Kummer map of an elliptic curve

The map is a connecting homomorphism from the multiplication exact sequence. Its image is constrained by the local conditions defining the Selmer group of an elliptic curve.

The multiplication-by-$n$ sequence gives an injective connecting mapLocal conditions restrict its image to the finite $n$-Selmer group.

For an isogeny of elliptic curves $\varphi :E\to E^{\prime }$ over a number field $K$, this finite subgroup of $H^1(K,E[\varphi ])$ consists of Galois cohomology classes whose restriction at every place comes from $E^{\prime }(K_v)/\varphi E(K_v)$. It fits intowhere the last group is the kernel of the induced map on Tate–Shafarevich groups. Thus local computations bound the rational-point quotient without necessarily determining it. For multiplication by $n$, this recovers the Selmer group of an elliptic curve.

Coverings surviving all local solubility tests need not have rational points. The quotient of an isogeny Selmer group by the rational-point image is a kernel in the Tate–Shafarevich group. A nonzero class therefore passes every local test but contributes no rational point to a search. This explains why two-isogeny descent can leave a gap between a certified upper bound and found lower bound for the rank of an elliptic curve. Higher descents address finer obstructions; see Tom Fisher's paper on higher descents with rational two-torsion.

Let $E:y^2=f(x)$ with $f$ monic and separable, and let $T=(\theta ,0)$. Over $K$ containing $\theta$, the map ${\alpha }_T(P)=[x(P)-\theta ]$ for $P\ne O,T$, completed by ${\alpha }_T(O)=1$ and ${\alpha }_T(T)=[f^{\prime }(\theta )]$, is a group homomorphism to the square-class group of a field. A nonvertical chord has the product of its three $x-\theta$ values a square; a chord through $T$ instead has the product of the other two values equal to $f^{\prime }(\theta )$. These identities include tangencies.

When an elliptic curve has full rational 2-torsion, its Kummer map embeds $E(\mathbb{Q})/2E(\mathbb{Q})$ into a product of square-class groups. Valuation and local-solubility conditions reduce the image to finitely many classes and thereby bound the rank of the Mordell-Weil group.

When ${\mu }_n\subset K$, the Kummer pairing sends $\sigma \in \mathrm{Gal}(L/K)$ and $[x]\in K^{\ast }/(K^{\ast }{)}^n$ to $\sigma (\sqrt[n]{x})/\sqrt[n]{x}\in {\mu }_n$. Its elliptic analogue sends $\sigma$ and $[P]\in E(K)/nE(K)$ to $\sigma (Q)-Q\in E[n]$ for $nQ=P$.

Suppose an elliptic curve $E/K$ has all four geometric 2-torsion points rational. For $P\in E(K)$ choose $Q\in E(\overline{K})$ with $2Q=P$. The function $c_P(\sigma )=\sigma (Q)-Q$ takes values in $E[2]$ and is a group homomorphism from the absolute Galois group to $E[2]$: its usual cocycle law becomes additivity because the action on $E[2]$ is trivial. Replacing $Q$ by another half, or $P$ by $P+2R$ with $R\in E(K)$, leaves it unchanged. Adding chosen halves proves additivity in $P$. Its kernel in $E(K)/2E(K)$ is zero, since $c_P=0$ precisely when the chosen half is rational. This constructs the injective map directly, without assuming any theorem of two-descent on an elliptic curve or finite generation of the Mordell-Weil group.

Let $E:y^2=(x-\alpha )(x-\beta )(x-\gamma )$ with distinct algebraic integers in a number field $K$. For a rational point $P$, its halving field is unramified outside the finite set $S$ of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At $v\notin S$, the elliptic-curve discriminant is a unit, so there is good reduction. A half of the reduction of $P$ is defined over a finite extension of the residue field. Pass to the corresponding unramified extension of the completion and lift this half using the Hensel lemma. The doubling error lies in the formal kernel of a minimal Weierstrass equation. Since two is a unit, prime-to-residue-characteristic multiplication on a formal group corrects this error uniquely. Every half differs by rational 2-torsion, so all halves are unramified at $v$. The halving cocycle with rational two-torsion has image of order at most four, making the halving field a Galois extension of degree at most four. The bounded-degree extensions with restricted ramification are finite in number; there are only finitely many homomorphisms from their finite Galois groups to $E[2]$. Thus $E(K)/2E(K)$ is finite, without assuming any form of the Mordell-Weil theorem.

For distinct integral roots $\alpha ,\beta ,\gamma$ with $m=\max (|\alpha |,|\beta |,|\gamma |)$, the finite set in unramified halving fields for a split cubic contains only primes at most $2m$. A quadratic Galois character unramified away from $s$ primes is represented by a signed square-free integer supported on those primes, giving at most $2^{s+1}$ characters. The halving cocycle with rational two-torsion has two independent coordinates in $\mathbb{Z}/2\mathbb{Z}$, so the quotient has at most $2^{2s+2}$ elements. Since $m\ge 1$ and there are at most $m$ primes at most $2m$, this gives $2^{2m+2}$. The bound is deliberately crude but depends only on the root size.

For a finite Galois extension $L/K$, the kernel ofinjects into the finite group $H^1(\mathrm{Gal}(L/K),{\mu }_n(L))$. Likewise, the kernel of $E(K)/nE(K)\to E(L)/nE(L)$ injects into $H^1(\mathrm{Gal}(L/K),E[n](L))$ and is finite.

For a number field $K$ and a finite set $S$ of finite primes,

The S-unramified power class group of a number field is finite. For $[a]\in K(S,n)$, the fractional ideal of $a$ away from $S$ is an $n$th power ${\mathfrak{a}}^n$. Sending $[a]$ to $[\mathfrak{a}]$ gives an exact sequenceThe S-unit group is finitely generated, and the localized ideal class group is a quotient of the finite ordinary ideal class group. Both end groups are finite. This is the arithmetic finiteness needed in the Kummer-theoretic proof of the weak Mordell-Weil theorem.

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.

For a degree-three isogeny whose kernel becomes ${\mu }_3$ under the Weil pairing, the Kummer connecting map embeds the quotient by the isogeny into $K^{\times }/(K^{\times }{)}^3$. Valuations outside the primes of bad reduction restrict its image to an S-unramified power class group.