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First compute the Mordell-Weil group rank by two-isogeny descent. HereThe two-isogeny formula gives $\varphi :E\to E^{\prime }$. If $u=1+i$, then $u^4=-4$, andis an isomorphism over $K$. Its $x$-coordinate multiplier $u^2$ is a square in $K$.

Use the two-torsion square-class homomorphismFor any prime ideal of the Gaussian integers, if $v(x)>0$, then $x^2+i$ is a unit and $2v(y)=v(x)$; if $v(x)<0$, the term $x^3$ has strictly smallest valuation and $2v(y)=3v(x)$. Hence every valuation of $x$ is even. Since the Gaussian integers form a principal ideal domain, dividing by a square leaves a unit. Their units are $\{1,-1,i,-i\}$, whose square classes are $\{1,i\}$ because $-1=i^2$ and $i$ is not a square in $K$. For the latter assertion, $(a+bi{)}^2=i$ with $a,b\in \mathbb{Q}$ would imply $a^2=b^2$ and $2ab=1$, hence $a^2=1/2$, impossible for rational $a$. ThusBoth classes occur, at $O$ and $T$. This is the unit square-class bound for two-isogeny descent.

Define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly, with ${\alpha }^{\prime }((0,0))=[-4i]=[i]$. Since $\theta$ multiplies nonexceptional $x$-coordinates by a square, and preserves the exceptional classes as well, its image is also $\{1,[i]\}$. The standard kernel identities in two-isogeny descent areHere $\hat{\varphi }$ is the dual isogeny and $\hat{\varphi }\varphi =[2]$. These identities can be checked directly from the formulas: $x(\hat{\varphi }(X,Y))=(Y/(2X){)}^2$, and conversely a square $x$-coordinate lets the quadratic equation for a preimage be solved using the curve equation. For example, if $x=r^2\ne 0$, the equation for $X$ is $X^2-(4x+2a)X+(a^2-4b)=0$, whose discriminant is $16(x^2+ax+b)=16(y/r{)}^2$; the $Y$-coordinate then follows from the dual formula. The exceptional points satisfy the same completed square-class criterion.

The two-isogeny index formula over a number field keeps track of a small kernel factor:In this case $\ker \hat{\varphi }=\{O,T^{\prime }\}$, where $T^{\prime }=(0,0)$, and $T^{\prime }\notin \varphi E(K)$ because ${\alpha }^{\prime }(T^{\prime })=[i]\ne 1$. Thus $\delta =2$, and the index is $2\cdot 2/2=2$. There is only one nonzero rational 2-torsion point on $E$: the other two would require $x^2=-i$, and $-i$ has the same nonsquare class as $i$. The Mordell-Weil theorem now givesso $r=0$.

It remains to identify all torsion, rather than merely the rank. The elliptic-curve discriminant is $-64i^3$, so the primes $(3)$ and $(2-i)$ have good reduction, with residue characteristics three and five. By the supplied point-count information their reduction groups have orders that are powers of two. The reduction of torsion points on an elliptic curve is injective on prime-to-residue-characteristic torsion. Every odd-primary torsion subgroup therefore injects into a group of two-power order at at least one of these two primes, and must be zero. All torsion is two-primary.

Finally, if a point had order four, its double would be $T$. The elliptic-curve addition formula givesFor $2P=T$ the denominator is nonzero, so $x^2=i$, impossible in $K$. A point of higher two-power order would have a multiple of order four, so it too is excluded. Thus the only torsion points are $O,T$. Together with rank zero,

Logarithms on sufficiently deep units. Suppose $\mathrm{char}K=0$, let $\ell$ be the residue characteristic, and put $e=v(\ell )$. For every integer $r>e/(\ell -1)$ the p-adic logarithm and p-adic exponential function, evaluated in $K$, converge on the required domains:The bounds $v(m!)=e{v}_{\ell }(m!)\le e(m-1)/(\ell -1)$ and $v(m)\le e(m-1)/(\ell -1)$ show that, for $x,y\in {\pi }^r{\mathcal{O}}_K$, every term beyond the linear one has valuation strictly greater than $v(x)$ or $v(y)$. The series therefore converge and preserve these lattices; in particular $\exp (y)\in 1+{\pi }^r{\mathcal{O}}_K$. Their formal composition and addition identities are valid by convergence, giving inverse continuous group homomorphismsThus the logarithm isomorphism on deep principal units isThe right-hand group is torsion-free.

Roots of unity in ${\mathbb{Q}}_p$. For odd $p$, take $\ell =p,e=1,r=1$. The principal unit group $1+p{\mathbb{Z}}_p$ is torsion-free by the logarithm isomorphism on deep principal units. The Teichmuller representative splitting leaves precisely the $p-1$ roots in ${\mu }_{p-1}$.

For $p=2$, take $r=2$. Each odd unit is uniquely $\pm u$ with $u\in 1+4{\mathbb{Z}}_2$, since its residue modulo four is either one or minus one. This latter group is torsion-free. ConsequentlyThis computes the roots of unity in the p-adic numbers.

The units of ${\mathbb{Q}}_2(i)$. Set $\pi =1-i$. Its polynomial $T^2-2T+2$ is Eisenstein, soThe Eisenstein polynomial facts justifying this are stated in the next solution. Write $U_s=1+{\pi }^s{\mathcal{O}}_K$. Reduction gives ${\mathcal{O}}_K^{\times }=U_1$, and the successive quotients of principal-unit groupshave order two for $s\ge 1$. Hence $[U_1:U_3]=4$. The four roots in ${\mu }_4=\{1,i,-1,-i\}$ have distinct images modulo $U_3$, sinceThey exhaust the quotient and intersect $U_3$ trivially. Since $r=3>2/(2-1)$, the logarithm isomorphism on deep principal units gives $U_3\cong ({\mathcal{O}}_K,+)$. Therefore the unit decomposition of the 2-adic Gaussian field isIn particular these four elements are all its roots of unity.

Quadratic extensions. A nonzero element of $K$ is uniquely a power of $\pi$ times a unit, so $K^{\times }\cong {\pi }^{\mathbb{Z}}\times {\mathcal{O}}_K^{\times }$. Squaring on the decomposition above acts as squaring on ${\mu }_4$ and multiplication by two on the additive ring ${\mathcal{O}}_K={\mathbb{Z}}_2\oplus {\mathbb{Z}}_{2}i$. Thus the square-class group of the 2-adic Gaussian field isBy quadratic extensions from square classes, in characteristic different from two the nontrivial square classes classify quadratic extensions $K(\sqrt{a})$: two such extensions are $K$-isomorphic exactly when $a/b$ is a square. There are therefore $\boxed{2^4-1=15}$ quadratic extensions up to $K$-isomorphism.

Write $T=(\theta ,0)$, so $f(\theta )=0$, $K=\mathbb{Q}(\theta )$, and $f^{\prime }(\theta )\ne 0$. Let $[a]$ denote the class of $a\in K^{\ast }$ modulo squares. The required two-torsion square-class homomorphism isThe last value is needed in the domain only if $T\in E(\mathbb{Q})$. Every displayed nonexceptional value is nonzero, because an affine point with $x=\theta$ necessarily has $y=0$ and equals $T$.

Consider a nonvertical line $y=mx+c$ whose three intersection points have $x$-coordinates $x_1,x_2,x_3$, counted with multiplicity. Since $f$ is monic,If none of these points is $T$, evaluating at $\theta$ givesand therefore the product of their $\alpha$-values is $1$ in the square-class group. Tangencies are included by repeated factors. Since negation preserves the $x$-coordinate, this identity says $\alpha (P+Q)=\alpha (P)\alpha (Q)$.

If the line passes through $T$, put $f(x)=(x-\theta )g(x)$ and write the line as $y=m(x-\theta )$. The other two intersection abscissae $x_1,x_2$ satisfyEvaluating at $\theta$ now givesThus $\alpha (T)\alpha (P_1)\alpha (P_2)=[f^{\prime }(\theta ){]}^2=1$, as required. This covers a tangent at a different point whose third intersection is $T$ as well.

The tangent at $T$ is vertical. For a vertical chord or tangent the two affine intersections are $P,-P$, so their product of square classes is $\alpha (P{)}^2=1$; in particular $\alpha (T{)}^2=1$ agrees with $2T=O$. Finally, adding $O$ changes neither side. These cases exhaust the chord-and-tangent group law, proving that the displayed map is a group homomorphism.

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.

The valuation contributes one factor $C_2$. The unit decomposition of the 2-adic Gaussian field contributes ${\mu }_4/{\mu }_4^2\cong C_2$ and the additive quotient ${\mathbb{Z}}_2[i]/2{\mathbb{Z}}_2[i]\cong C_2^2$. Thus there are sixteen square classes and fifteen quadratic extensions over this field.

Suppose the ring of integers of a number field is a principal ideal domain, $a,b$ are integral, and $b$ is a unit. On $y^2=x(x^2+ax+b)$ every nonzero $x$-coordinate has even valuation at every finite prime: positive valuation gives $2v(y)=v(x)$, and negative valuation gives $2v(y)=3v(x)$. Thus the two-torsion square-class homomorphism takes values among unit square classes. For the Gaussian integers, their unit group modulo squares has two elements, represented by $1,i$. This can reduce a two-isogeny descent to a very small calculation even though the field contains no ordering.

For odd $p$, a p-adic unit is a square exactly when its residue is a square, by the Hensel lemma. Its unit square-class group is $C_2$. For $p=2$, a unit is a square exactly when it is congruent to one modulo eight, by the strong form of Hensel lemma. The four square classes are represented by $1,3,5,7$, giving $C_2\times C_2$.