Erdős 699: the two-descent class of the marked elliptic point

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This is a retained earlier research note. The current reading guide gives the latest status and distinguishes completed proofs from remaining cases.

The elliptic curves · The irreducible cubic

The marked point arising from a hypothetical bad triple has an explicit two-descent class. In many cases this class is already nontrivial over $\mathbb Q_2$. This prevents halving the point; it does not prevent the point, or the hypothetical triple, from existing. Global nonexistence and global finiteness remain unproved.

1. The class in the cubic field

Retain the notation from the elliptic note:

$$P(z)=az^3+bz^2+ez+f,\qquad P(1)=2^m,\qquad m=h-t,\quad \epsilon\equiv m\pmod2,\quad D=2^\epsilon.$$

Here $a,b,e,f$ are positive integers, $P$ is irreducible, and $P'(1)$ is odd. On $E_D:y^2=DP(z)$ the marked point is

$$Q=(1,2^{(m+\epsilon)/2}).$$

Let $\alpha$ be a root of $P$ and $K=\mathbb Q(\alpha)$. The integral monic model obtained by $X=Da z$, $Y=Da y$ is

$$Y^2=X^3+DbX^2+D^2aeX+D^3a^2f.$$

Its corresponding cubic root is $\theta=Da\alpha$. The usual Kummer map has kernel $2E_D(\mathbb Q)$ and sends $(X,Y)$ to the square class of $X-\theta$. Consequently

$$\delta(Q)=[Da(1-\alpha)]\in K^\times/K^{\times2}.\tag{1}$$

This description of the descent map is given in Michael Stoll's explanation of explicit two-descent.

We can express (1) in the field defined by $H(w)=w^3-kw+2q$. Put $C=c/2^t$ and let $\beta$ be the root corresponding to $\alpha$. The earlier change of variable gives

$$\alpha=\frac{2^t\beta-j}{2^t\beta+c},\qquad 1-\alpha=\frac{2^m u}{\beta+C}.$$

Since $D2^m$ is a rational square, (1) becomes

$$\boxed{\delta(Q)=[au(\beta+C)].}\tag{2}$$

Indeed, it first becomes $[au/(\beta+C)]$, and the two representatives differ by the square $(\beta+C)^2$. The leading coefficient is

$$a=\frac{c(c-1)(c-2)}{2^t u(n-1)(n-2)},$$

so the identity for $H(-C)$ gives

$$N_{K/\mathbb Q}(\beta+C)=-H(-C)=2^{2m}u^3a.$$

In particular,

$$N_{K/\mathbb Q}\bigl(au(\beta+C)\bigr) =2^{2m}a^4u^6.\tag{3}$$

Thus the square-norm condition in two-descent is automatic.

2. A local obstruction to halving

Theorem. Suppose either $t\ge3$, or $t\ge2$ and $m$ is odd. Then $Q\notin2E_D(\mathbb Q_2)$, and hence $Q\notin2E_D(\mathbb Q)$.

Proof. For $t\ge2$, the exact coefficient valuations give $a,f$ odd and $b,e$ even. Thus

$$P(z)\equiv z^3+1=(z+1)(z^2+z+1)\pmod2.$$

These factors are separable and coprime. It follows that $K\otimes\mathbb Q_2\cong\mathbb Q_2\times L$, where $L/\mathbb Q_2$ is the unramified quadratic extension. Let $\alpha$ now denote the root belonging to $L$. Both $\alpha$ and $1-\alpha$ are units.

If $m$ is odd, then $D=2$ and $Da(1-\alpha)$ has valuation one in $L$. It is not a square, so (1) obstructs halving.

Suppose instead that $m$ is even and $t\ge3$. Put $J=j/2^t$, so $J+C=2^m u$. Direct reduction of the coefficient formulas modulo four gives

$$a\equiv C/u,\qquad f\equiv J/u,\qquad b\equiv e\equiv0\pmod4.$$

Since $m\ge2$, this yields $P(z)\equiv a(z^3-1)\pmod4$. The element $\alpha-1$ is a unit in $L$, so

$$\alpha^2+\alpha+1\equiv0\pmod4.$$

If $a(1-\alpha)$ were a square, a square root would reduce to $\alpha$ modulo two: in the residue field $\mathbb F_4$, $1-\alpha=\alpha^2$, and squaring is bijective. Its square modulo four would therefore be $\alpha^2$. However,

$$a(1-\alpha)-\alpha^2 \equiv(a+1)+(1-a)\alpha \equiv\begin{cases}2,&a\equiv1\pmod4,\\ 2\alpha,&a\equiv3\pmod4. \end{cases}$$

Both are nonzero modulo four. Thus the class in (1) is nontrivial over $L$, proving the theorem. $\square$

This theorem identifies a nonzero class in the Mordell--Weil group modulo two. It supplies no upper bound on the rank or on $n$. No assertion about halving is made here for the remaining cases.

3. Why the transported class is not an obstruction

The same class has a rational-point representative on a twist of the monic cubic. Namely,

$$E':y^2=-auH(w)$$

has the point $Q'=(-C,2^m a u^2)$, because $H(-C)=-2^{2m}u^3a$. If $B=-au$, the monic coordinates for this curve are $X=Bw$, $Y=By$, with cubic root $B\beta$. Its Kummer class at $Q'$ is exactly

$$[B(-C-\beta)]=[au(\beta+C)].$$

Thus transporting (2) through the shared two-torsion does not produce a class lacking a rational-point representative. The twist factor $au$ varies with the original parameters. Neither (3) nor the local theorem controls the remaining Selmer classes or the varying cubic fields. An additional uniform restriction is needed to turn this descent into nonexistence or a globally finite exceptional set.