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For a nonsingular Short Weierstrass form $y^2=x^3+ax+b$ over a number field, the unordered pair of coordinates $x(P+Q),x(P-Q)$ depends only on $x(P),x(Q)$. As a binary quadratic's roots, that pair is represented in ${\mathbb{P}}^2$ by the coefficientswith polynomial $C{T}^2-BT+A$. Homogenizing gives three forms of bidegree $(2,2)$ on ${\mathbb{P}}^1\times {\mathbb{P}}^1$. They have no common zero. Off the diagonal $C\ne 0$; on the finite diagonal $x_1=x_2=t$, $B=4f(t)$ and $A=f^{\prime }(t{)}^2-8tf(t)$ for $f(t)=t^3+at+b$, so simultaneous vanishing would make a multiple root of $f$. At $(\infty ,\infty )$, the homogeneous $A$ is nonzero. Therefore they define a global morphism, also when a sum or difference is the identity.

For a Short Weierstrass form $y^2=x^3+Ax+B$, use $c_4=-48A$ and $\Delta =-16(4A^3+27B^2)$. Both curves have $A=-1$, so $c_4=48=3$ in ${\mathbb{F}}_5$. For $E_1$, $B=0$ gives ${\Delta }_1=64=4$; for $E_2$, $B=1$ gives ${\Delta }_2=-368=2$. ThusThe j-invariant of an elliptic curve is preserved by every isomorphism over the algebraic closure. Since these values differ, the two curves are not geometrically isomorphic, even though the preceding part gives a degree-two elliptic isogeny.

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

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

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

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

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

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

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

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

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.

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 $(x,y)\mapsto (u^{2}x,u^{3}y)$.

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.