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Suppose an elliptic curve $E/K$ has all four geometric 2-torsion points rational. For $P\in E(K)$ choose $Q\in E(\overline{K})$ with $2Q=P$. The function $c_P(\sigma )=\sigma (Q)-Q$ takes values in $E[2]$ and is a group homomorphism from the absolute Galois group to $E[2]$: its usual cocycle law becomes additivity because the action on $E[2]$ is trivial. Replacing $Q$ by another half, or $P$ by $P+2R$ with $R\in E(K)$, leaves it unchanged. Adding chosen halves proves additivity in $P$. Its kernel in $E(K)/2E(K)$ is zero, since $c_P=0$ precisely when the chosen half is rational. This constructs the injective map directly, without assuming any theorem of two-descent on an elliptic curve or finite generation of the Mordell-Weil group.

For an elliptic curve over a number field, its Hasse-Weil L-function is the product of the local factors determined by its Tate module. At a good finite place, the factor is $(1-a_{v}N{v}^{-s}+N{v}^{1-2s}{)}^{-1}$, with $a_v=Nv+1-\#E(\kappa (v))$. At a bad place use the inertia-invariant subspace and an auxiliary prime different from the residue characteristic. For a CM elliptic curve with all endomorphisms defined over the base, this factors as the product of the two conjugate Hecke L-functions.

For naive logarithmic $x$-height on an elliptic curve, $h_x(P+Q)+h_x(P-Q)=2h_x(P)+2h_x(Q)+O(1)$ uniformly. The unordered pair of sum and difference on the $x$-line defines a bidegree-$(2,2)$ morphism to ${\mathrm{Sym}}^2{\mathbb{P}}^1$, yielding the estimate. Applying it to $2^{n}P,2^{n}Q$ and taking the defining limit gives the exact identity for the canonical height of an elliptic curve.

The group ${\mathrm{Hom}}_K(E_1,E_2)$ consists of morphisms of elliptic curves defined over $K$ that preserve the group operation, with pointwise addition. Its zero element is the constant map to the identity. Every nonzero element is an isogeny of elliptic curves: its image is a closed connected subgroup of the target, so nonconstancy makes it the whole curve, and a nonconstant morphism of smooth projective curves is finite. The degree of an isogeny extends to this group by assigning degree zero to the zero homomorphism.

This sheaf consists of holomorphic functions vanishing at $p$ on opens containing $p$, and all holomorphic functions on opens not containing $p$. Evaluation gives $0\to {\mathcal{I}}_p\to {\mathcal{O}}_M\to {\mathbb{C}}_p\to 0$, with a skyscraper sheaf as quotient. On an elliptic curve, $H^0({\mathcal{I}}_p)=0$, $H^1({\mathcal{I}}_p)\cong \mathbb{C}$, and all higher cohomology vanishes.

Elliptic curves isogenous over a finite field have the same number of rational points: the isogeny intertwines Frobenius, and multiplicativity of degree gives equal degrees for $1-\pi$. Their groups need not be isomorphic. Over ${\mathbb{F}}_7$, $y^2=x^3-x$ and $y^2=x^3+4x$ are two-isogenous and have groups $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/8\mathbb{Z}$, respectively.

An isogeny between elliptic curves is a nonconstant morphism preserving their group identities; it is finite and surjective with finite kernel.

An isogeny is a nonconstant morphism of elliptic curves preserving their identity points. It is a finite surjective group homomorphism.

At a point of a smooth algebraic curve, its local ring is a discrete valuation ring. A local parameter is a generator $t$ of its maximal ideal, equivalently an element of valuation one. It measures the order of a zero of a function or pole by expressing functions as powers of $t$ times units. Over the complex numbers its analytic counterpart is a local coordinate vanishing simply at the point. For an elliptic curve in a Weierstrass equation of an elliptic curve, $t=-x/y$ is the standard parameter at the identity.

The nonzero even theta constants and the Jacobi abstruse identity give $\lambda (-1/\tau )=1-\lambda (\tau )$ and $\lambda (\tau +1)=\lambda (\tau )/(\lambda (\tau )-1)$. It is invariant under the principal congruence subgroup of level two. This parameter puts the associated complex elliptic curve in Legendre form $y^2=x(x-1)(x-\lambda )$; its Klein j-invariant is $256(1-\lambda +{\lambda }^2{)}^3/[{\lambda }^2(1-\lambda {)}^2]$.

For a fixed rational point $R$ on an elliptic curve over a number field, the addition formula gives $h_x(P-R)\le 2h_x(P)+O_R(1)$. The duplication map has degree four, so $h_x(2Q)=4h_x(Q)+O(1)$. Combining them gives the displayed contraction. A finite set of representatives modulo twice the point group makes the constants uniform. Points above a fixed height threshold descend by a definite amount; Northcott theorem makes the terminal bounded-height set finite. This converts the Weak Mordell-Weil theorem into the full finite-generation theorem.

For an elliptic curve over an algebraically closed field, ${\mathrm{Pic}}^0(E)\cong E(k)$ via $P\mapsto [P-O]$. This group is infinite and divisible by any prime $\ell$ different from the characteristic, since $[\ell ]:E\to E$ has nonzero derivative and is a surjective morphism of projective curves. A finitely generated abelian group divisible by $\ell$ has zero free rank and is finite. Hence $E(k)$, and therefore $\mathrm{Cl}(E)$, cannot be finitely generated. This proof applies to countable algebraically closed fields as well.

An elliptic curve in characteristic zero admits no nonconstant rational map of projective varieties from the projective line. Over $\mathbb{C}$, write its equation as $y^2={\prod }_{j=1}^3(x-e_j)$ with distinct $e_j$. Substituting $x=u/v$ with coprime polynomials gives $(y{v}^2{)}^2=v{\prod }_j(u-e_{j}v)$. These four pairwise coprime factors must all be squares. The polynomial pencil with four square members forces $u,v$ constant, ruling out a nonconstant rational parametrization of an algebraic curve.

The modular group is $S{L}_2(\mathbb{Z})$ acting on the complex upper half-plane byIts central element $-I$ acts trivially on points, so the effective group is $PS{L}_2(\mathbb{Z})$. It is generated by $S:\tau \mapsto -1/\tau$ and $T:\tau \mapsto \tau +1$: the Euclidean algorithm reduces the bottom row of a matrix using these operations until it is a translation. The relations $S^2=(ST{)}^3=1$ hold in the projective group.

A standard fundamental domain of the modular group isTo see that each orbit meets it, use $\mathrm{Im}(\gamma \tau )=\mathrm{Im}\tau /|c\tau +d{|}^2$. Choose a coprime integer pair $(c,d)$ minimizing the nonzero lattice modulus $|c\tau +d|$; a modular matrix with that bottom row maximizes the imaginary part. Translate to the central strip. If the result had modulus below one, inversion would further increase its imaginary part, a contradiction. Interior representatives are unique: for a point in the interior, $|c\tau +d|>1$ when $c\ne 0$. For $|c|\ge 2$ use $\mathrm{Im}\tau >\sqrt{3}/2$; for $|c|=1$ use the strip and the strict unit-circle bound. Thus such a matrix lowers the imaginary part, which would contradict the same fact applied to its inverse if both representatives were in the interior. Matrices with $c=0$ are translations, and cannot move one interior strip point to another.

The vertical sides are identified by $T$, and the circular halves by $S$. The exceptional points with elliptic stabilizers of the modular group are represented by $i$ and $\rho =e^{2\pi i/3}$, of projective stabilizer orders two and three. The rational boundary points together with infinity form one modular cusp orbit. In the hyperbolic metric the region has area ${\int }_{-1/2}^{1/2}(1-x^2{)}^{-1/2}dx=\pi /3$.

A useful subgroup is the level-two principal congruence subgroup,Reduction modulo two is onto $S{L}_2({\mathbb{F}}_2)$, which has six elements and permutes the three nonzero vectors. Hence its kernel has index six, and the quotient is $S_3$. In the projective action, $\Gamma (2)$ is torsion-free. Indeed finite-order nonidentity modular elements have trace zero or $\pm 1$; matrices in $\Gamma (2)$ have even trace, and trace zero is impossible because $a,d$ are odd while $b,c$ are even: $ad-bc=1$ with $d=-a$ would imply $bc=-a^2-1\equiv 2(\mathrm{mod}4)$. The subgroup has three modular cusp classes, represented by $\infty ,0,1$, each of width two. Parity of a primitive pair distinguishes these classes, and elementary level-two row operations reduce each class to its representative. Six translates of $\mathcal{F}$ supply a fundamental region. Its compactified level-two modular curve is obtained by adjoining these three modular cusp points.

Invariant holomorphic functions arise naturally from theta constants. In the radian convention of Question 4, put ${\Theta }_j(\tau )={\theta }_j(0,\tau )$, and defineHere ${\Theta }_2={\sum }_{n\in \mathbb{Z}}{q}^{(n+1/2{)}^2}$. The theta-constant inversion and translation laws, obtained by Gaussian Poisson summation and by reindexing, interchange ${\Theta }_2,{\Theta }_4$ under $S$ with the common factor $\sqrt{-i\tau }$, leave ${\Theta }_3$ fixed up to that factor, and under $T$ interchange ${\Theta }_3,{\Theta }_4$ while multiplying ${\Theta }_2$ by $e^{\pi i/4}$. Together with the Jacobi abstruse identity ${\Theta }_3^4={\Theta }_2^4+{\Theta }_4^4$, this givesThese transformations generate the six permutations of $0,1,\infty$. In particular $T^2$ and $S{T}^2{S}^{-1}$ fix $\lambda$; with $-I$, these generate $\Gamma (2)$. Thus the modular lambda function is invariant under this subgroup, but not under the whole modular group. Its even theta constants do not vanish on $\mathbb{H}$, so it is holomorphic there and avoids $0,1,\infty$. Near infinity, $\lambda =16q+O(q^2)$, with $q=e^{\pi i\tau }$, which is the width-two modular cusp coordinate. The two transformation laws give modular cusp values one at zero and infinity at one, with the same simple local orders. Thus $\lambda$ has exactly one simple pole on the compactified level-two quotient and no pole in the half-plane. A meromorphic map with one simple pole has degree one, so it identifies that quotient with the sphere and the open quotient with $\mathbb{C}\setminus \{0,1\}$. This proves the coordinate assertion and describes all three modular cusps.

Symmetrizing this subgroup invariant produces a full modular-invariant function:Substitution shows that this rational expression is unchanged by both $\lambda \mapsto 1-\lambda$ and $\lambda \mapsto \lambda /(\lambda -1)$, hence by $S$ and $T$. It is the Klein j-invariant. Its denominator does not vanish on $\mathbb{H}$, so it is holomorphic there; at the modular cusp it has $j\sim q^{-2}$, a simple pole in $Q=e^{2\pi i\tau }$. It therefore remains nonconstant without being a bounded holomorphic function on the compactified quotient. In fact a holomorphic invariant extending holomorphically to the compactified modular curve would be constant by the maximum principle.

This also connects invariance to elliptic curves: changing an oriented basis of the lattice $\mathbb{Z}+\tau \mathbb{Z}$ gives precisely the modular action after a rescaling. On the compact full modular quotient, $j$ has just the one simple modular cusp pole, so its map to the sphere has degree one. It therefore classifies the resulting homothety of complex lattices classes, while the modular lambda function retains a level-two labeling of the branch points in the Legendre model $y^2=x(x-1)(x-\lambda )$. Finally, a modular form of nonzero weight obeys a transformation law with a factor $(c\tau +d{)}^k$, not plain invariance; ratios of equal-weight forms instead give weight-zero modular functions. The distinction between forms, invariant holomorphic functions on the open half-plane, and their modular cusp behavior is essential to this construction.

Let $O=(0:1:0)$ be the point at infinity of the Weierstrass equation of an elliptic curve. It is the identity element for the elliptic curve group law. For an affine rational point $P=(x,y)$, the inverse isIndeed, the two affine intersections with the vertical line have these ordinates, since their sum as roots of the quadratic in $y$ is $-a_{1}x-a_3$. This reflection is valid for a generalized Weierstrass equation of an elliptic curve in every characteristic of a field.

For distinct $P,Q$, draw their line; for $P=Q$, draw the tangent line. Let $R$ be its third intersection with the elliptic curve, counted with intersection multiplicity. Then the sum is the reflection of the third intersection:A vertical line has third intersection $O$, so $P+(-P)=O$. If a tangent has third intersection $O$, then $2P=O$. Define addition with $O$ by $P+O=P$; these conventions include all exceptional cases of the affine construction.

The construction is defined over the coefficient field: after two known intersections, the remaining root is rational over that field. It gives a commutative group operation. To see why it is associative, use the elliptic curve group law from Riemann-Roch: the map $P\mapsto [P-O]$ identifies the elliptic curve with its degree-zero divisor class group. A line has intersection divisor $P+Q+R$ linearly equivalent to $3O$, the divisor of the line at infinity. Thus $[P-O]+[Q-O]+[R-O]=0$. The geometric sum agrees with addition of divisor classes, which is associative.

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

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

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

Use two-isogeny descent with $a=-7$, $b=12$, so the partner elliptic curve is $E^{\prime }:Y^2=X^3+14X^2+X$. On $E$, a negative $x$ would make $x(x-3)(x-4)$ negative. Hence the prime-support bound in two-isogeny descent restricts the square class image to $\{1,2,3,6\}$. All four occur: the identity gives $1$, and the rational points $(2,2)$, $(0,0)$, $(6,6)$ give respectively $2$, $12\equiv 3$, and $6$ in ${\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$. Thus $|\alpha E(\mathbb{Q})|=4$.

For the partner, $b^{\prime }=1$, so its square class image is contained in $\{1,-1\}$. A realization of $-1$ would yield coprime integers $U,V$, not both zero, satisfyingIf exactly one of $U,V$ is odd, the right side is $3$ modulo four, which is not a square. If both are odd, $U^4\equiv V^4\equiv 1(\mathrm{mod}16)$ and $14U^2{V}^2\equiv 14(\mathrm{mod}16)$, so the right side is $12$ modulo sixteen, again impossible. Both even is excluded by coprimality. This proves the negative unit obstruction for the fourteen-coefficient isogeny cover; the identity realizes $1$, so $|{\alpha }^{\prime }{E}^{\prime }(\mathbb{Q})|=1$.

The two-isogeny rank formula now gives $2^r=4\cdot 1/4=1$. Consequently this is the rank-zero elliptic curve with roots zero three and four:The displayed nonzero-ordinate points are therefore torsion points of an elliptic curve, rather than evidence of positive rank. In fact doubling either $(2,2)$ or $(6,6)$ gives $(4,0)$, so both have order four.

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:

An elliptic curve over $\mathbb{Q}$ has good reduction at $p$ if it admits an integral Weierstrass equation of an elliptic curve over ${\mathbb{Z}}_p$ whose reduced projective cubic is a smooth algebraic curve over ${\mathbb{F}}_p$. Equivalently, its Minimal Weierstrass equation has a unit elliptic-curve discriminant, or its smooth proper model is an elliptic curve over the valuation ring. The existence of an appropriate integral model matters: a nonminimal equation can have discriminant divisible by $p$ even when the curve has good reduction. Equivalently,