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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.

Start with an elliptic curve having a rational point $T$ of order $2$. Move that point to $(0,0)$ and take an integral modelPut $b^{\prime }=a^2-4b$ andThe two-isogeny formula and its dual areThey extend to the projective curves, with kernels $\{O,T\}$ and $\{O,T^{\prime }\}$, where $T^{\prime }=(0,0)\in E^{\prime }$. Substitution verifies their target equations, and the elliptic-curve addition formula verifies $\hat{\varphi }\varphi =[2]$ and $\varphi \hat{\varphi }=[2]$. The division by $4,8$ in the dual identifies the twice-transformed curve, with coefficients $4a,16b$, with $E$.

Define the two-torsion square-class homomorphismsQuestion 1(ii) proves these are homomorphisms, since $f^{\prime }(0)=b$. Their kernels areFor completeness, the first coordinate of the dual is $Y^2/(4X^2)$, so a nonexceptional dual image has square $x$-coordinate. Conversely, if $x=u^2\ne 0$, the preimage equation iswhose discriminant is $16(x^2+ax+b)=16(y/u{)}^2$. The roots $X=2x+a\pm 2y/u$ are rational and nonzero. Choosing $Y=\pm 2uX$ gives a point on $E^{\prime }$, and a choice of sign makes its dual image exactly $(x,y)$. The point $T$ has a rational dual preimage precisely when $b$ is a square, as seen from the roots $a\pm 2\sqrt{b}$ of the nonzero two-torsion polynomial on $E^{\prime }$. The identity is already a dual image. Applying this argument to the transformed curve proves the second kernel assertion as well, because scaling an $x$-coordinate by $4$ does not change its square class.

The two-isogeny descent is finite because an image square class has a signed square-free integer representative dividing $b$. Indeed, if $\ell \nmid b$ and $v_{\ell }(x)>0$, the other factor $x^2+ax+b$ is a unit, so $v_{\ell }(y^2)=v_{\ell }(x)$ forces an even valuation. If $v_{\ell }(x)<0$, the term $x^2$ dominates that factor and $v_{\ell }(y^2)=3v_{\ell }(x)$, again forcing even valuation. Thus odd valuations can occur only at primes dividing $b$. The same argument applies to $b^{\prime }$ on $E^{\prime }$.

For each candidate signed squarefree divisor $d\mid b$, put $x=d(U/V{)}^2$, with coprime integers $U,V$. The curve equation becomes the two-isogeny descent quarticA solution with $UV\ne 0$ yields $x=d{U}^2/V^2$ and $y=dUN/V^3$. Conversely, every point in that class gives such a primitive integer solution, since a rational square root of an integer is integral. The boundary solutions $U=0$ and $V=0$ account respectively for the classes of $T$ and $O$. Rational solutions prove that a class occurs; a real or congruence obstruction excludes it. Merely finding local solutions everywhere does not automatically prove a rational solution.

To extract the rank, write $A=\alpha (E(\mathbb{Q}))$ and $A^{\prime }={\alpha }^{\prime }(E^{\prime }(\mathbb{Q}))$. The Mordell-Weil theorem givesThe isogeny factorization gives the index $|A|$ for the first quotient by $\hat{\varphi }{E}^{\prime }$. For the remaining index, apply $\hat{\varphi }$ to $E^{\prime }/\varphi E$. Its kernel has size $2/e$, whereThe equality follows because $\hat{\varphi }(\varphi P)=2P$. Consequentlyand cancellation yields the two-isogeny rank formulaThis formula is valid whether there is just one rational nonzero two-torsion point or all three.

For the first curve, $a=-1,b=1$, and the isogenous curve isThe only candidate square classes on $E$ are $1,-1$. Since $x^2-x+1>0$ for every real $x$, a real affine point has $x\ge 0$; the exceptional torsion class is $[b]=1$. ThusOn $E^{\prime }$, the candidates are $1,-1,3,-3$. They all occur: the identity gives $1$, $T^{\prime }$ gives $-3$, $(3,6)$ gives $3$, and $(-1,2)$ gives $-1$. Hence $|A^{\prime }|=4$, andThe product in the rank formula is $1\cdot 4=4$, not $1\cdot 2$; all four classes on the companion curve are essential.

For the second curve, $a=5,b=-6$, andThe candidate classes on $E$ are $\{\pm 1,\pm 2,\pm 3,\pm 6\}$. The identity, $T$, $(2,4)$ and $(-3,6)$ show thatThis is a subgroup of order $4$. Its other coset is $\{3,6,-1,-2\}$, so it suffices to exclude the representative $3$.

The modulo-eight obstruction to a two-isogeny descent class uses the corresponding two-isogeny descent quartic,If both $U,V$ are odd, its right side is $6$ modulo $8$. If $U$ is odd and $V$ even, it is $3$ or $7$ modulo $8$. If $U$ is even and $V$ odd, it is $6$ or $2$ modulo $8$. These are all primitive parity cases, and none is a square modulo $8$. Hence the class $3$ is impossible. Since $A$ is a subgroup, every class in its coset is impossible, and thereforeOn the companion curve, the only candidates are $\pm 1,\pm 7$. Its quadratic factor is $(X-5{)}^2+24>0$, so every nonzero real affine $X$ is positive. The exceptional value $[49]$ is $1$, and the point $(7,14)$ supplies the class $7$. ThusThe rank formula gives $2^r=4\cdot 2/4=2$, soEvery included class has an explicit rational representative and every excluded coset has a proved real or congruence obstruction, so these are exact ranks rather than bounds obtained from a point search.

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.

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.

Galois cohomology and the finite division quotient. Let $K$ be a number field, $G_K=\mathrm{Gal}(\overline{K}/K)$ with its profinite topology, and $M$ a discrete continuous $G_K$-module. Its Galois cohomology is continuous group cohomology. In degrees zero and one,Thus a crossed homomorphism measures the failure of a choice to be Galois-invariant, and changing the choice changes the cocycle by a coboundary. A short exact sequence of such modules gives a long exact sequence in group cohomology. The connecting map sends an invariant element to the cocycle obtained by lifting it and comparing its Galois translates.

For an elliptic curve $E/K$ and $n\ge 2$, multiplication by $n$ is surjective on $E(\overline{K})$ and has kernel $E[n]$. The Kummer exact sequence of an elliptic curve therefore givesExplicitly, choose $nQ=P$ and set $\kappa (P)(\sigma )=\sigma Q-Q$. Replacing $Q$ changes this by a coboundary; replacing $P$ by $P+nR$, $R\in E(K)$, does not change it. If its class is zero, some torsion translate of $Q$ is Galois-invariant, so $P\in nE(K)$. This proves injectivity of the Kummer map of an elliptic curve.

The whole group $H^1(K,E[n])$ need not be finite. The crucial restriction is ramification. Let $S$ contain the finite primes of bad reduction of an elliptic curve and those dividing $n$. At $v\notin S$, good reduction and prime-to-residue-characteristic multiplication imply that every division point of a local point is in an unramified extension, by the formal-group lifting argument. Thus every global Kummer class restricts trivially to inertia there.

Choose a finite Galois extension $L/K$ containing $E[n]$ and the $n$th roots of unity, and enlarge $S$ to include its ramified primes. Over $L$, the torsion module is constant and can be identified with ${\mu }_n^2$. By Kummer theory,For classes unramified outside the primes $S_L$ above $S$, valuations outside $S_L$ must be divisible by $n$. Their restriction images therefore lie in $L(S_L,n{)}^2$, the square of an S-unramified power class group. Its finiteness can be seen without assuming the Mordell-Weil theorem: the finiteness of S-unramified Kummer classes follows fromThe S-unit group is finitely generated by the Dirichlet unit theorem and its valuations at $S_L$. The localized ideal class group is a quotient of the ordinary ideal class group of $L$, which is finite by the finiteness of the ideal class group. The Minkowski bound for ideal classes supplies an integral ideal of bounded norm in each class, and only finitely many integral ideals have bounded norm.

The kernel of restriction from $H^1(K,E[n])$ to $H^1(L,E[n])$ is also finite: a cocycle trivial after restriction can, after subtracting a coboundary, be made to factor through the finite group $\mathrm{Gal}(L/K)$. There are only finitely many maps from that group to the finite module $E[n]$. This is the finite-extension kernel of a Kummer map mechanism. Consequently the possible Kummer classes form a finite set, provingThis is the Weak Mordell-Weil theorem. Imposing membership in the local Kummer image at every place refines the finite unramified collection to the n-Selmer group. Its relation with the obstruction to a globally rational point iswhere the last group is the $n$-torsion of the Tate–Shafarevich group. Thus local solubility gives a finite computable upper bound but can leave a genuine global obstruction.

Heights and finite generation. For the other essay, normalize the absolute values on a field $K$ so that the product formula holds. The Absolute logarithmic Weil height of $[x_0:x_1]\in {\mathbb{P}}^1(K)$ isAt the infinite places the local degrees are one or two and the usual real or complex modulus is used. The product formula makes this independent of scaling; it is also unchanged by extending $K$. Over $\mathbb{Q}$, using coprime integral coordinates gives $h([u:v])=\log \max (|u|,|v|)$, the naive height on the projective line. The Northcott theorem says that points of bounded height and bounded field degree form a finite set; in particular this holds over the fixed field $K$. One way to see the finiteness is the height-Mahler measure formula: for an algebraic number of degree $d$, the coefficients of its primitive minimal integer polynomial are bounded by binomial factors times $\exp (dh)$. Bounds on $d$ and $h$ therefore leave only finitely many integer polynomials and hence only finitely many algebraic numbers.

Set $h_E(P)=h(x(P))$ for $P\ne O$, and $h_E(O)=0$. Since $x:E\to {\mathbb{P}}^1$ has degree two, bounded $h_E$ gives only finitely many points of $E(K)$. The height growth under a morphism of the projective line states that a degree-$d$ morphism gives $h(f(t))=dh(t)+O(1)$. The upper bound follows from its homogeneous coordinate polynomials; the lower bound follows from their having no common zero, using a resultant identity at each place. Applied to the degree-four duplication map, it gives a uniform constant $C$ withConsequently $4^{-j}{h}_E(2^{j}P)$ is Cauchy: the absolute difference of consecutive terms is at most $C{4}^{-(j+1)}$. Define the canonical height of an elliptic curve byIt is nonnegative and satisfies $\hat{h}(2P)=4\hat{h}(P)$. The factor one-half is the conventional normalization for the divisor $(O)$, since the first-coordinate height belongs to $2(O)$.

The addition-divisor identity also gives the bounded-error height identityuniformly in $P,Q$. This is the height form of the identity for the sum and difference pullbacks of the line bundle associated to $2(O)$; it includes points where the affine addition formulas have a zero denominator. Apply it to $2^{j}P,2^{j}Q$, divide by $2\cdot 4^j$, and take the limit. Thus $\hat{h}$ satisfies the parallelogram law, and its polarization is a bilinear form, the height pairing. Nonnegativity implies the Cauchy-Schwarz inequality for that pairing, by evaluating its nonnegative quadratic polynomial on $mP+nQ$ for all integers $m,n$ and approximating real ratios by rational numbers. In particularThese properties do not presuppose finite generation. Bounded canonical height gives a finite set by the bounded difference and the Northcott theorem. Torsion points have canonical height zero. Conversely, if $\hat{h}(P)=0$, all $2^{j}P$ have bounded naive height, so two coincide; their difference makes $P$ torsion. Thus the height detects precisely the free part, once finite generation is proved.

Use the Weak Mordell-Weil theorem with $n=2$ and choose finitely many coset representatives $R_i$. Write $P=2Q+R_i$, and put $H={\max }_i\hat{h}(R_i)$. The height inequality yieldsIteration enters the finite set $\{P:\hat{h}(P)\le H+1\}$, since after $j$ steps the height is at most $H+2^{-j}\max (\hat{h}(P)-H,0)$. Unwinding $P=2Q+R_i$ expresses every point using that finite set and the $R_i$. This proves the canonical-height proof of Mordell-Weil finite generation and henceThe Fundamental theorem of finitely generated abelian groups supplies this decomposition. The Mordell-Weil theorem combines a finite quotient from arithmetic Galois cohomology with a contracting height descent; either ingredient alone would not establish finite generation.

Evaluation is surjective as a sheaf morphism: near $p$, any prescribed value is supplied by a constant holomorphic function. We therefore have a short exact sequence of sheaveswhere ${\mathcal{I}}_p$ is the point ideal sheaf and ${\mathcal{O}}_X$ the sheaf of holomorphic functions. By part (c), the skyscraper sheaf is flasque, so $H^0(X,{\mathbb{C}}_p)=\mathbb{C}$ and $H^i(X,{\mathbb{C}}_p)=0$ for $i>0$.

On the connected compact elliptic curve, the maximum modulus principle makes every global holomorphic function constant. Consequently $H^0(X,{\mathcal{O}}_X)=\mathbb{C}$, and evaluation on global sections is an isomorphism. Also $H^1(X,{\mathcal{O}}_X)\cong \mathbb{C}$: by Serre duality for compact complex manifolds, its dual is $H^0(X,K_X)$, generated by the nowhere-zero form $dz$ descended from $\mathbb{C}$. Higher cohomology of ${\mathcal{O}}_X$ vanishes by the Dolbeault theorem in complex dimension one.

The long exact sequence in sheaf cohomology thus beginsIt follows that

The Hasse theorem for elliptic curves states that every elliptic curve $E/{\mathbb{F}}_q$ satisfiesLet $\pi$ be the Frobenius isogeny and put $a=\mathrm{tr}(\pi )$. Its fixed points are exactly $E({\mathbb{F}}_q)$, so separability of $1-\pi$ givesThe degree of an isogeny is a nonnegative quadratic form on the endomorphism ring, and for all integers $m,n$,If $a^2>4q$, this quadratic polynomial has two real roots and takes a negative value at some rational $m/n$ between them, hence after clearing denominators at some integer pair $(m,n)$. Therefore $a^2\le 4q$, and substituting $a=q+1-\#E({\mathbb{F}}_q)$ proves the bound. This is the degree-form proof of the Hasse bound.

If $p\mid D$, square-freeness gives the cuspidal reduction $y^2=x^3$. Its nonsingular group is isomorphic to the additive group ${\mathbb{F}}_p$, so it is cyclic of order $p$.

Suppose $p\nmid D$. The reduction is an elliptic curve. Since $p\equiv 3(\mathrm{mod}4)$, $-1$ is a quadratic nonresidue. Pairing $x$ with $-x$ in the quadratic-character sum and usingshows that the two contributions cancel. Therefore $\#\tilde{E}({\mathbb{F}}_p)=p+1$. The three roots $0,D,-D$ are distinct, so the full 2-torsion is rational. A cyclic group has at most two elements killed by $2$, hence $\tilde{E}({\mathbb{F}}_p)$ is noncyclic. Thus

Kummer theory begins with the exact sequenceGalois cohomology and Hilbert theorem 90 identifyso cyclic extensions of exponent dividing $n$ are described by adjoining $n$th roots when $K$ contains ${\mu }_n$.

For an elliptic curve $E/K$, multiplication by $n$ gives the Kummer exact sequence of an elliptic curveIts connecting homomorphism is the injective Kummer map of an elliptic curvewhere $nQ=P$. Passing to the finite division field of an elliptic curve $K(E[n])$ makes $E[n]$ constant. Rational functions whose divisors are $n(T)-n(O)$ then express the classes through finitely many elements of $K(E[n]{)}^{\times }/K(E[n]{)}^{\times n}$.

Let $S$ contain the primes above $n$, the primes of bad reduction of an elliptic curve and the finitely many primes introduced by these functions. The local theory of good reduction shows that every Kummer class coming from $E(K)$ is unramified outside $S$, so it lies in an S-unramified power class group. Such a group is finite: valuations outside $S$ vanish modulo $n$, the ideal class group is finite, and the Dirichlet unit theorem makes the group of $S$-units modulo $n$th powers finite. The kernel created by passing to $K(E[n])$ is finite by finite-group Galois cohomology. HenceThis is the Kummer-theoretic proof of the weak Mordell-Weil theorem. Combining it with the height descent lemma proves the Mordell-Weil theorem: $E(K)$ is a finitely generated abelian group.

A minimal algebraic surface is a smooth projective surface containing no exceptional curve of the first kind, namely no smooth rational curve of self-intersection $-1$. An abelian surface contains no rational curve because every morphism from ${\mathbb{P}}^1$ to an abelian variety is constant. It therefore has no $(-1)$-curve and is minimal.

For every $n\ge 2$, the Minimal Hirzebruch surface ${\mathbb{F}}_n$ is a rational minimal surface. These surfaces are pairwise nonisomorphic: the negative section is the unique irreducible curve of negative self-intersection and has square $-n$, so an isomorphism would recover $n$. Thus there are infinitely many nonisomorphic minimal rational surfaces.

A K3 surface is a smooth projective surface $X$ with $K_X\cong {\mathcal{O}}_X$ and $H^1(X,{\mathcal{O}}_X)=0$. For a smooth curve $C\subset X$ of geometric genus $g$, the adjunction formula givesand hence

An elliptic surface is a smooth projective surface with a morphism to a smooth curve whose generic fiber is a smooth genus-one curve. If $E$ is an elliptic curve, projectionis an elliptic fibration. Its canonical bundle is pulled back from $K_{{\mathbb{P}}^1}$, so every positive pluricanonical space vanishes and the Kodaira dimension is $-\infty$. This supplies the requested negative-Kodaira-dimension example.

For an elliptically fibered K3 surface, choose a smooth quartic $X\subseteq {\mathbb{P}}^3$ containing a line $L$. The canonical bundle of a smooth projective hypersurface formula makes $K_X$ trivial, and the standard cohomology sequence gives $H^1(X,{\mathcal{O}}_X)=0$, so $X$ is K3. The pencil of planes through $L$ cuts $X$ into $L$ plus a residual plane cubic. The residual linear system $|H-L|$ is basepoint-free, has square zero and defines a morphism $X\to {\mathbb{P}}^1$ whose generic fiber is a smooth plane cubic. This is the Elliptic K3 surface from a quartic containing a line.

A surface of general type is a smooth projective surface of Kodaira dimension two. Let $B\subseteq {\mathbb{P}}^2$ be a smooth plane curve of degree eight and letbe the degree-two cover branched along $B$. The branch-cover canonical-bundle formula givesThis divisor is ample, so $X$ is a double plane of general type and $\pi$ is the required finite morphism of degree two.

For an endomorphism $\varphi$ of an elliptic curve, define its trace of an elliptic-curve endomorphism byPolarizing the quadratic form $\mathrm{deg}$ shows that this is the integer for whichwhere $\hat{\varphi }$ is the dual isogeny. ConsequentlyAlsoThe left side is $[\mathrm{tr}({\varphi }^2)]$, so

Let $q(\varphi )=\mathrm{deg}\varphi$ for nonzero elements of the homomorphism group of elliptic curves, and $q(0)=0$. For an elliptic curve $E$, the degree-two coordinate $x$ on a Weierstrass model has a double pole at its identity and identifies a generic point only with its inverse. Thus on $E\times E$,where the two diagonals are $P=Q$ and $P=-Q$. The resulting line bundle isomorphism is $a^{\ast }{\mathcal{O}}_E(O)\otimes s^{\ast }{\mathcal{O}}_E(O)\cong {\mathrm{pr}}_1^{\ast }{\mathcal{O}}_E(O{)}^{\otimes 2}\otimes {\mathrm{pr}}_2^{\ast }{\mathcal{O}}_E(O{)}^{\otimes 2}$, for the sum and difference maps $a,s$. Pull it back along $(\varphi ,\psi )$ and take degrees. This gives the degree parallelogram law, even when one homomorphism equals plus or minus the other: bundle restriction remains valid in these cases.

It implies $q(n\varphi )=n^{2}q(\varphi )$ by the second-difference recursion, and polarization gives the symmetric bilinear form $B(\varphi ,\psi )=(q(\varphi +\psi )-q(\varphi -\psi ))/4$. To check additivity, the parallelogram law gives $B(x+z,y)+B(x-z,y)=2B(x,y)$; interchange $x,z$ and use oddness to get $B(x+z,y)-B(x-z,y)=2B(z,y)$. Adding proves bilinearity. Since a nonzero isogeny has positive degree, $q$ is positive on every nonzero element. On every finite-dimensional rational span the matrix of $B$ is rational. Positivity on rational vectors implies nonnegativity on real vectors by density; a nontrivial kernel of a rational matrix would contain a nonzero rational vector. Hence the real extension is positive definite. This proves the positive-definite quadratic form property, rather than only a parallelogram identity.

For an elliptic curve over ${\mathbb{F}}_p$, put $N=\#E({\mathbb{F}}_p)$ and $a=p+1-N$. The trace of the square of an elliptic-curve endomorphism gives $\mathrm{tr}({\pi }^2)=a^2-2p$ for the Frobenius isogeny. Therefore $\#E({\mathbb{F}}_{p^2})=p^2+1-a^2+2p=N(2p+2-N)$.

If $\ell$ is a prime different from the characteristic, multiplication by $\ell$ on an elliptic curve has degree ${\ell }^2$ and nonzero differential. It is a separable isogeny, so its geometric kernel has ${\ell }^2$ points, all killed by $\ell$. This gives the displayed group over an algebraic closure; the rational subgroup over the base field may be smaller.

For an elliptic curve with good reduction over ${\mathbb{Q}}_2$, let $E_1$ be the reduction kernel and $E_2$ its formal subgroup with parameter in $4{\mathbb{Z}}_2$. The deep logarithm subgroup of a formal group makes $E_2$ torsion-free, while $[E_1:E_2]=2$. Thus the torsion of the reduction kernel has order at most $2$. Combined with the Hasse theorem for elliptic curves, rational torsion of order $16$ forces bad reduction at $2$.

The degree parallelogram identity for endomorphisms of an elliptic curve gives $d_{n+1}+d_{n-1}=2d_n+2$ for $d_n=\mathrm{deg}[n]$, with $d_0=0$ and $d_1=1$. Induction gives $d_n=n^2$ in every characteristic. A zero pullback on differentials therefore does not mean that multiplication is the zero morphism.

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

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