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A division point above $P$ is a point $Q$ over an algebraic closure satisfying $[N]Q=P$, for a specified nonzero integer $N$. Two such points differ by an $N$-torsion point of an elliptic curve. If the characteristic of a field does not divide $N$, there are $N^2$ distinct division points; in other characteristics the scheme fibre still has length $N^2$, but its geometric points can be fewer. The unramified division torsors at good primes control their coordinate fields over local fields.

An inflection point of a smooth plane cubic is a point where its tangent line has intersection multiplicity three. Once one inflection point is chosen as the identity, the nine geometric inflection points are exactly the 3-torsion points.

Use reduction of torsion points on an elliptic curve, which is injective on torsion of order prime to the residue characteristic. Any two-primary torsion point of an elliptic curve must inject into the group of order seven at three, so is zero. Any three-primary torsion point of an elliptic curve injects into the group of order five at two, so is zero. For a prime $q\ne 2,3$, its primary torsion injects into both reduced groups and would have order dividing both five and seven, so is also zero. Decomposing the cyclic group generated by any torsion point of an elliptic curve into its primary parts shows that $E(\mathbb{Q})$ has no nonidentity torsion at all. This is the two-prime bound on rational elliptic torsion, with the residue-characteristic exceptions handled explicitly. Since $P=(0,0)$ is not $O$,

The Lutz–Nagell theorem says that for a nonsingular short equation $y^2=x^3+ax+b$ with $a,b\in \mathbb{Z}$, every nonidentity rational torsion point of an elliptic curve $(x,y)$ has $x,y\in \mathbb{Z}$, and either $y=0$ orThis is a necessary condition for torsion, not a converse.

To prove integrality, fix a prime $q$. If $v_q(x)<0$, the $x^3$ term dominates the right side of the equation. Hence $2v_q(y)=3v_q(x)$, so $v_q(x)=-2s$, $v_q(y)=-3s$ for some $s\ge 1$. The parameter $t=-x/y$ has valuation $s$ and identifies the point with the formal neighbourhood of the identity, namely the formal group of an elliptic curve evaluated on $q{\mathbb{Z}}_q$.

For odd $q$, this group is a torsion-free group by the logarithm argument of Question 2. For $q=2$, use the special short-model structure. Negation sends $t$ exactly to $-t$, so the formal multiplication series $[2](t)$ is odd and has integral coefficients:If $s=v_2(t)\ge 1$, the first term has valuation $s+1$ and all the others have valuation at least $3s>s+1$. Thus $v_2([2](t))=s+1$, and no iteration of doubling kills a nonzero point. Multiplication by an odd integer has unit linear coefficient and also cannot kill it. This proves the torsion-free formal subgroup for a short Weierstrass equation, including at two. Consequently a rational torsion point of an elliptic curve cannot have $v_q(x)<0$ at any prime. Its $x$ is integral, and the equation then makes its rational $y$ integral too.

For the divisibility conclusion suppose $y\ne 0$. Then $2P\ne O$ and is also a rational torsion point of an elliptic curve, so $x(2P)$ is integral. Its tangent slope $\lambda =f^{\prime }(x)/(2y)$ satisfies $x(2P)={\lambda }^2-2x$. Thus ${\lambda }^2\in \mathbb{Z}$; a rational number with integral square is itself integral. In particular $y^2\mid f^{\prime }(x{)}^2$. Reduce the supplied polynomial identity modulo $y^2=f(x)$: both terms on the left are divisible by $y^2$, so the right side $4a^3+27b^2$ is too. This is the divisibility proof in the Nagell–Lutz theorem.

For the specific curve, $4a^3+27b^2=3^7{5}^4$. The computed equality $2P_1=-P_1$ makes $P_1$ a point of exact order three. The coordinate $y(P_3)=17$ has square not dividing $3^7{5}^4$, so $P_3$ has infinite order. To decide $P_2$, use the already computed point $P_1+P_2=(15,-60)$. Its $y$-coordinate has even square, which cannot divide the odd number $3^7{5}^4$, so that sum has infinite order. Since $P_1$ is torsion, $P_2$ must have infinite order as well. $P_1$ has order three; $P_2$ and $P_3$ both have infinite order.

Kummer theory studies extracting roots through Galois actions, and on an elliptic curve the corresponding operation is division of points. The bridge is a cocycle: different Galois conjugates of a chosen division point differ by torsion. Controlling the ramification of these cocycles is what proves the Weak Mordell-Weil theorem.

Classically, for a characteristic-zero field $K$, the sequenceis exact. Hilbert's theorem 90 gives $H^1(K,{\overline{K}}^{\times })=0$, so Galois cohomology identifiesThe class of $a$ is represented by $\sigma \mapsto \sigma (\sqrt[m]{a})/\sqrt[m]{a}$. If ${\mu }_m\subset K$, the module is constant and these are continuous characters into ${\mu }_m$; the associated fields are cyclic Kummer extensions of degree dividing $m$. Several such classes give abelian extensions of exponent dividing $m$. Without the roots-of-unity hypothesis the cohomological description still holds, but the action on ${\mu }_m$ must not be discarded.

For an elliptic curve over a number field $K$, multiplication by $m\ge 2$ gives the exact sequence of Galois modulesThe map is onto because it is a nonconstant isogeny over an algebraically closed field. For $P\in E(K)$ choose $Q\in E(\overline{K})$ with $mQ=P$, and setIt satisfies $\delta (P{)}_{\sigma \tau }=\delta (P{)}_{\sigma }+\sigma \delta (P{)}_{\tau }$, the Galois 1-cocycle identity. Changing $Q$ by a torsion point of an elliptic curve changes the cocycle by a coboundary. If the class is zero, a corresponding change of $Q$ makes it Galois-fixed, so $P\in mE(K)$. Conversely a rational division point gives the zero class. Thus the Kummer map of an elliptic curve is injective on the quotient, and the cohomology sequence givesThis is the Kummer exact sequence of an elliptic curve. The last group parametrizes the appropriate torsor classes. It is important that the entire middle group need not be finite; an unrestricted Kummer class can ramify at arbitrarily many primes.

The Weak Mordell-Weil theorem states $E(K)/mE(K)$ is finite for every number field $K$ and every $m\ge 1$. The case $m=1$ is trivial. To prove the remaining cases, choose a finite set $S$ containing the primes of bad reduction and the primes dividing $m$. At a prime outside $S$, the elliptic curve extends to a smooth proper group scheme over the valuation ring, and $[m]$ is finite étale. Properness extends a local rational point to a section; its division fibre is a finite étale torsor over that ring. Therefore its Kummer class is unramified. This is unramified division torsors at good primes, and shows that ${\delta }_m(E(K))$ lies in first Galois cohomology unramified outside a finite set.

Let $L=K(E[m])$, enlarged if necessary to contain ${\mu }_m$, and include in $S$ the finitely many primes ramified in $L/K$. This is a finite Galois extension. Over $L$ the torsion module is constant, isomorphic to $(\mathbb{Z}/m\mathbb{Z}{)}^2$, and so to two copies of ${\mu }_m$. Restricting a cocycle to $G_L$ therefore reduces it to two classical Kummer classes unramified outside the primes above $S$. Such a class is represented by $a\in L^{\times }$ satisfyingWrite this power-class group as $L(S_L,m)$. The implication follows locally because, away from residue characteristic dividing $m$, the valuation of an unramified $m$th-root extension must be divisible by $m$.

Its finiteness comes from two standard arithmetic finiteness results, rather than from an assertion that all of $L^{\times }/(L^{\times }{)}^m$ is finite. The valuations outside $S_L$ associate to $a$ an ideal whose $m$th power is principal. There is an exact sequenceThe left group is finite by the finite generation of the S-unit group, and the right group is finite by finiteness of the ideal class group. This proves finiteness of S-unramified Kummer classes. Hence there are only finitely many restricted cocycles over $L$. The kernel of restriction from $K$ is contained in $H^1(\mathrm{Gal}(L/K),E[m])$, which is finite because both group and module are finite. The restricted cohomology group over $K$ is therefore finite, and its injected subgroup $E(K)/mE(K)$ is finite too. This completes a Kummer-theoretic proof of the weak Mordell-Weil theorem.

For arithmetic computation one adds local conditions. The Selmer group of an elliptic curve consists of classes in $H^1(K,E[m])$ whose restriction at every completion $K_v$ lies in the image of the corresponding local Kummer map. They are unramified outside the same finite set, so this group is finite. The failure of a everywhere locally soluble torsor to have a global point is measured by the Tate–Shafarevich group, defined as the kernel of the local restriction map on $H^1(K,E)$. The resulting exact sequence isThus the Selmer group of an elliptic curve gives a computable upper bound for the quotient; actual rational points supply lower bounds. Equality need not follow just from local solubility. Finiteness of this fixed torsion subgroup does not prove finiteness of the whole Tate–Shafarevich group.

Finally, weak finiteness alone is not finite generation: the additive group $\mathbb{Q}$ has trivial quotients by multiplication and is not finitely generated. The additional ingredient for the Mordell-Weil theorem is height descent. Choose representatives $P_i$ of $E(K)/mE(K)$ and write $P=P_i+mQ$. The quadratic growth of the canonical height of an elliptic curve, together with bounds for translation by the finitely many $P_i$, makes $Q$ have smaller height whenever $P$ is sufficiently large. Northcott theorem gives only finitely many points of bounded height, and repeated descent yields a finite generating set. This explains the roles of Kummer theory, local information and heights without conflating the weak and full theorems.

Use the congruent number elliptic curve in the equivalent coordinatesThe rational point $P=(25,100)$ lies on it, since $25^3-225\cdot 25=10000$. By the preceding rational torsion of a congruent number curve result, every rational torsion point of an elliptic curve has $y=0$ or is $O$. Thus $P$ is not a torsion point of an elliptic curve, and its positive multiples give infinitely many distinct rational points with nonzero $y$.

For any such point $(x,y)$, putThe identity $(x^2-225{)}^2+(30x{)}^2=(x^2+225{)}^2$ proves that these positive rational numbers are the sides of a right triangle. Their area isAll three sides are nonzero because a point with $y\ne 0$ has $x\notin \{0,15,-15\}$. For $P$ the construction gives $(a,b,c)=(4,15/2,17/2)$.

It remains to ensure that infinitely many points do not describe only finitely many triangles. Given the ordered positive pair $b,c$, set $r=c/b$. Then $X=|x|$ satisfiesThere are at most two possible $X$, then at most two signs of $x$ and two signs of $y$. Thus each ordered triangle has at most eight preimages; allowing interchange of its legs still gives a finite number. There are infinitely many distinct rational right triangles of area $15$. This is the infinitely many rational right triangles from a nontorsion point principle.

Let $K^{\mathrm{ur}}$ be the maximal unramified extension of $K$, with residue field $\overline{k}$. First show that every geometric solution of $nQ=P$ lies in $E(K^{\mathrm{ur}})$.

The multiplication-by-n morphism on the reduced elliptic curve is surjective over $\overline{k}$. Choose $\overline{Q}$ with $n\overline{Q}=\overline{P}$, defined over some finite finite field extension $k^{\prime }/k$. Let $L/K$ be the corresponding finite unramified extension. By the surjectivity of good reduction over a local field, lift $\overline{Q}$ to $Q_0\in E(L)$. Then $P-n{Q}_0\in E_1(L)$. The multiplication isomorphism of a formal group law gives a unique $R\in E_1(L)$ with $nR=P-n{Q}_0$, so $Q_0+R$ is an $n$-division point of $P$ defined over $L$.

Taking $P=O$ gives every element of $E[n]$ over $K^{\mathrm{ur}}$: reduction is injective on this torsion point of an elliptic curve group because $[n]$ is injective on the formal kernel, and each reduced $n$-torsion point has the unique corrected lift just constructed. Every other division point differs from one lift by an element of $E[n]$. Hence the entire field $K([n{]}^{-1}P)$ is unramified, not merely a field containing one choice of $Q$.

To bound the composite uniformly in $P$, put $f=[K(E[n]):K]$. This division field of an elliptic curve is unramified. The Galois group of $K^{\mathrm{ur}}/K$ is procyclic, generated topologically by its Frobenius element $\sigma$, and ${\sigma }^f$ fixes every element of $E[n]$. If $nQ=P\in E(K)$, thenIteration gives ${\sigma }^{jf}Q=Q+jT$. In particular ${\sigma }^{nf}Q=Q$. Thus every such $Q$, for every $P$ together, lies in the unique unramified extension of degree $nf$. If $M$ is the composite of all these fields, the uniform unramified division field over a local field givesThe prime-to-$p$ hypothesis is essential to the inverse-series argument; no assertion about arbitrary $p$-division is being made.

Let $T$ have order $m>0$. Property (ii) gives $\hat{h}(T)=0$, sinceIt also givesThereforefor every rational torsion point of an elliptic curve. The construction in part (i), quadraticity in part (ii), and this identity establish existence; the bounded-difference and doubling argument in part (i) establishes uniqueness.

For $E:y^2=x^3-x$ over ${\mathbb{F}}_3$, each of $x=0,1,2$ gives the single affine point with $y=0$. Including $O$ gives $\#E({\mathbb{F}}_3)=4$, so the Frobenius trace is $a=3+1-4=0$. Therefore the Frobenius isogeny of an elliptic curve $\pi$ satisfiesOn the $11$-torsion, this reads ${\pi }^2=[-3]=[8]$. The element $8\in {\mathbb{F}}_{11}^{\times }$ has order ten and $8^5=-1$. Henceand no smaller positive power of $\pi$ is the identity on $E[11]$. A division field of an elliptic curve over a finite field has degree equal to the order of Frobenius on the torsion module, so

Let $E/{\mathbb{Q}}_p$ have good reduction, and choose an integer $m>1/(p-1)$. Its filtration of elliptic-curve points over a local field has a torsion-free group $E_m$ identified by the formal logarithm with $p^m{\mathbb{Z}}_p$. The quotient group $G=E({\mathbb{Q}}_p)/E_m$ is finite. For $g\in G$ of order of a group element $N$, choose a lift $P$. There is a torsion point of an elliptic curve in this coset exactly when the displayed condition holds. Indeed, it is equivalent to the existence of $Q\in E_m$ with $[N]Q=[N]P$. Then $P-Q$ is killed by $N$; conversely every torsion representative has order $N$, since $E_m$ is torsion-free. The logarithmic division and the formal group exponential give the unique such $Q$. Changing the lift changes the logarithm by an element of $N{p}^m{\mathbb{Z}}_p$, so the criterion is well-defined. This finite-coset test includes residue-characteristic torsion; only the prime-to-p torsion lifts uniquely at good reduction are automatic.