Two-isogeny descent

Here descent is an arithmetic calculation with square classes and an isogeny of elliptic curves, rather than the gluing of objects along a covering.

For $E:y^2=x^3+a{x}^2+bx$ and its two-isogenous curve $E^{\prime }:y^2=x^3-2a{x}^2+(a^2-4b)x$, square-class maps on $E(\mathbb{Q})$ and $E^{\prime }(\mathbb{Q})$ determine the Mordell-Weil rank.

For this elliptic curve, the two-isogeny formula gives the partner $E^{\prime }:Y^2=X^3+14X^2+X$. The prime-support bound in two-isogeny descent and positivity of nonzero rational $x$ restrict the first square class image to $1,2,3,6$. The identity and the rational points $(2,2),(0,0),(6,6)$ realize these classes. The partner image is contained in $\{1,-1\}$, and the negative unit obstruction for the fourteen-coefficient isogeny cover excludes $-1$. The two-isogeny rank formula therefore gives $2^r=4\cdot 1/4=1$.

The two-torsion square-class homomorphism on $E$ has candidate classes $\{1,-1,17,-17\}$ by the prime-support bound in two-isogeny descent. The identity, $(0,0)$ and $(-1,4)$ realize $1,-17,-1$, so multiplicativity gives all four. The two-isogeny formula companion $E^{\prime }:y^2=x^3+68x$ has only positive candidate classes $\{1,2,17,34\}$; $(0,0)$ realizes $[68]=17$ and $(2,12)$ realizes $2$. Thus both images have order four, and the two-isogeny rank formula gives $2^r=4\cdot 4/4=4$. The rank of an elliptic curve is therefore two, without any unresolved covering class.

For $E:y^2=x^3-2x$, prime-support bounds allow image classes $\{1,-1,2,-2\}$. The rational points $(0,0)$ and $(2,2)$ realize $-2$ and $2$, and the two-torsion square-class homomorphism is multiplicative, so all four occur. Its companion $E^{\prime }:y^2=x^3+8x$ has only positive candidate classes $1,2$, both realized by the identity and $(0,0)$. The two-isogeny rank formula gives $2^r=4\cdot 2/4=2$, so the rank is one. The point $(2,2)$ has infinite order: its double is $(9/4,-21/8)$, contradicting the integrality of torsion coordinates in the Nagell–Lutz theorem if it were torsion.

For $E:y^2=x^3+x$, the two-torsion square-class homomorphism has image $\{1\}$: prime-support bounds allow only $\pm 1$, and every nonzero affine $x$ is positive. Its two-isogeny formula companion $E^{\prime }:y^2=x^3-4x$ has possible image classes $\{1,-1,2,-2\}$, all realized by its rational two-torsion. The two-isogeny rank formula gives $2^r=1\cdot 4/4=1$, so the rank is zero.

The two-isogeny formula gives the partner $E^{\prime }:y^2=x^3-16x^2+92x$. The two-torsion square-class homomorphism image for $E$ is $\{1,-7\}$: prime support leaves only $\{1,-1,7,-7\}$, the modulo-sixteen obstruction for the negative unit descent class excludes $-1$, and the subgroup property excludes $7$. For $E^{\prime }$, $(x-8{)}^2+28>0$ forces nonnegative $x$, so the image is contained in $\{1,2,23,46\}$. The two-torsion point realizes $23$ and the rational point $(18,48)$ realizes $2$, giving all four classes. The two-isogeny rank formula yields $2^r=2\cdot 4/4=2$, hence rank one. The dual isogeny sends $(18,48)$ to $(16/9,116/27)$; this statement does not assert that the resulting point is a generator of the full free part.

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 $(1,3)$.

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 the two-isogeny formula $\varphi :E\to E^{\prime }$ over a number field, the two-torsion square-class homomorphisms have kernels $\hat{\varphi }{E}^{\prime }(K)$ and $\varphi E(K)$. ConsequentlyIndeed $2E(K)=\hat{\varphi }\varphi E(K)$, and the surjection from $E^{\prime }(K)/\varphi E(K)$ to $\hat{\varphi }{E}^{\prime }(K)/2E(K)$ has kernel $\ker \hat{\varphi }/(\ker \hat{\varphi }\cap \varphi E(K))$. Combined with the Mordell-Weil theorem, the index is $2^r\#E(K)[2]$, where $r$ is the rank of an abelian group. This formula prevents an erroneous extra factor of two in an isogeny descent.

For $E:y^2=x(x^2+ax+b)$, set $b^{\prime }=a^2-4b$ and $E^{\prime }:Y^2=X(X^2-2aX+b^{\prime })$. The isogeny of elliptic curves with kernel $\{O,(0,0)\}$ is $\varphi (x,y)=(x+a+b/x,y(1-b/x^2))$. Its dual is $\hat{\varphi }(X,Y)=((X-2a+b^{\prime }/X)/4,Y(1-b^{\prime }/X^2)/8)$. The maps extend over their exceptional affine points and satisfy $\hat{\varphi }\varphi =[2]$.

For integral $a,b$ with $b(a^2-4b)\ne 0$, every square class in the image of the two-isogeny descent map on $y^2=x(x^2+ax+b)$ has a signed square-free representative supported on primes dividing $b$. At a prime not dividing $b$, a positive valuation of $x$ must be even from the curve equation; a negative valuation is even because the cubic leading term determines $2v(y)=3v(x)$. Thus its image has at most $2^{\nu (b)+1}$ elements. Apply this also to the two-isogenous curve and use the square-class index formula for two-isogeny descent to obtain $r\le \nu (b)+\nu (a^2-4b)$, where $\nu$ counts distinct prime divisors of the nonzero absolute value.

For $E:y^2=x(x^2+ax+b)$ and $E^{\prime }:y^2=x(x^2-2ax+a^2-4b)$ over $\mathbb{Q}$, define the square-class homomorphisms by $\alpha (x,y)=[x]$, $\alpha (O)=1$, $\alpha ((0,0))=[b]$, and similarly ${\alpha }^{\prime }$ with $b^{\prime }=a^2-4b$. Their kernels are the images of the dual degree-two isogenies of elliptic curves. If $r$ is the rank of an abelian group of $E(\mathbb{Q})$, then $2^r=\#\alpha (E)\#{\alpha }^{\prime }(E^{\prime })/4$. To check the factor four, put $\delta =[\ker \hat{\varphi }:\ker \hat{\varphi }\cap \varphi E]$. The index of $2E$ in $E$ is $\#\alpha (E)\#{\alpha }^{\prime }(E^{\prime })/\delta$. If $b^{\prime }$ is a square, $\delta =1$ and $\#E(\mathbb{Q})[2]=4$; otherwise $\delta =2$ and $\#E(\mathbb{Q})[2]=2$. In either case $\delta \#E(\mathbb{Q})[2]=4$.

For a square class $d$ dividing $b$, the associated quartic covering isIt has a nontrivial rational point exactly when $d$ occurs in the square-class image attached to $E:y^2=x(x^2+ax+b)$.

This quartic covering in a two-isogeny descent has no solution with coprime integers $U,V$ not both zero. If precisely one is odd, the right side is $3$ modulo four. If both are odd, their fourth powers are one modulo sixteen and $14U^2{V}^2=14$ modulo sixteen, so the right side is twelve modulo sixteen. Neither residue is a square. Coprimality excludes the remaining parity case. Consequently the negative unit square class is absent from the two-torsion square-class homomorphism for $Y^2=X^3+14X^2+X$.

For $y^2=x^3+a{x}^2+bx$ with $a\equiv 8$ and $b\equiv 9(\mathrm{mod}16)$, the square class $-1$ in a two-isogeny descent would require coprime integers $u,v$ and an integer $N$ with $N^2=-u^4+a{u}^2{v}^2-b{v}^4$. If $u$ is odd and $v$ even, the right side is $15$ modulo $16$; if $u$ is even and $v$ odd, it is $7$; if both are odd, it is $14$. None is a square modulo $16$. Therefore $-1$ does not occur in the two-torsion square-class homomorphism image.

For $E_5^{\prime }:y^2=x^3+100x$, the two-isogeny descent classes $2$ and $10$ have no rational covering points. Their primitive integral quartics are respectively $N^2=2U^4+50V^4$ and $N^2=10(U^4+V^4)$, with $\gcd (U,V)=1$.

For the first equation, reduction modulo $5$ forces $5\mid U$ and $5\mid N$, since $2$ is not a square modulo $5$ and a nonzero fourth power is $1$. Put $U=5U_1,N=5N_1$. Dividing by $25$ gives $N_1^2=50U_1^4+2V^4$, again a nonsquare modulo $5$ because $5\nmid V$. For the second equation, a primitive pair has $U^4+V^4$ nonzero modulo $5$: it is $1$ or $2$. The right side has five-adic valuation exactly $1$, impossible for a square. Since $(5,25)$ lies on $E_5^{\prime }$, its square-class image is exactly $\{1,5\}$. Combined with the rank bound for a prime congruent-number elliptic curve computation, the two-isogeny rank formula yields rank exactly $1$ for $E_5:y^2=x^3-25x$.

This quartic has no primitive integral solution. If $5\nmid U$, reduction modulo five makes $W^2\equiv 2$, impossible. If $5\mid U$, coprimality gives $5\nmid V$ and $5\mid W$; dividing by twenty-five then gives $(W/5{)}^2\equiv 2V^4\equiv 2(\mathrm{mod}5)$, again impossible. It excludes the square class two in two-isogeny descent on $Y^2=X^3+100X$. The point $(5,25)$ supplies class five, so the subgroup property excludes class ten as well.

The two-isogeny descent quartics associated with square classes $2$ and $-2$ on $Y^2=X^3+24X^2+64X$ have no rational points. A rational solution can be scaled to integral $W,U,V$ with coprime integers $U,V$. If $U$ is odd, reduction modulo sixteen gives $2$ or $10$ for the positive class, and $14$ or $6$ for the negative class; none is a quadratic residue. If $U=2u$ is even, then $V$ is odd and the right side is $32(\pm u^4+3u^2{V}^2\pm V^4)$, with the same sign on the two signed terms. The bracket is odd, so its 2-adic valuation is exactly five, impossible for a nonzero square. These two parity cases exhaust primitive solutions.

If $p\equiv 3(\mathrm{mod}8)$, this quartic covering in a two-isogeny descent has no primitive integral solution. Opposite parities of $U,V$ give $N^2\equiv 2(\mathrm{mod}8)$. If both are odd, their fourth powers are one modulo sixteen and $p^2\equiv 9(\mathrm{mod}16)$, giving $N^2\equiv 20(\mathrm{mod}32)$. Neither residue is a square. This excludes the class $2$ on $Y^2=X^3+4p^{2}X$.

For the curve $Y^2=X^3+4p^{2}X$ with $p\equiv 3(\mathrm{mod}4)$, the classes $p$ and $2p$ in two-isogeny descent lead to $N^2=p(U^4+4V^4)$ and $N^2=2p(U^4+V^4)$. A primitive solution forces $p\mid N$ and $p\nmid V$. The resulting congruences make $-1$ a square modulo $p$, a contradiction. Thus these coverings have no rational points.

A proposed square class in a two-isogeny descent can be excluded by testing its quartic modulo $8$. For $N^2=3U^4+5U^2{V}^2-2V^4$ with $\gcd (U,V)=1$, both variables odd give residue $6$; odd-even gives $3$ or $7$; even-odd gives $6$ or $2$. None is a square modulo $8$, excluding class $3$ and, by the subgroup property of the descent image, its entire coset.