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Translate the rational point of order two to $(0,0)$ and complete the square. After clearing denominators by admissible scaling, writeThe companion curve is $E^{\prime }:Y^2=X(X^2-2aX+a^2-4b)$. The two-isogeny formula gives maps $\varphi :E\to E^{\prime }$ and $\psi :E^{\prime }\to E$ satisfying $\psi \varphi =[2]$.

For the Weak Mordell-Weil theorem in this setting, use the two-torsion square-class homomorphismand its counterpart ${\alpha }^{\prime }$ on $E^{\prime }$. These are homomorphisms: a nonvertical line intersecting the cubic in three points has the product of their $x$-coordinates a square, by comparison of the constant term of the intersection cubic. For a line through $(0,0)$, the product of the other two coordinates is $b$, explaining the special value there. The kernels are $\psi E^{\prime }(\mathbb{Q})$ and $\varphi E(\mathbb{Q})$, respectively, as follows by solving the elliptic isogeny preimage quadratic and testing its discriminant for a square.

The prime-support bound in two-isogeny descent makes both images finite. Indeed at a prime not dividing $b$, a negative valuation of $x$ makes the leading cubic term dominate, giving $2v(y)=3v(x)$; a positive valuation makes $x^2+ax+b$ a unit, giving $2v(y)=v(x)$. In either case $v(x)$ is even. Thus the square class of $x$ is supported only on primes dividing $b$, together with a possible sign. The same argument uses $b^{\prime }=a^2-4b$ on $E^{\prime }$. Since $2E=\psi \varphi E$, the two finite elliptic isogeny quotients imply that $E(\mathbb{Q})/2E(\mathbb{Q})$ is finite.

Finite generation still needs a descent on height. Let $h_x$ be the logarithmic height of the $x$-coordinate. The duplication formula has a degree-four rational $x$-coordinate map with coprime numerator and denominator. Polynomial height estimates, with the resultant supplying the lower bound, give the height growth under a morphism of the projective line relation $h_x(2P)=4h_x(P)+O(1)$. Thus the convergent telescoping limitdefines the canonical height of an elliptic curve, without assuming finite generation. It is nonnegative, satisfies $\hat{h}(2P)=4\hat{h}(P)$ and the height parallelogram identity, and differs from $h_x/2$ by a bounded amount. Consequently its bounded subsets of rational points are finite by the Northcott theorem.

Choose the finite list $R_i$ of representatives modulo $2E(\mathbb{Q})$ and let $H={\max }_i\hat{h}(R_i)$. For $P=2Q+R_i$, the canonical height pairing inequality givesRepeated halving brings the height below $H+1$. There are only finitely many such terminal points, and recovering $P$ expresses it as an integer combination of those points and the $R_i$. This proves the canonical-height proof of Mordell-Weil finite generation, so

The propositional completeness theorem states that $\Gamma \models \varphi$ if and only if $\Gamma ⊢\varphi$. Soundness follows by checking the axioms against truth valuations and noting that modus ponens rule preserves truth. For completeness, use the deduction theorem for propositional logic: if $\Gamma \cup \{p\}⊢q$, then $\Gamma ⊢p\Rightarrow q$. In the usual negation rules this also says that a contradiction from $\Gamma \cup \{p\}$ proves $\Gamma ⊢\neg p$.

First construct a maximal consistent set extending any consistent set $\Delta$. Enumerate all formulas, possible because the primitive propositions are countable. Starting with ${\Delta }_0=\Delta$, add formula ${\psi }_n$ at step $n$ if consistency is preserved; otherwise add $\neg {\psi }_n$. This latter addition is consistent: if both extensions gave contradictions, the deduction theorem for propositional logic would prove both $\neg {\psi }_n$ and $\neg \neg {\psi }_n$ from ${\Delta }_n$, a contradiction already in ${\Delta }_n$. The union $M$ is consistent because a finite proof uses only finitely many newly added premises. It decides every formula: exactly one of $\psi ,\neg \psi$ belongs to $M$. It is deductively closed, since an omitted derivable formula would have its negation in $M$, contradicting consistency.

Assign a valuation $v(p)=1$ exactly when the primitive proposition $p$ lies in $M$. Structural induction proves the maximal consistent set truth lemma $v(\psi )=1$ if and only if $\psi \in M$. Negation follows from the decision property. For implication, if $\psi \Rightarrow \chi \in M$ and $\psi \in M$, modus ponens rule gives $\chi \in M$. Conversely, if $\chi \in M$, the elementary implication axioms give $\psi \Rightarrow \chi \in M$. If $\psi \notin M$, then $\neg \psi \in M$, and the usual contradiction rule proves $\psi \Rightarrow \chi$. These two alternatives match the truth table. The other connectives, if primitive, are handled by their corresponding rules or defined from negation and implication. Thus every consistent set has a valuation satisfying it.

Now suppose $\Gamma /⊢\varphi$. If $\Gamma \cup \{\neg \varphi \}$ were inconsistent, the deduction theorem for propositional logic would give $\Gamma ⊢\neg \neg \varphi$. Here the third axiom $\neg \neg \varphi \Rightarrow \varphi$ is essential: it would give $\Gamma ⊢\varphi$, a contradiction. Hence that extension is consistent and has a satisfying valuation. This valuation satisfies $\Gamma$ and falsifies $\varphi$, proving the contrapositive of completeness. In this organization of the proof, the explicit classical double-negation elimination occurs in this last consistency step; the construction and truth lemma use the ordinary negation and contradiction rules, not an unmentioned identification of $\neg \neg \varphi$ with $\varphi$.

The propositional compactness theorem says that a set of formulas is satisfiable exactly when each finite subset is satisfiable. If it were not satisfiable, completeness would make it inconsistent. Its finite contradiction proof uses only a finite subset of premises; soundness makes that subset unsatisfiable. This contradicts the assumed finite satisfiability and proves compactness.

For a finite extension of fields complete for a Non-Archimedean absolute value, use the unique extended field absolute value. The extension is unramified when its residue extension is a separable field extension andEquivalently, it has ramification index one, a separable field extension of residue fields, and no defect. For general nondiscrete valued fields, merely saying that the ramification index is one is insufficient; the displayed degree equality is part of the unramified condition.

Put $n=[L:K]=[k_L:k_K]$. Since $\bar{x}$ generates the residue extension, $1,\bar{x},\dots ,{\bar{x}}^{n-1}$ are linearly independent over $k_K$. Their lifts $1,x,\dots ,x^{n-1}$ are linearly independent over $K$: a nonzero relation could be divided by a coefficient of largest field absolute value, giving an integral relation whose reduction has at least one nonzero coefficient, contrary to residue independence. They therefore form a $K$-basis of $L$.

For any $y={\sum }_{j=0}^{n-1}{a}_j{x}^j$, choose $a_r$ of largest field absolute value. All coefficients of $y/a_r$ belong to ${\mathcal{O}}_K$, and at least one is a unit. Its reduction is a nonzero linear combination of the residue basis, so $y/a_r$ is a unit of ${\mathcal{O}}_L$. This proves the residue-basis norm formula for an unramified extensionHence $y\in {\mathcal{O}}_L$ exactly when every $a_j\in {\mathcal{O}}_K$. We concludeThe last equality also uses the fact that every polynomial in the integral element $x$ is integral. The proof requires neither a uniformizer nor discrete valuation, and applies to every residue generator specified in the question.

Every nonzero P-adic number has a unique decomposition $z=p^{m}u$ with $m\in \mathbb{Z}$ and $u\in {\mathbb{Z}}_p^{\times }$. The group homomorphism kernel of reduction from the unit group onto ${\mathbb{F}}_p^{\times }$ is $1+p{\mathbb{Z}}_p$, the group of principal units. Each nonzero residue class has a unique lift $\omega$ satisfying ${\omega }^{p-1}=1$, by the simple-root Hensel lemma proved in question 3. Uniqueness makes these lifts multiplicative. They form the cyclic group ${\mu }_{p-1}$, and every unit has a unique factorizationThus ${\mathbb{Q}}_p^{\times }=p^{\mathbb{Z}}\times {\mu }_{p-1}\times (1+p{\mathbb{Z}}_p)$.

For odd $p$, the p-adic logarithm and p-adic exponential function give inverse group isomorphisms $1+p{\mathbb{Z}}_p\leftrightarrow p{\mathbb{Z}}_p$. To verify the convergence domain, for $t\in p{\mathbb{Z}}_p$ the p-adic logarithm terms have valuations $n{v}_p(t)-v_p(n)\to \infty$, and the exponential terms have valuations $n{v}_p(t)-v_p(n!)\to \infty$, since $v_p(n!)\le (n-1)/(p-1)$ and $p>2$. In both series every term after the linear one has strictly larger valuation than the linear term. Consequently $\log (1+t)\in p{\mathbb{Z}}_p$ and $\exp (t)\in 1+p{\mathbb{Z}}_p$. The formal identities $\log (vw)=\log v+\log w$ and $\exp (\log v)=v$, $\log (\exp t)=t$ hold on these convergent domains, as follows by multiplying the convergent series or passing to their formal identities termwise.

Choose a generator $\zeta$ of ${\mu }_{p-1}$. An explicit isomorphism isThe inverse reads off the valuation, the root-of-unity unit factor, and $p^{-1}\log (v)$. Both directions are continuous with the standard product topology, so this is also a topological group isomorphism. The odd-prime condition is essential for using all of $1+p{\mathbb{Z}}_p$ as the p-adic logarithm-isomorphism domain.

We first prove the simple-root Hensel lemma directly. Put $x_1=x$. Suppose $x_n\equiv x(\mathrm{mod}\pi )$ and $f(x_n)\equiv 0(\mathrm{mod}{\pi }^n)$. The derivative remains a unit since its reduction agrees with $f^{\prime }(x)$. Choose $t_n\in {\mathcal{O}}_K$ whose residue solvesand set $x_{n+1}=x_n+{\pi }^n{t}_n$. Polynomial expansion gives $f(x_{n+1})=f(x_n)+{\pi }^n{t}_n{f}^{\prime }(x_n)+{\pi }^{2n}B$ with $B\in {\mathcal{O}}_K$. Since $2n\ge n+1$, the new value vanishes modulo ${\pi }^{n+1}$. The sequence is Cauchy because $v(x_{n+1}-x_n)\ge n$, and completeness supplies $y\in {\mathcal{O}}_K$ with $y\equiv x(\mathrm{mod}\pi )$. Continuity gives $f(y)=0$.

For uniqueness, if $y,z$ are two roots in that residue class, thenThe factor in parentheses is a unit, since $f^{\prime }(z)$ is a unit and $y-z\in \pi {\mathcal{O}}_K$. Thus $y=z$. This proves existence and uniqueness of the specified lifted root.

Now suppose the residue field has $q=p^r$ elements and put $e=v(p)$. Every element of $k_K^{\times }$ is a simple root of $T^{q-1}-1$. The Hensel lemma therefore lifts its $q-1$ roots uniquely to a subgroup of the roots of unity in $K$. These are the Teichmuller representatives and give all the prime-to-$p$ torsion. To verify the latter assertion, if a root of unity $u\equiv 1$ has order $m$ prime to $p$, write $u=1+t$. In $u^m-1=mt+\left(\genfrac{}{}{0ex}{}{m}{2}\right)t^2+\cdots$, the first term has valuation $v(t)$ and all later terms have greater valuation, so the sum cannot vanish unless $t=0$. Reduction is therefore injective on prime-to-$p$ torsion, whose order divides $q-1$.

Any root of unity has a unique decomposition into a $p$-power-order factor and a prime-to-$p$ factor, by splitting its finite order using coprime powers. For a primitive $p^a$th root $\zeta$, $a\ge 1$, reduction in characteristic $p$ gives $\bar{\zeta }=1$. The cyclotomic identityshows that all factors have the same valuation: $({\zeta }^j-1)/(\zeta -1)=1+\zeta +\cdots +{\zeta }^{j-1}$ reduces to $j\ne 0$. Consequently the valuation of a primitive p-power root of unity isThe left side is a positive integer. Hence $p^{a-1}(p-1)\le e$, so the possible $p$-power orders are bounded. Combining this bound with the prime-to-$p$ result bounds all root orders by one integer, proving finiteness. A finite multiplicative subgroup of a field is cyclic, and it contains exactly the full prime-to-$p$ subgroup of order $q-1$. ThusIf $e<p-1$, even a primitive $p$th root would violate the valuation formula, so $s=0$.

At the boundary $e=p-1$, assume first that $p$ is odd. Take $K_{+}={\mathbb{Q}}_p({\zeta }_p)$. The polynomial ${\Phi }_p(1+T)=p+\left(\genfrac{}{}{0ex}{}{p}{2}\right)T+\cdots +T^{p-1}$ is Eisenstein, so $K_{+}/{\mathbb{Q}}_p$ is totally ramified of degree $p-1$ and $v_{K_{+}}(p)=p-1$. It contains a primitive $p$th root, while the valuation formula excludes primitive $p^2$th roots. Thus $s=1$.

For the other example, take $K_0={\mathbb{Q}}_p(\pi )$ with ${\pi }^{p-1}=p$. The polynomial $T^{p-1}-p$ is Eisenstein, giving the same ramification index and residue field ${\mathbb{F}}_p$. If a primitive $p$th root belonged to $K_0$, the valuation formula would give ${\zeta }_p-1=u\pi$ with $u$ a unit. Substitute this into ${\Phi }_p(1+({\zeta }_p-1))=0$ and divide by $p$. All intermediate terms reduce to zero, leaving $1+{\bar{u}}^{p-1}=0$ in ${\mathbb{F}}_p$. But ${\bar{u}}^{p-1}=1$, contradicting $2\ne 0$. Thus $s=0$ in $K_0$. This proves both cases of roots of unity at the tame ramification boundary for odd primes.

The printed boundary claim needs the qualification $p>2$. When $p=2$, every characteristic-zero field already contains the root $-1$ of order two, so $s=0$ is impossible. If $v(2)=1$, the valuation formula excludes order four and higher, forcing $s=1$. For example $K={\mathbb{Q}}_2$ has exactly this behavior. This is a genuine exception to the final request as printed, rather than an omitted construction.

Choose compatible primitive roots of unity ${\zeta }_{p^{n+1}}$, with ${\zeta }_{p^{n+2}}^p={\zeta }_{p^{n+1}}$. Use the slightly faster-converging series permitted by “or otherwise”:The terms have field absolute value $p^{-2n}$, so the partial sums are a Cauchy sequence in ${\overline{\mathbb{Q}}}_p$. We show that an algebraic limit would contradict the growth of cyclotomic extension degrees. The faster weights make the strict conjugate-distance estimate work uniformly also at $p=2$.

Put $L_N={\mathbb{Q}}_p({\zeta }_{p^{N+1}})$. Its degree is $p^N(p-1)$: the translated cyclotomic polynomial ${\Phi }_{p^{N+1}}(1+T)$ has constant term $p$ and reduces modulo $p$ to $T^{p^N(p-1)}$, so is Eisenstein. To verify the reduction, write it as ${\sum }_{j=0}^{p-1}(1+T{)}^{j{p}^N}$ and use characteristic $p$ to reduce this to ${\sum }_{j=0}^{p-1}(1+T^{p^N}{)}^j=T^{p^N(p-1)}$. All roots lie in $L_N$, and its Galois automorphisms are ${\sigma }_a(\zeta )={\zeta }^a$ for units $a$ modulo $p^{N+1}$.

For a nonidentity automorphism, put $r=v_p(a-1)$, choosing a representative with $0\le r\le N$. It fixes the terms with $n<r$. For $n\ge r$, the root ${\zeta }_{p^{n+1}}^{a-1}$ has order $p^{n+1-r}$, so the root-of-unity valuation formula, with $v_p(p)=1$, givesThis is strictly increasing as $n$ increases: successive differences are $2-p^{-(n-r+1)}>0$. Thus the first changed term is the unique term of smallest valuation and cannot cancel. ConsequentlyIn particular no nonidentity automorphism fixes $A_N$, so ${\mathbb{Q}}_p(A_N)=L_N$. Also the distance from $A_N$ to every distinct conjugate is at least $p^{-2N-1/(p-1)}$.

Suppose the Cauchy sequence converged to $A\in {\overline{\mathbb{Q}}}_p$. The tail estimate givesApply Krasner's lemma with $\alpha =A_N$, $\beta =A$. It forces $L_N={\mathbb{Q}}_p(A_N)\subseteq {\mathbb{Q}}_p(A)$ for every $N$. The right-hand side has finite degree, while $[L_N:{\mathbb{Q}}_p]=p^N(p-1)$ is unbounded, a contradiction. ThereforeThis proves algebraic closure of the p-adic numbers is not complete by an explicit algebraic Cauchy sequence and a fully quantified conjugate-separation argument.

The valuation ring is $\mathcal{O}=\{x\in K:|x|\le 1\}$, and $\mathfrak{m}=\{x\in K:|x|<1\}$ is an ideal by the ultrametric inequality. Its complement inside $\mathcal{O}$ consists exactly of elements of field absolute value one, whose inverses also lie in $\mathcal{O}$. Every proper ideal must avoid units and hence lie in $\mathfrak{m}$. Therefore $\mathfrak{m}$ is the unique maximal ideal: $\mathcal{O}$ is a local ring. Also its fraction field is $K$, since every nonzero element outside the ring has its inverse inside it.

If $z\in K$ is integral over $\mathcal{O}$, it satisfies a monic equation $z^n+a_{n-1}{z}^{n-1}+\cdots +a_0=0$ with $|a_i|\le 1$. Were $|z|>1$, the leading term would have strictly larger field absolute value than every other term, so the ultrametric inequality would prevent cancellation to zero. Thus $z\in \mathcal{O}$, proving the valuation ring is integrally closed.

For the ideal criterion, assume the valuation is nontrivial and write $v=-\log |\cdot |$. If its value group is discrete, normalize it to $\mathbb{Z}$ and choose a uniformizer $\pi$ of valuation one. In any nonzero ideal $I$, the nonnegative integer valuations have a least value $m$. Choose $a\in I$ of that value. For every $b\in I$, $v(b/a)\ge 0$, so $b\in a\mathcal{O}$. Hence $I=(a)=({\pi }^m)$ and $\mathcal{O}$ is a principal ideal domain.

Conversely, if $\mathcal{O}$ is a principal ideal domain, its nonzero maximal ideal is $(\pi )$. For every element $x$ of positive valuation, $x\in \mathfrak{m}$ implies $x=\pi y$ with $y\in \mathcal{O}$, so $v(x)\ge v(\pi )>0$. Thus $\gamma =v(\pi )$ is the least positive value. For any value $t$, subtract $\lfloor t/\gamma \rfloor \gamma$. The remainder belongs to the value group and lies in $[0,\gamma )$, so is zero. The value group is therefore $\gamma \mathbb{Z}$, and the valuation is discrete. We have proved the principal ideal criterion for a nontrivial rank-one valuation ring.

The nontriviality qualification matters under the standard definition of discrete valuation, which requires value group isomorphic to $\mathbb{Z}$. With the trivial field absolute value, $\mathcal{O}=K$ is still a principal ideal domain, but the value group is zero and there is no uniformizer. Thus the literal unrestricted equivalence has a field exception. If “discrete” instead includes the trivial value group as a discrete subgroup of $\mathbb{R}$, that exceptional case satisfies the equivalence too.

Arithmetic heights measure arithmetic complexity. For a reduced rational number $x=a/b$, its multiplicative height is $H(x)=\max (|a|,|b|)$ and its logarithmic height is $h(x)=\log H(x)$. Over a number field $K$, use the absolute projective heightNormalize the absolute values on a field by the product formula. That formula makes arithmetic height independent of the chosen homogeneous representative, and the normalization makes it independent of enlarging $K$. The Northcott theorem says that algebraic points of bounded degree and bounded arithmetic height form a finite set. For rational numbers this follows immediately by bounding the numerator and denominator. In a fixed number field, the same finiteness follows by bounding the coefficients of the minimal polynomial using its conjugates and the arithmetic height.

For an elliptic curve put $h_x(P)=h([x(P):1])$ and $h_x(O)=0$. The two points above a given $x$ show that bounded $h_x$ gives finitely many $K$-rational points. More importantly, multiplication expands this arithmetic height. To see the mechanism, choose a short Weierstrass equation of an elliptic curve $y^2=x^3+Ax+B$ over $K$. Doubling givesAfter homogenization the numerator and denominator are quartics with no common zero. At infinity the numerator is nonzero; at a root $t$ of the cubic the numerator becomes $(3t^2+A{)}^2$, nonzero by nonsingularity. For a degree-$d$ morphism of the projective line, fixed homogeneous coefficients give the upper arithmetic height bound $h(f(x))\le dh(x)+C$. A nonzero resultant supplies the reverse bound: at each place it bounds the common cancellation of the two homogeneous values, with a nontrivial constant needed at only finitely many places. Applying these bounds to duplication gives a constant $C$ independent of $P$ withThis includes points mapping to $O$, by the projective formulation.

The canonical height removes the bounded errors. With the conventional factor one half, defineThe successive terms before multiplying by one half differ by at most $C{4}^{-n-1}$. The limit exists, and summing this geometric bound givesHence bounded canonical height still gives finitely many points over the fixed number field.

To obtain the pairing, the quadratic nature comes from a uniform arithmetic height parallelogram estimate, not just from doubling. If $x=x(P)$ and $z=x(Q)$, the unordered coordinates $x(P+Q),x(P-Q)$ are the roots of the binary quadratic with coefficientsThese formulas follow by adding and multiplying the two chord expressions with slopes $(y(P)-y(Q))/(x-z)$ and $(y(P)+y(Q))/(x-z)$. Their homogenizations have bidegree $(2,2)$ and no simultaneous zero. Off the diagonal $C_2\ne 0$; on the finite diagonal $x=z=t$, $C_1=4f(t)$ and $C_0=f^{\prime }(t{)}^2-8tf(t)$ for $f(t)=t^3+At+B$, so simultaneous vanishing would contradict nonsingularity. At the double point at infinity the homogeneous $C_0$ is nonzero. Local coefficient norms, or the corresponding resultant bounds, therefore giveThe arithmetic height of a binary quadratic's coefficient vector differs by $O(1)$ from the sum of the heights of its two roots. At finite places this follows from multiplicativity of the maximum coefficient norm of products; at infinite places the fixed-degree norms are comparable. ThusApply this to $2^{n}P,2^{n}Q$, divide by $2\cdot 4^n$ and take limits. The errors vanish and give the exact height parallelogram identity. Polarization consequently makesa symmetric bilinear pairing. Nonnegativity of $\hat{h}(aP+bQ)$ for integers $a,b$, and density of rational ratios, give the Cauchy-Schwarz inequality. In particular $\hat{h}(P-Q)\le 2\hat{h}(P)+2\hat{h}(Q)$. Also $\hat{h}(mP)=m^2\hat{h}(P)$ for integers $m$. Torsion points of an elliptic curve have canonical height zero because their doubling orbit is finite; conversely, if a point has canonical height zero, all its integer multiples have bounded naive height, and finiteness forces two multiples to coincide. Thus canonical height vanishes exactly on torsion.

There is an arithmetic finiteness input before height descent: $E(K)/2E(K)$ is finite, the Weak Mordell-Weil theorem. Here is its mechanism. The multiplication exact sequence gives an injective Kummer map $E(K)/2E(K)\to H^1(K,E[2])$: choose a half-point $Q$ and use the cocycle $\sigma \mapsto \sigma Q-Q$; changing $Q$ changes a coboundary, and a trivial class means a half-point can be chosen over $K$. Let $L=K(E[2])$, a Finite Galois extension. Over $L$ the torsion module is $(\mathbb{Z}/2\mathbb{Z}{)}^2$, so Kummer theory identifies its first Galois cohomology with two copies of $L^{\times }/L^{\times 2}$.

Only classes unramified outside a fixed finite set $S$ occur. Include the primes over $2$, bad-reduction primes and the ramified primes of $L/K$. At a good prime away from $2$, multiplication by two on the proper smooth elliptic model is finite étale morphism, so the torsor of halves of a point is unramified. In square-class language, valuations outside $S$ are even. Such classes form a finite group: write the principal ideal of a representative as a square times an ideal supported on $S$. The ideal class gives an element of the finite two-torsion of the localized ideal class group; after fixing that class, ambiguity is an $S$-unit modulo squares. The $S$-unit theorem makes the latter finite. The group kernel of restriction from $K$ to $L$ is contained in $H^1(\mathrm{Gal}(L/K),E[2])$, which is finite because both the group and module are finite. This proves the weak theorem, without assuming finite generation of $E(K)$.

Finally choose representatives $R_1,\dots ,R_s$ of $E(K)/2E(K)$ and put $M={\max }_i\hat{h}(R_i)$. Every $P$ has the form $2Q+R_i$, andIf $\hat{h}(P)>M+1$, this reduces canonical height by more than one half. Iterate until a point of canonical height at most $M+1$ is reached. There are only finitely many such points, by Northcott and the bounded difference of heights. Reconstructing $P=2Q+R_i$ at every stage proves that these finitely many small points and the finitely many representatives generate $E(K)$. ThereforeThe canonical pairing is positive definite on the free part. It turns rank computations into an arithmetic lattice problem, supplies independence tests for proposed generators, and bounds searches once a arithmetic height bound is available. Weak finite-quotient finiteness alone would not prove this result: the arithmetic height contraction is the step that converts it into finite generation.

Here $F$ is the usual one-dimensional commutative formal group law. Normalize the valuation $v$ by $v(\pi )=1$, and put $e=v(p)$. We show explicitly that for any integer $m>e/(p-1)$,This subgroup will have finite index in $F(\pi {\mathcal{O}}_K)$.

Set $a(T)={\partial }_{Y}F(T,0)$. Since $F(X,Y)=X+Y+$ terms of total degree at least two, $a(T)\in 1+T{\mathcal{O}}_K[[T]]$. Its inverse has integral coefficients, say $a(T{)}^{-1}={\sum }_{j\ge 0}{c}_j{T}^j$ with $c_0=1$. Define the formal logarithm over $K$ byDifferentiate associativity $F(F(X,Y),Z)=F(X,F(Y,Z))$ with respect to $Z$ at zero. This gives $a(F(X,Y))={\partial }_{Y}F(X,Y)a(Y)$, soEvaluating at $Y=0$ proves the formal identity $\ell (F(X,Y))=\ell (X)+\ell (Y)$.

The coefficients satisfy $v(c_{n-1}/n)\ge -e{v}_p(n)$. Thus the series converges on $\pi {\mathcal{O}}_K$, since for $v(T)\ge 1$ the valuations of its terms are at least $n-e{v}_p(n)\to \infty$. More quantitatively,Indeed, if $k=v_p(n)$ then $n\ge p^k\ge 1+k(p-1)$. If $v(T)\ge m>e/(p-1)$, every nonlinear term has valuation strictly greater than $v(T)$. Write $h(T)=\ell (T)-T$. For $S,T\in {\pi }^m{\mathcal{O}}_K$, use $T^n-S^n=(T-S){\sum }_{j=0}^{n-1}{T}^{n-1-j}{S}^j$ to obtainThe extra valuation is integral and positive because $(n-1)m-e{v}_p(n)>0$ for all $n\ge 2$.

For any $b\in {\pi }^m{\mathcal{O}}_K$, solve $\ell (T)=b$ by iterating $T\mapsto b-h(T)$. This map preserves the complete ball ${\pi }^m{\mathcal{O}}_K$ and improves differences by at least one valuation step. The contraction mapping theorem gives a unique solution. Therefore $\ell$ is a bijection of this ball onto itself, and the formal identity, now evaluated in convergent series, makes it a group isomorphism. Multiplication by ${\pi }^{-m}$ then identifies the additive target with ${\mathcal{O}}_K$.

Integral coefficients in $F$ and its formal inverse ensure that every $F({\pi }^j{\mathcal{O}}_K)$ is a subgroup. Reduction modulo ${\pi }^m$ respects the group law; two parameters represent the same coset of $F({\pi }^m{\mathcal{O}}_K)$ exactly when they are congruent modulo ${\pi }^m$. If the residue field has $q_K$ elements, the quotient consequently has $q_K^{m-1}$ elements. This proves both the finite-index assertion and the claimed additive structure, with an explicit depth that also works in ramified extensions and when $p=2$.

To compute the rank of an elliptic curve, translate the rational point of order two to $(0,0)$ and put the curve in the form $y^2=x^3+a{x}^2+bx$, with $a,b\in \mathbb{Z}$ after clearing denominators and $b(a^2-4b)\ne 0$. Its two-isogenous partner and the elliptic isogeny areThe maps extend over their exceptional points. The dual isogeny isThe two-torsion square-class homomorphism is $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\times }/{\mathbb{Q}}^{\times 2}$, defined by $\alpha (O)=1$, $\alpha ((0,0))=[b]$, and $\alpha ((x,y))=[x]$ otherwise. Define ${\alpha }^{\prime }$ similarly on $E^{\prime }$ with $b^{\prime }=a^2-4b$.

Here are the ingredients behind the descent procedure. For a generic chord $y=mx+c$, the cubic of intersection coordinates has root product $c^2$. As the third intersection is the negative of the group sum and negation leaves $x$ unchanged, this proves the square-class homomorphism property. For a line through $(0,0)$, the two nonzero intersection coordinates have product $b$, so the assigned class $[b]$ gives the same multiplicativity. A vertical line has two equal coordinates and contributes their square, and $O$ has identity class. These observations also cover tangencies and the exceptional sums. Its kernel is $\hat{\varphi }(E^{\prime }(\mathbb{Q}))$. Indeed, the displayed formula gives $x(\hat{\varphi }(X,Y))=(Y/(2X){)}^2$. Conversely, if a point on $E$ has $x=r^2\ne 0$, the possible preimage coordinates satisfyIts elliptic-curve discriminant is $16(r^4+a{r}^2+b)=16y^2/r^2$, a rational square. A root $X$ and $Y=2rX$, with the sign chosen appropriately, give a preimage. The identity is always in the image. The point $(0,0)$ is in the image exactly when $b$ is a square: its nonzero preimage coordinates are $X=a\pm 2\sqrt{b}$, with $Y=0$. Thus the exceptional points satisfy precisely the same kernel statement. Interchanging the curves similarly gives $\ker {\alpha }^{\prime }=\varphi (E(\mathbb{Q}))$.

A prime not dividing $b$ can occur only with even valuation in an $x$-coordinate. If its valuation is positive, the factor $x^2+ax+b$ is a unit and $2v(y)=v(x)$; if negative, $x^2$ dominates that factor and $2v(y)=3v(x)$. Thus square classes in $\mathrm{im}\alpha$ have signed square-free representatives dividing $b$, giving finitely many candidates. To test a candidate $d$, put $x=d(u/v{)}^2$ and $y=duN/v^3$. The curve equation becomesA nontrivial rational solution, including the appropriate limiting points, is exactly a realization of the class. Clear denominators and arrange $\gcd (u,v)=1$ for congruence tests. Do the same for $E^{\prime }$ with $b^{\prime }$.

The factor in the two-isogeny rank formula can also be checked directly. Let $A=E(\mathbb{Q})$, $B=E^{\prime }(\mathbb{Q})$ and $t=\#A[2]$. The index of $\hat{\varphi }(B)$ in $A$ is $\#\mathrm{im}\alpha$, while the map from $B/\varphi (A)$ onto $\hat{\varphi }(B)/2A$ has kernelThe numerator has order two. Moreover $\ker \hat{\varphi }\cap \varphi (A)=\varphi (A[2])$, since $\hat{\varphi }\varphi P=2P$, and the latter has order $t/2$. Hence this quotient has order $4/t$, andThe Mordell-Weil theorem gives $[A:2A]=2^{r}t$, where $r$ is the rank. Cancelling $t$ yieldsIn practice, local congruence and real-solubility tests remove many candidate quartics, giving an upper bound, while rational point searches realize classes and give a lower bound. These often coincide. Local solubility alone is not a proof of a rational point: an everywhere locally soluble covering can still fail globally, so a remaining gap must be reported rather than treated as an exact rank.

For the curve at hand, $a=8$, $b=-7$, $b^{\prime }=92$, andThe first image is contained in $\{1,-1,7,-7\}$ and contains $1,-7$, the latter from the point $(0,0)$. To exclude $-1$, its quartic would require coprime $u,v$ withModulo $16$, an odd fourth power is $1$ and an even fourth power is $0$. If $u$ is odd and $v$ even the right side is $15$; if $u$ is even and $v$ odd it is $7$; if both are odd it is $14$. None is among the square residues $0,1,4,9$. Thus $-1$ is impossible. The image is a subgroup, so $7$ is also impossible: its product with the known class $-7$ would be $-1$. We obtain $\mathrm{im}\alpha =\{1,-7\}$.

For $E^{\prime }$, its quadratic factor is $X^2-16X+92=(X-8{)}^2+28>0$. Every nonzero rational $X$-coordinate is therefore positive. Prime support leaves only $1,2,23,46$ as possible classes; $1$ and $23=[92]$ are supplied by $O$ and the two-torsion point. The explicit point $(18,48)$ is on $E^{\prime }$ becauseIts class is $[18]=2$. Multiplying classes supplies $46$, so $\mathrm{im}{\alpha }^{\prime }=\{1,2,23,46\}$. The two-isogeny rank formula now provesAs an additional concrete point on the original curve, the dual isogeny sends $(18,48)$ to $(16/9,116/27)$. The descent calculation proves the rank; it does not by itself certify that this point generates the full free quotient.

A rational point of order $2$ can be moved to $(0,0)$, giving an integral modelIts two-isogeny descent partner and the two-isogeny formula areThe maps extend over the identity and $(0,0)$, have these two points as their respective kernels, and direct substitution and the doubling formula give $\hat{\varphi }\varphi =[2]$ and $\varphi \hat{\varphi }=[2]$.

Define the two-torsion square-class homomorphisms byand similarly ${\alpha }^{\prime }$ for $E^{\prime }$, using $b^{\prime }$. To see the homomorphism property, intersect a line $y=mx+c$ with $E$. Its three $x$-coordinates have product $c^2$, since the resulting monic cubic has constant term $-c^2$. The third point is the negative of the sum of the first two, and negation preserves $x$, so the square classes multiply as required. If the line passes through $(0,0)$, the other two coordinates have product $b$, giving exactly the prescribed exceptional value. Vertical lines and repeated intersections give the identity and tangent cases.

We also verify the kernel needed for descent:For a point on $E^{\prime }$, the first coordinate of its image under $\hat{\varphi }$ is $Y^2/(4X^2)$, a square. Conversely, if $P=(s^2,y)$ has square $x$-coordinate, solving $x(\hat{\varphi }(X,Y))=s^2$ givesIts discriminant is $16(s^4+a{s}^2+b)=(4y/s{)}^2$, so it has rational nonzero roots $X=a+2s^2\pm 2y/s$. Taking $Y=\pm 2sX$ gives a point on $E^{\prime }$; choose the signs so its image has second coordinate $y$. The exceptional point $(0,0)$ lies in the image exactly when $b$ is a square, as seen from the points with $Y=0$, $X=a\pm 2\sqrt{b}$. This proves the first kernel formula. Applying the same argument to $E^{\prime }$ and then rescaling the second partner by $(X,Y)\mapsto (X/4,Y/8)$ proves the second.

There are only finitely many candidate square classes. For a rational point write $x=M/e^2$ and $y=N/e^3$, with $\gcd (M,e)=1$. These denominator forms follow because a negative prime valuation of $x$ makes the leading cubic term the unique lowest-valuation term, forcing that valuation to be even. ThenIf a prime divides $M$ but not $b$, the second factor is a unit at that prime, so its valuation in $M$ is even. Thus every class in $\alpha (E)$ has a signed squarefree representative $d\mid b$. This is the prime-support bound in two-isogeny descent; the analogous candidates for ${\alpha }^{\prime }(E^{\prime })$ divide $b^{\prime }$.