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For an elliptic curve over ${\mathbb{F}}_q$, where $q$ is a prime power, the Hasse theorem for elliptic curves isWe prove it from the degree form on elliptic-curve endomorphisms, including the required degree facts.

A nonzero homomorphism of elliptic curves is a finite surjective morphism; its degree is the degree of the induced extension of function fields. Set the degree of the zero homomorphism to $0$. The degree of an isogeny is also the degree of the pullback of any point divisor: locally, a finite map of smooth curves gives a module of rank equal to the function-field degree, and its fibre length is the sum of the local multiplicities. For a composition, the tower law for function fields givesIf the isogeny of elliptic curves is separable, its pullback of a nonzero invariant differential on an elliptic curve is nonzero. Translation invariance implies that its differential is nonzero at every point, so each fibre point has multiplicity $1$. Every fibre is a translate of the kernel. ThereforeFor an inseparable isogeny, multiplicities must be retained; counting geometric kernel points alone would be incorrect.

Here is a divisor proof of the degree parallelogram law. On $E\times E$, let $\Delta$ be the diagonal and ${\Delta }^{-}$ the locus $P=-Q$. A Weierstrass $x$-function has a double pole at $O$, and the equation $x(P)=x(Q)$ describes $P=Q$ or $P=-Q$. ConsequentlyThis remains valid in characteristic $2$ using a general Weierstrass equation of an elliptic curve: the $x$-map still has degree $2$ and its two points are exchanged by negation. The divisor identity gives an identity of line bundles. Pull it back by $P\mapsto (\alpha (P),\beta (P))$ and take degrees. The diagonal is the inverse image of $O$ under subtraction and the antidiagonal under addition, soUsing line bundles makes this pullback argument valid even when $\alpha =\beta$, $\alpha =-\beta$ or one map is zero, when directly substituting into the rational function would fail.

Let $d_n=\mathrm{deg}[n]$. Applying the degree parallelogram law to $[n]$ and $[1]$ gives $d_{n+1}+d_{n-1}=2d_n+2$, with $d_0=0,d_1=1$. Induction givesMore generally, the same recurrence and polarization show that the degree restricted to the subgroup generated by two endomorphisms is a quadratic form: its mixed coefficient is determined by their sum or difference. This can be checked directly by the second-difference recurrence in each integer variable and the parallelogram identity in the two diagonal directions.

Let $\pi$ be the $q$-power Frobenius isogeny of an elliptic curve. It has degree $q$: it is purely inseparable, has one geometric point in each fibre, and sends a local parameter at the rational point $O$ to its $q$th power, giving fibre multiplicity $q$. Its pullback of any differential is zero. On the other hand, the standard addition rule for an invariant differential on an elliptic curve givesHence $1-\pi$ is separable, and its kernel consists exactly of the points fixed by Frobenius, namely $E({\mathbb{F}}_q)$. Put $N=\#E({\mathbb{F}}_q)$ and $t=q+1-N$. Thenand the quadratic-form calculation gives, for all $m,n\in \mathbb{Z}$,If $t^2>4q$, the real polynomial $X^2-tX+q$ is negative on an open interval. That interval contains a rational $n/m$ with $m\ne 0$, contradicting the displayed nonnegativity after multiplication by $m^2$. Thus $t^2\le 4q$, which is precisely the Hasse theorem for elliptic curves.

Heights measure arithmetic size, and their finiteness and growth properties turn finite divisibility information into finite generation. For an elliptic curve $E$ over a number field $K$, the resulting Mordell-Weil theorem isThe distinction between this statement and the Weak Mordell-Weil theorem, which only asserts that $E(K)/mE(K)$ is finite, is exactly where heights enter.

For a rational point of the projective line, choose coprime integers $a,b$ and define the logarithmic projective heightFor a fixed number field, its absolute logarithmic version iswith the usual normalized absolute values. The product formula makes this independent of scaling $(a,b)$, and the local degree formula makes it independent of enlarging the number field. Heights are nonnegative and are useful precisely because they control numerator and denominator together, unlike an ordinary real absolute value.

The Northcott theorem states that algebraic points of bounded degree and bounded height form a finite set. Over $\mathbb{Q}$ this is immediate from the coprime numerator-denominator description. In general, bounded degree and height bound the coefficients of the primitive minimal polynomial: their elementary symmetric functions are bounded by its Mahler measure, which is the appropriate power of the multiplicative height. There are only finitely many such integer polynomials.

On a Weierstrass equation of an elliptic curve, define the naive logarithmic height $h_x(P)=h(x(P))$ and $h_x(O)=0$. Since an $x$-coordinate has at most two points above it, Northcott theorem implies that only finitely many points of $E(K)$ have bounded $h_x$.

Duplication descends to a rational map of the projective $x$-line of degree $4$. More generally, if a morphism $\varphi :{\mathbb{P}}^1\to {\mathbb{P}}^1$ has degree $d$, thenuniformly in $z$. Write $\varphi$ as two homogeneous degree-$d$ polynomials without common zeros. Bounding their coefficients gives the upper estimate at each place. A nonzero resultant, or homogeneous Bezout identities, prevents simultaneous cancellation and gives the lower estimate; only finitely many places contribute a nonzero constant. Applied to duplication, this gives the height growth under a morphism of the projective line estimate

Define the canonical height of an elliptic curve byThe factor $1/2$ is a convention making it the height for the degree-one symmetric divisor $O$. If $a_n=4^{-n}{h}_x(2^{n}P)$, then $|a_{n+1}-a_n|\le C_E{4}^{-(n+1)}$, so the limit exists by a geometric-series estimate. Summing that estimate also provesThus bounded canonical height still gives only finitely many $K$-points, and $\hat{h}(2P)=4\hat{h}(P)$ exactly.

The addition formula on the quotient by negation gives the uniform approximate parallelogram identityThis height parallelogram identity follows as follows. One way to obtain it is to express the unordered pair of their $x$-coordinates as a morphism ${\mathbb{P}}^1\times {\mathbb{P}}^1\to {\mathrm{Sym}}^2{\mathbb{P}}^1$ of bidegree $(2,2)$; the height of that unordered pair is the sum of the two heights, up to a bounded constant. It is also the height identity for the symmetric divisor $O$. Apply this identity to $2^{n}P,2^{n}Q$, divide by $2\cdot 4^n$, and take limits. The errors disappear, givingTogether with evenness and $\hat{h}(O)=0$, the recurrence implies $\hat{h}(mP)=m^2\hat{h}(P)$ for every integer $m$. The associated canonical height pairingis bilinear. Nonnegativity of the height implies its Cauchy-Schwarz inequality, by applying it to integer combinations and then rational approximations. In particular,The height vanishes precisely at torsion points: torsion gives a finite set of multiples, whereas if $\hat{h}(P)=0$, every multiple of $P$ has bounded naive height and therefore the set of multiples is finite by Northcott theorem.

Now apply the Weak Mordell-Weil theorem with $m=2$, proved cohomologically in the other essay. Choose a finite set $R$ of representatives for $E(K)/2E(K)$, and put $H={\max }_{R_i\in R}\hat{h}(R_i)$. Write an arbitrary point as $P=2Q+R_i$. ThenIf $\hat{h}(P)>H+1$, this decreases the height by more than $1/2$. Repeat until the current point has height at most $H+1$. There are only finitely many such points. Unwinding the equations $P=2Q+R_i$ shows that $E(K)$ is generated by this finite bounded-height set together with $R$. This is the height descent lemma, and proves the Mordell-Weil theorem.

Once finite generation is established, the structure theorem for finitely generated abelian groups gives the displayed decomposition. The canonical height pairing is positive definite on the real vector space of the free part: a null direction would, by approximation with integer combinations of generators, give infinitely many lattice points of bounded height, contradicting Northcott theorem. Its determinant is the regulator of an elliptic curve, and its geometry gives practical tools for searching for generators and comparing independence. Thus heights supply both the finiteness mechanism in the proof and the quadratic size function used in arithmetic computations.

For a nonzero isogeny of elliptic curves $\varphi :E_1\to E_2$, its degree of an isogeny is the degree of the induced extension of function fields. Since $\varphi$ commutes with negation, its first coordinate is a rational function of $x_1$ alone. Write it in lowest terms asThe coordinate maps $x_i:E_i\to {\mathbb{P}}^1$ are finite morphisms of degree two. Multiplying function field degrees in the commutative diagram $x_2\circ \varphi =R\circ x_1$ gives the degree of an isogeny from its x-coordinate map:This is the total degree, so the argument also covers an inseparable isogeny of elliptic curves. Counting distinct points in the kernel of an isogeny would give only its separable degree. The zero group homomorphism is assigned degree zero separately.

On a Short Weierstrass form $y^2=x^3+ax+b$ in characteristic of a field other than two, the elliptic-curve addition formula givesThere is no cancellation: at a root $r$ of $f(x)=x^3+ax+b$, the numerator equals $f^{\prime }(r{)}^2$, which is nonzero because the elliptic-curve discriminant is nonzero. The numerator has degree four and the denominator degree three. HenceFor completeness, the same answer holds in characteristic two, where the displayed short equation is not a nonsingular model. On a general Weierstrass equation of an elliptic curve usewhere $b_2=a_1^2+4a_2$, $b_4=a_1{a}_3+2a_4$, $b_6=a_3^2+4a_6$, and $b_8=a_1^2{a}_6+4a_2{a}_6-a_1{a}_3{a}_4+a_2{a}_3^2-a_4^2$. The resultant of this numerator and denominator is ${\Delta }^2$. Their resultant is therefore nonzero in every nonsingular case; even when the denominator drops degree, the numerator retains degree four.

Work first where all three coordinates are finite and the joining line is not vertical. By the chord-and-tangent group law, the third intersection of $y=\lambda x+\mu$ with the elliptic curve is $-(P_1+P_2)$, whose first coordinate is also $x_3$. Substitution gives the monic polynomialIts roots are $x_1,x_2,x_3$, counted with intersection multiplicity. Comparing its coefficients with the elementary symmetric polynomials provesThis symmetric-coordinate identity for elliptic-curve addition is an identity of rational functions. It includes tangent cases by specialization. Literally assigning finite $s_i$ when a point is $O$ would be meaningless; such cases must be interpreted on the projective curve. A nonsingular equation in the printed short form has characteristic different from two.

Here is how the identity yields the quadratic form for the degree of an isogeny. With $u=x(P)$, $v=x(Q)$, write $s=u+v$ and $q=uv$. As an equation for $t=x(P+Q)$, it becomesBoth $x(P+Q)$ and $x(P-Q)$ satisfy it, because replacing $Q$ by $-Q$ does not change $v$. Thus its two branches are the sum and difference maps. Their common denominator is the square of $u-v$. The following divisor on an algebraic curve calculation keeps track of cancellations and exceptional points rigorously, instead of assuming degrees of displayed numerators always add.

On $E_2\times E_2$, equality of the two first coordinates means $Q=P$ or $Q=-P$. Each coordinate has a double pole at $O$. Consequently the principal divisor identity iswhere $D_{+}$ is the diagonal and $D_{-}$ is the graph of negation. Equivalently, the two zero divisors are the inverse images of $O$ under difference and sum, exactly the two branches identified above. Their generic multiplicities are one; the divisor identity also accounts for their intersections at 2-torsion points.

For nonzero isogenies of elliptic curves $\varphi ,\psi :E_1\to E_2$ with $\varphi \ne \pm \psi$, pull this identity back by $(\varphi ,\psi )$. Taking degrees of the resulting principal divisor on $E_1$ gives the divisor proof of the degree parallelogram law:If one map is zero, this is immediate. If $\varphi =\pm \psi$, it follows from $\mathrm{deg}([2]\circ \varphi )=4\mathrm{deg}\varphi$. This also proves the parallelogram law for a general Weierstrass equation of an elliptic curve in characteristic two: the same divisor identity uses the quotient by negation, even though the particular symmetric-coordinate formula above is unavailable.

Put $q(\varphi )=\mathrm{deg}\varphi$ and $q(0)=0$. Applying the parallelogram law to $n\varphi ,\varphi$ gives the recurrence $q((n+1)\varphi )+q((n-1)\varphi )=2q(n\varphi )+2q(\varphi )$, and hence $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$. Its polarization $B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$ is symmetric and integer-valued. For fixed $\psi$, the parallelogram law makes $B(\cdot ,\psi )$ odd and gives $B(x+y,\psi )+B(x-y,\psi )=2B(x,\psi )$. Interchanging $x,y$ and adding proves additivity, since integers have no two-torsion. Thus $B$ has bilinearity, and $q$ is a positive quadratic form on $\mathrm{Hom}(E_1,E_2)$, with $q(\varphi )>0$ whenever $\varphi \ne 0$.

An elliptic curve has good reduction over $K$ if it admits an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. Equivalently, the reduction of its Minimal Weierstrass equation is a nonsingular cubic over the residue field $k$. Its smooth integral projective model is a group scheme with the point $O$ as identity.

Choose homogeneous coordinates in the valuation ring ${\mathcal{O}}_K$ with at least one coordinate a unit, and reduce them modulo its maximal ideal ${\mathfrak{m}}_K$. This defines $\mathrm{red}:E(K)\to \tilde{E}(k)$. The chord-and-tangent group law extends over the smooth integral model, so reduction is a group homomorphism. For any point of $\tilde{E}(k)$, choose an affine projective chart containing it. Smoothness means some partial derivative of its equation is nonzero in $k$. Lift the other coordinate arbitrarily and apply the Hensel lemma to the coordinate with unit derivative. This constructs a point of $E(K)$ above the prescribed point, proving the surjectivity of good reduction over a local field.

The kernel of reduction of an elliptic curve consists precisely of points reducing to $O$. The parameter $t=-x/y$ at $O$ gives a bijection between this kernel and ${\mathfrak{m}}_K$: the integral formal-coordinate series solve the equation uniquely near $O$. Addition in these coordinates is the formal group of an elliptic curve, not ordinary addition in ${\mathfrak{m}}_K$. ThusHere the group operation on the last expression is its formal group law.

For this integral Weierstrass equation of an elliptic curve, computeIf an odd prime divides $m$ or $m^2-1$, then $c_4$ is a unit there and $\Delta$ is not. The j-invariant of an elliptic curve consequently has negative valuation. Since good reduction implies an integral j-invariant, such a prime cannot be a prime of good reduction, regardless of whether the displayed equation is minimal. Conversely, if it divides neither factor, this equation already has unit discriminant.

At five the good possibilities have $m\equiv \pm 2$, so $m^2\equiv 4$. At seven they have $m^2\equiv 2$ or $4$. Direct elliptic-curve point count over a finite field gives the following nonzero ordinates, together with the three zero-ordinate points and $O$:At either prime, torsion-freeness of the formal group over Qp for odd p makes reduction injective on all of $E(\mathbb{Q}{)}_{\mathrm{tors}}$, so its order is at most eight. Using only prime-to-$p$ injectivity here would leave an unjustified possible $p$-primary component.

All three nonzero 2-torsion points $(0,0),(-1,0),(-m^2,0)$ are rational. The rational point $P=(m,m(m+1))$ lies on the curve, since $m(m+1)(m+m^2)=m^2(m+1{)}^2$. For an equation $y^2=x(x^2+ax+b)$ the elliptic-curve addition formula givesHere $a=1+m^2$, $b=m^2$, so $2P=(0,0)$ and $P$ has order four. The point $(-1,0)$ is not in $⟨P⟩$, whose only nonzero point of order two is $(0,0)$. Therefore $P$ and $(-1,0)$ generate a subgroup of order eight isomorphic to $\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. The upper bound proves the rational torsion in the family x times x plus one times x plus m squared:

Translate the rational 2-torsion point to $(0,0)$ and clear denominators to obtain the integral Weierstrass equation of an elliptic curve. Nonsingularity requires $b\ne 0$ and $b^{\prime }=a^2-4b\ne 0$. The two-isogeny descent usesand its dual isogenyThese formulas extend across their missing affine points as degree-two isogenies of elliptic curves, with kernels $\{O,(0,0)\}$ on their respective curves; substitution gives $\psi \varphi =[2]$ and $\varphi \psi =[2]$.

Define homomorphisms to the square-class group of a field ${\mathbb{Q}}^{\times }/({\mathbb{Q}}^{\times }{)}^2$ byFor ordinary intersections with a line, the product of the three first coordinates is the square of its intercept, proving the homomorphism identity; tangent cases use the same product with multiplicities. For $T=(0,0)$, direct addition gives $x(P+T)=b/x(P)$, so $\alpha (P+T)=[b]\alpha (P)$. Also $\alpha (P)\alpha (-P)=1$ and $\alpha (T{)}^2=1$, which handle inverse points and $T+T=O$. These verify the exceptional values as well. Their kernels are $\psi E^{\prime }(\mathbb{Q})$ and $\varphi E(\mathbb{Q})$. One can check the first assertion directly: if $x=s^2\ne 0$, solving for the first coordinate of a preimage under $\psi$ giveswith rational corresponding ordinate. Conversely, on $E^{\prime }$ one has ${\psi }_x=Y^2/(4X^2)$, a square. The exceptional point $(0,0)$ has a rational preimage exactly when $b$ is a square. The other kernel assertion follows identically, since the double companion curve is isomorphic to $E$ by scaling its coordinates by four and eight.

We now justify the rank factor in the square-class index formula for two-isogeny descent, rather than forgetting a torsion correction. Put $A=E(\mathbb{Q})$, $B=E^{\prime }(\mathbb{Q})$, andThe preimage of $\psi \varphi A$ under $\psi$ is $\varphi A+\ker \psi$, soThe Mordell-Weil theorem gives $[A:2A]=2^r\#A[2]$, where $r$ is the rank of an abelian group of $A$. If $b^{\prime }$ is a square, $\#A[2]=4$ and $(0,0)\in \varphi A$, so $\delta =1$. If it is not a square, $\#A[2]=2$ and $\delta =2$. Hence $\delta \#A[2]=4$ in both cases, and

To bound these two images, let $\ell \nmid b$ and take a point with $x\ne 0$. If $v_{\ell }(x)>0$, then $x^2+ax+b$ is a unit, so $2v_{\ell }(y)=v_{\ell }(x)$ is even. If $v_{\ell }(x)<0$, the leading term $x^3$ uniquely has least valuation in the equation, so $2v_{\ell }(y)=3v_{\ell }(x)$ and again $v_{\ell }(x)$ is even. Thus every image class has a signed square-free integer representative supported on the prime divisors of $b$. The exceptional class $[b]$ also has that property. There are at most $2^{\nu (b)+1}$ such classes. Applying the same reasoning to $E^{\prime }$ gives $\#{\alpha }^{\prime }(B)\le 2^{\nu (b^{\prime })+1}$. The prime-support bound in two-isogeny descent is thereforeHere $\nu$ counts the distinct prime divisors of the absolute value of a nonzero integer; $\nu (1)=\nu (-1)=0$. The nonsingularity hypotheses exclude the otherwise undefined case $\nu (0)$.

For the unheaded practical procedure, enumerate signed square-free divisors $d$ of $b$, and of $b^{\prime }$ on the companion curve. A class $d$ is represented by a point of $E$ exactly when its quartic covering in a two-isogeny descenthas a rational solution with $(U,V)\ne (0,0)$. For $UV\ne 0$, reconstruct $x=d{U}^2/V^2$, $y=dUW/V^3$; substitution verifies the equivalence. Solutions with $V=0$ or $U=0$ account for the identity class or the class of $(0,0)$ respectively. Clearing denominators allows integral $U,V,W$ with $\gcd (U,V)=1$.

Test these finitely many coverings over the real numbers and local fields, especially at two and the primes dividing $b{b}^{\prime }$, to exclude impossible classes. Search the survivors for rational points, close the witnessed classes under multiplication, and use the index formula once both images are determined. Equivalently, local solubility gives a descent upper bound and independent rational points, certified for example by their canonical height of an elliptic curve pairing, give a lower bound. When the bounds coincide the rank is determined. This procedure often succeeds, but local solubility alone does not prove global solubility: a nonzero Tate–Shafarevich group can leave surviving coverings without rational points. In that case the computation gives a rigorous bound rather than a falsely certified exact rank.

Choose a point $Q\in E$. Since $g(E)=1$, a nonzero regular differential has an effective divisor of degree $2g-2=0$, so $K_E\sim 0$. The Riemann-Roch theorem therefore givesChooseThe rational functions $x$ and $y$ have their only poles at $Q$, of orders two and three respectively.

The seven functionsbelong to the six-dimensional Riemann-Roch space $L(6Q)$, so they satisfy a nontrivial linear relation. Comparing pole orders shows that the coefficients of both $x^3$ and $y^2$ are nonzero. After rescaling, the relation has the Weierstrass formThe divisor $3Q$ is very ample, so its complete linear system embeds $E$ in ${\mathbb{P}}^2$ and the displayed relation cuts out its image; this is the Weierstrass construction from Riemann-Roch in genus one.

Because $\mathrm{char}k\ne 2$, completing the square removes the terms linear in $y$, and because $\mathrm{char}k\ne 3$, translating $x$ removes the quadratic term. Thus the equation becomesSince $k$ is algebraically closed, the cubic factors asSmoothness of $E$ says that the elliptic-curve discriminant is nonzero, equivalently the three roots are distinct. Hence

The Weierstrass equation of an elliptic curve over a field $K$ has the general formwith nonzero elliptic-curve discriminant. Admissible change of Weierstrass coordinates relates two such equations defining isomorphic pointed curves over $K$:where $u\in K^{\times }$ and $r,s,t\in K$.

If $\mathrm{char}K\ne 2,3$, completing the square and translating $x$ simplify every equation. Short Weierstrass form isTwo short equations are $K$-isomorphic precisely when, for some $u\in K^{\times }$,the isomorphism from the first curve to the second is $(x,y)\mapsto (u^{2}x,u^{3}y)$.

Because $\varphi$ is a separable isogeny,The degree-zero divisorcorresponds under $E_1\simeq {\mathrm{Pic}}^0(E_1)$ to the sum of all elements of the finite abelian group $E_1[\varphi ]$. Pairing every $T$ with $-T$ shows that this sum is the sum of the elements in $E_1[\varphi ]\cap E_1[2]$. It vanishes when that intersection has one element and also when it has four elements, since the sum of the four elements of $(\mathbb{Z}/2\mathbb{Z}{)}^2$ is zero. The principal divisor criterion on an elliptic curve therefore gives a rational function $g$ with $\mathrm{div}(g)=D$.

The divisor $D$ is invariant under $[-1]$, so $[-1{]}^{\ast }g/g$ has zero divisor and is constant. Applying $[-1]$ twice shows that this constant has square one; hence $[-1{]}^{\ast }g=\pm g$.

For a short Weierstrass equation of an elliptic curvethe multiplication-by-two isogeny hasThus one may take $g=y$, and $[-1{]}^{\ast }y=-y$. For multiplication by three, the third division polynomial of an elliptic curvevanishes simply at the eight nonzero points of $E[3]$ and has a pole of order eight at $O_E$. Hence $g={\psi }_3(x)$ has divisor $[3{]}^{\ast }(O_E)-9(O_E)$ and satisfies $[-1{]}^{\ast }g=g$. Both signs occur.

An admissible change of Weierstrass equation of an elliptic curve replaces the $x$-coordinate by a degree-one rational function, so the two logarithmic heights differ by $O(1)$. Dividing this bounded difference by $4^r$ in the defining limit shows that the canonical height is unchanged. It is intrinsic to the elliptic curve and the chosen point.

Apply the Riemann-Roch theorem to the divisors $nO$. Since the genus is one and the canonical divisor is trivial, $\ell (nO)=n$ for $n\ge 1$. ChooseThen $x$ and $y$ have exact pole orders two and three at $O$. The seven functionslie in the six-dimensional space $L(6O)$, so they satisfy one relation. Comparing pole orders and completing squares and cubes gives a nonsingular Weierstrass equation of an elliptic curveThe functions $1,x,y$ define the morphism $P\mapsto (x(P):y(P):1)$ away from $O$. Their pole orders show that it extends with $O\mapsto (0:1:0)$. It has degree one and is therefore an isomorphism of smooth projective curves.

A Minimal Weierstrass equation for $E/{\mathbb{Q}}_p$ is a Weierstrass equation of an elliptic curve with coefficients in ${\mathbb{Z}}_p$ whose discriminant has minimum $p$-adic valuation among all integral equations for $E$ related by admissible changes of variables.

Let $P\in E({\mathbb{Q}}_p)$ and choose projective coordinates $[X:Y:Z]$ with $X,Y,Z\in {\mathbb{Z}}_p$ and at least one coordinate a unit. Reducing the coordinates modulo $p$ givesMultiplying the primitive coordinates by a unit does not alter this point, so this defines the reduction map $E({\mathbb{Q}}_p)\to \bar{E}({\mathbb{F}}_p)$.

There are infinitely many rational pairs $(u,v)$ with $u^3+v^3=7$. The corresponding Weierstrass equation of an elliptic curve is $y^2=x^3-21168$, and $(u,v)=(2,-1)$ maps to $P=(84,756)$. The primes $5$ and $11$ are good, with point counts $6$ and $12$ because cubing is bijective in their residue fields. Reduction of torsion points on an elliptic curve bounds rational torsion by order dividing $6$. The flexes of a diagonal cubic exclude three-torsion for $d=-7$, while $d/2=-7/2$ is not a cube and hence excludes two-torsion. Thus $P$ has infinite order.

All nonzero multiples of $P$ map to distinct rational affine pairs, since the only rational point at infinity of the diagonal cubic is its identity. The inverse map is $u=(y+252)/(6x)$ and $v=(252-y)/(6x)$.

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.

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.

Let $E$ be a genus one curve and choose $O\in E$. Since a canonical divisor is linearly equivalent to zero, Riemann--Roch gives $\ell (nO)=n$ for $n\ge 1$. Choose $x\in L(2O)\setminus L(O)$ and $y\in L(3O)\setminus L(2O)$. Their pole orders at $O$ are two and three. The seven functionsare linearly dependent because $\ell (6O)=6$, and comparison of pole orders puts the relation in Weierstrass form. The complete linear system $|3O|$ is very ample and realizes this relation as a smooth plane cubic. In characteristic other than two and three, completing the square and translating $x$ produce a Short Weierstrass form.