $d=0$. In a preadditive category, vanishing of the categorical cokernel of $g$ implies that $g$ is epic: any $u,v$ with $ug=vg$ have $u-v$ factor through the zero categorical cokernel. By hypothesis $g$ is therefore a categorical cokernel, and it is also monic. If $g=\mathrm{coker}a$, monicity and $ga=0$ force $a=0$; a categorical cokernel of a zero arrow is an isomorphism. Thus $g$ is invertible, making $m=kg$ normal, a contradiction. Consequently normal monomorphisms cannot be closed under composition in any such category; the explicit pair $k,l$ witnesses the failure.

Use the standard theorem on reduction of torsion points on an elliptic curve: at a prime of good reduction, reduction is injective on torsion whose order is prime to that prime. Reduction at $2$ implies that every odd-primary part of the rational torsion subgroup has order dividing $3$. Thus its only possible odd prime is $3$, and its $3$-primary part has order at most $3$. Reduction at $5$ injects the $2$-primary part into a group of order $9$, so that part is trivial. No other primary part can occur.

Consequently the torsion subgroup has order at most three. Since $P_1$ has exact order three,This argument uses only prime-to-residue-characteristic injectivity, and therefore avoids an unjustified claim of full torsion injectivity at the prime $2$.

The Hessian criterion for flexes of a plane cubic applied to the defining polynomial givesConsequently the inflection points of a plane cubic on this curve are exactly its intersections with the three coordinate lines. Choose $r\in \overline{\mathbb{Q}}$ with $r^3=d$, and a primitive cube root of unity $\zeta$. The nine distinct flexes of a diagonal cubic areWith the chosen flex as identity, these points are precisely $E_d[3]$. A tangent cutting the Weil divisor $3P$ gives $3(P-O)=0$; conversely $3(P-O)=0$ makes the hyperplane Weil divisor equivalent to $3P$, which supplies a tangent with triple contact. This is the three-torsion points are flexes of a plane cubic correspondence.

On $Z=0$ only $O$ is rational, since the other cube roots of unity are not rational. On each other coordinate line a rational point exists exactly when $d$ is a rational cube, and then exactly one does. ThereforeIn particular the rational torsion subgroup cannot contain two independent points of order $3$.

One can also determine the whole rational torsion of a diagonal cubic. First it is killed by $6$. To see this, use the Weierstrass equation of an elliptic curve from (iii). At primes $\ell \equiv 2(\mathrm{mod}3)$ outside the finite set containing $2,3$ and the prime divisors of the numerator and denominator of $d$, the cube map permutes ${\mathbb{F}}_{\ell }$, sowhere the quadratic character is extended by $\chi (0)=0$, and the character sum vanishes after substituting $t=x^3-432d^2$. Reduction of torsion points on an elliptic curve injects the prime-to-$\ell$ part of rational torsion into this finite group. This injection follows because multiplication by an integer prime to $\ell$ is invertible on the formal group of an elliptic curve, and hence its reduction kernel has no such torsion.

For any prime $q>3$, choose a good $\ell \equiv 2(\mathrm{mod}3)$ with $\ell \equiv 1(\mathrm{mod}q)$; the Dirichlet theorem on primes in arithmetic progressions and the Chinese remainder theorem supply such primes outside the finite bad set. Then $q\nmid \ell +1$, excluding $q$-torsion. Similarly good $\ell \equiv 5(\mathrm{mod}12)$ exclude order $4$, and good $\ell \equiv 11(\mathrm{mod}18)$ exclude order $9$. Thus every rational torsion point has order dividing $6$.

The inversion formula in (i) shows that the only possible nonidentity rational point of order $2$ has $X=Y$ and $Z\ne 0$. It exists precisely when $(X/Z{)}^3=-d/2$, that is, when $d/2$ is a rational cube. There is at most one such point. This condition and $d$ being a cube cannot hold together, since $2$ is not a rational cube. Combining it with the three-torsion calculation gives the full answer:

Set $d=-7$. The affine rational pair $(u,v)=(2,-1)$ gives $P=(84,756)$ on the elliptic curve $y^2=x^3-21168$. It is a point of infinite order: in the rational torsion of a diagonal cubic classification, neither $-7$ nor $-7/2$ is a rational cube, so this curve has trivial rational torsion subgroup.

There is also a short direct torsion check specific to this curve. The primes $5$ and $11$ are good and have point counts $6$ and $12$, respectively, because cubing is bijective in their finite fields. Reduction at these two primes bounds the entire rational torsion by order dividing $6$, including its possible $5$- and $11$-primary parts by reduction at the other prime. The flex calculation excludes order $3$, and the inversion calculation excludes order $2$. Hence $P$ is indeed nontorsion.

Every nonzero multiple $nP$ is distinct, and its inverse projective image has $Z\ne 0$: the only rational point at infinity on the diagonal cubic is its identity. Thus each yields a distinct rational pair. More explicitly, if $nP=(x_n,y_n)$ on the Weierstrass model, thenHere $x_n\ne 0$, since $x_n=0$ would require a rational square equal to $-21168$. For example the elliptic-curve addition formula gives $2P=(28,28)$, which produces $(u_2,v_2)=(5/3,4/3)$. This proves there are infinitely many rational pairs of cubes summing to seven.

Let $K$ be a number field and $E/K$ an elliptic curve. Heights measure the arithmetic size of points, and turn the finite quotient supplied by the Weak Mordell-Weil theorem into finite generation. For a projective point $[X:Z]\in {\mathbb{P}}^1(K)$ define the Absolute logarithmic Weil heightusing the usual normalized absolute values and local degrees $n_v=[K_v:{\mathbb{Q}}_v]$. The product formula makes this independent of scaling and of enlarging $K$. For a Short Weierstrass form put $h_x(P)=h([x(P):1])$ and $h_x(O)=h([1:0])=0$. Over $\mathbb{Q}$, if $x=A/B$ in lowest terms, $h(x)=\log \max (|A|,|B|)$.

The finiteness property is Northcott theorem: for bounded height and bounded degree over $\mathbb{Q}$ there are only finitely many algebraic numbers. One sees the mechanism from the height-Mahler measure formula. If $\alpha$ has degree $d\le D$ and height at most $H$, its primitive integral minimal polynomial has Mahler measure $e^{dh(\alpha )}\le e^{DH}$. Expanding its product of linear factors bounds every coefficient by $2^D{e}^{DH}$. Only finitely many integral polynomials satisfy this bound, and each has finitely many roots. Thus a height bound gives only finitely many $x$-coordinates in the fixed number field $K$, and each gives at most two points of $E$, besides $O$.

To understand multiplication, the doubling formula givesIts homogeneous numerator and denominator have degree four. They have no common projective zero: at infinity the numerator is nonzero, and at a root of $f(t)=t^3+at+b$ the numerator equals $f^{\prime }(t{)}^2$, since it is $f^{\prime }(t{)}^2-8tf(t)$. Nonsingularity excludes a common root of $f,f^{\prime }$. The coordinate doubling map is therefore a degree-four morphism of the projective line.

The general height growth under a morphism of the projective line is $h(\Phi (t))=dh(t)+O(1)$ for a fixed degree-$d$ morphism. Here is why the error is bounded independently of $t$. At a place $v$, normalize its homogeneous coordinates to have maximum norm $1$. The fixed homogeneous forms have uniformly bounded norm. They also have a positive lower bound: the normalized coordinate space is compact over $K_v$, and their common zero set is empty. Outside finitely many places their coefficients are integral and their reduction has no common zero, making both bounds $1$. Homogeneity and summation over places then give the asserted height estimate. Consequently there is a fixed $C$ with

Define the canonical height of an elliptic curve by the conventional normalizationThe successive approximants differ by at most $C/(2\cdot 4^{n+1})$, so the limit exists andThus canonical height retains Northcott finiteness, while replacing the bounded-error doubling rule by an exact one.

We also need the height parallelogram identity. It follows from the symmetric-square proof of the naive height parallelogram estimate, which we give explicitly using the supplied addition formulae. For $x_i=X_i/Z_i$, the unordered pair $x(P+Q),x(P-Q)$ is represented by the binary quadratic $C{T}^2-BTS+A{S}^2$, whereThese forms define the biquadratic morphism for elliptic sums and differences. They have no common zero. If $C=0$, the two projective coordinates agree. On the finite diagonal $x_1=x_2=t$, $B=4f(t)$ and $A=f^{\prime }(t{)}^2-8tf(t)$, so simultaneous vanishing would violate nonsingularity. At $(\infty ,\infty )$ the homogeneous $A$ is nonzero. This also handles $P=\pm Q$ and sums or differences equal to $O$, for which the affine fractions alone would be inappropriate.

The same local norm argument as above, now using bidegree $(2,2)$, givesOn the other hand the binary quadratic is the product of two linear factors for $x(P+Q)$ and $x(P-Q)$. At nonarchimedean places coefficient maximum norms multiply exactly: scale each factor to norm $1$ and reduce; the product of the two nonzero residue polynomials remains nonzero. This is multiplicativity of the nonarchimedean polynomial norm. At archimedean places the norms are comparable up to fixed constants, by the coefficient triangle inequality and compactness of normalized linear factors. ThereforeCombining the estimates proves the naive-height parallelogram relation with a uniform error. Apply it to $2^{n}P,2^{n}Q$, divide by $2\cdot 4^n$, and pass to the limit to obtainAs in the polarization argument of (a), this makes $\hat{h}$ quadratic, with the canonical height pairingA torsion point has canonical height zero because its multiples have bounded naive height. Conversely, if $\hat{h}(P)=0$, then all $2^{n}P$ have bounded naive height. Northcott finiteness makes two equal, giving $(2^j-2^i)P=O$ for some $j>i$. Thus canonical height vanishes exactly on torsion. In particular the torsion subgroup is finite, since the comparison with naive height bounds the heights of all its points.

We can now prove the Mordell-Weil theorem. The Weak Mordell-Weil theorem, as explained in Question 2(ii), makes $E(K)/2E(K)$ finite. Choose representatives $R_1,\dots ,R_s$ and put $M={\max }_i\hat{h}(R_i)$. Every $P$ can be written $P=2Q+R_i$. The parallelogram identity and nonnegativity giveIterating this height descent makes the height at step $n$ at most $M+2^{-n}\max (\hat{h}(P)-M,0)$, so after finitely many steps it is at most $M+1$. There are only finitely many points of that height by Northcott and the naive-height comparison. Reversing $P=2Q+R_i$ shows that every point is an integer combination of the finitely many representatives and this finite bounded-height set. ThereforeOn the resulting free part the canonical height pairing is positive definite. Its nonnegativity extends from rational vectors to real vectors by density. If a nonzero real vector $v$ were in its kernel, in a basis of the free part round $nv$ to integer vectors $z_n$. Their rounding errors are uniformly bounded, and $\hat{h}(z_n)=\hat{h}(z_n-nv)$ would be uniformly bounded. The vectors $z_n$ are unbounded and therefore include infinitely many distinct vectors, contradicting Northcott finiteness. Thus there is no real null vector. Heights consequently support quantitative point searches, tests of independence, and the regulator of an elliptic curve, in addition to providing the descent that proves finite generation.

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.