Erdős 699: the elliptic surface has exactly one rational section direction

---

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

The elliptic interpretation · The two-descent class

The elliptic surface has four independent sections over an algebraic closure, but only the known section survives over the rational function field for every row that could be bad. We determine that group exactly. This is a theorem about rational functions of the column index, not about all rational points on an individual specialized curve. It does not prove nonexistence or global finiteness of bad triples.

1. The family and the precise theorem

Fix an integer $n\ge8$ divisible by four, and regard $J$ as an indeterminate. Put $c=n-J$ and $d_0=(n-1)(n-2)$. Define

$$A=-\frac{3Jc^2(c-1)}{(n-1)^2(n-2)},\qquad B=\frac{2Jc^2(c-1)^2(n+J-2)}{(n-1)^3(n-2)^2}.$$

Consider the constant quadratic twist

$$\mathcal E_n:\quad Y^2=X^3+n^2AX+n^3B\quad\text{over }\mathbb Q(J).\tag{1}$$

It has the section

$$Q=\left(\frac{nc(c-1)}{n-1},\ \frac{n^2c(c-1)(c-2)}{d_0}\right).\tag{2}$$

Theorem. If $M=(n-2)/2$ is not a cube, then

$$\boxed{\mathcal E_n(\mathbb Q(J))=\mathbb Z Q.}\tag{3}$$

The geometric Mordell–Weil group is torsion-free of rank four. Its subgroup of rational sections has rank one, and $Q$ is a generator, not merely a point of finite index in that subgroup.

Here is the connection to the earlier normalization. Write $W_0=(n-1)(n-2)/6$ and

$$R(z)=\sum_{r=0}^3\binom J{3-r}\binom cr z^r,\qquad P_0=R/W_0.$$

The usual monic conversion and translation of $y^2=P_0(z)$ give $y^2=x^3+Ax+B$. Its twist by $n$ gives (1), with the point above $z=1$ becoming (2). At a hypothetical bad column $J=j$, the earlier curve is $y^2=DP_0(z)/(2^t u)$, where $n=2^hu$ and $D=2^\epsilon$, $\epsilon\equiv h-t\pmod2$. The ratio of the two twisting factors is

$$\frac{n}{D/(2^t u)}=2^{h+t-\epsilon}u^2,$$

a rational square. Thus (1) specializes to the specified curve from the elliptic note, up to rational isomorphism.

2. Singular fibers and geometric rank

We compute over an algebraic closure, where the constant twist by $n$ can be removed. For $E_0:y^2=x^3+Ax+B$, the discriminant is

$$\Delta_0= \frac{1728J^2c^4(J-1)(c-1)^3(c-2)^2} {(n-1)^5(n-2)^4}.\tag{4}$$

The finite singular fibers are as follows. The orders listed are in the local parameter at the indicated value of $J$.

These equations are minimal at every finite place. At infinity, use $s=1/J$, $x=s^{-2}\xi$, $y=s^{-3}\eta$. Since $A,B$ have degrees four and six and (4) has degree twelve with nonzero leading coefficient, the new discriminant is a unit at $s=0$. Thus infinity is a smooth fiber; there is no additional contribution there.

This is a rational elliptic surface with Picard number ten. The reducible fiber root lattices are $A_2\oplus A_1\oplus A_1$. Shioda–Tate gives

$$\operatorname{rank}E_0(\overline{\mathbb Q}(J)) =10-2-2-1-1=4.\tag{5}$$

The standard results used here are stated in Schütt–Shioda, Elliptic Surfaces, Corollary 6.13 and Sections 8.4 and 8.8. The height formulas and correction terms used below are in Section 11.8 and Table 4; the lattice discriminant identity is in Section 11.10.

3. Four explicit sections

Let $r_1,\ldots,r_4$ be the roots of

$$F_n(r)=(n-2)r^2(r^2+2r+3)-4r-2.\tag{6}$$

For each root $r$, define a section of (1) by

$$S_r=\left( \frac{nr^2c(c-1)}{n-1},\ \frac{n^2c(c-1)}{d_0} \left(\frac{3r^2J}{r^2+r+1}-r^3(n-2)\right) \right).\tag{7}$$

Substitution verifies (7), with the difference between the two sides divisible by $F_n(r)$ after clearing denominators. The denominator $r^2+r+1$ is nonzero for roots of (6). Also $r\ne0$, $r^2\ne1$, and distinct roots have distinct squares. For the last assertion, if $r$ and $-r$ were both roots, subtraction would give $(n-2)r^2=2$; their sum would then be nonzero for $n\ge8$.

We compute intersections using $E_0$ rather than its constant twist. The corresponding section has coordinates obtained from (7) by dividing $X$ by $n$ and $Y$ by $n\sqrt n$. The known section becomes

$$Q_0=\left(\frac{c(c-1)}{n-1}, \frac{\sqrt n\,c(c-1)(c-2)}{d_0}\right).$$

All these sections have polynomial coordinates of degrees at most two and three. They are disjoint from the zero section, including at infinity. Their intersections with the reducible fibers can be seen directly from the following local models.

At $c=s=0$, blow up by $x=s\xi$, $y=s\eta$. The exceptional lines are

$$\eta^2=\frac{4n}{d_0^2}.$$

Every $S_r$ meets the line $\eta=-2\sqrt n/d_0$, at $\xi=-r^2/(n-1)$. The section $Q_0$ meets the opposite line. These are the two nonidentity components of the type $\mathrm{IV}$ fiber.

At $c=1+s$, the same blowup gives the exceptional curve

$$\eta^2=-\frac{3\xi}{d_0} +\frac{2(2n-3)}{d_0^2}.$$

The sections $S_r$ meet it at $\xi=r^2/(n-1)$ and $Q_0$ at $\xi=1/(n-1)$. Thus all meet the nonidentity component of the type $\mathrm{III}$ fiber, at distinct points.

At $c=2$, the cubic is

$$y^2=\left(x-\frac2{n-1}\right)^2 \left(x+\frac4{n-1}\right).$$

The section $Q_0$ passes through its node and meets the nonidentity component after resolution. Each $S_r$ is a smooth point of the original fiber, because $r^2\ne1$, and meets the identity component.

Two distinct sections among the $S_r$ and $Q_0$ can intersect in the finite Weierstrass model only at $c=0$ or $c=1$: their $x$-coordinate differences are nonzero constants times $c(c-1)$. The blowups just described separate them. Their leading $x$-coefficients at infinity are also distinct, so no pair intersects there. Consequently all pairwise intersection numbers are zero.

4. Heights and the integral generator

Write $\langle\ ,\ \rangle$ for the geometric height pairing. Its diagonal formula here is

$$\langle T,T\rangle=2+2(T.O)-\sum_v\operatorname{contr}_v(T).$$

The maximal self-corrections from $\mathrm{IV},\mathrm{III},\mathrm I_2$ are respectively $2/3,1/2,1/2$. Therefore every nonzero section satisfies

$$\langle T,T\rangle\ge2-\frac23-\frac12-\frac12=\frac13.\tag{8}$$

In particular there is no nonzero torsion section. Our intersection calculations give

$$\langle Q,Q\rangle=\frac13,\qquad \langle Q,S_r\rangle=\frac16,$$

and, for the four sections from (7),

$$\langle S_{r_i},S_{r_j}\rangle= \begin{cases}5/6,&i=j,\\-1/6,&i\ne j.\end{cases}\tag{9}$$

For distinct sections, the $\mathrm{IV}$ correction is $2/3$ if they meet the same nonidentity component and $1/3$ if they meet opposite ones. This accounts for both cross-pairing formulas.

The matrix in (9) is $I_4-\frac16\mathbf1\mathbf1^{\mathsf T}$ and has determinant $1/3$. Thus these four sections are independent and span the geometric group after tensoring with $\mathbb Q$. The height of $\sum_iS_{r_i}-2Q$ is zero, so torsion-freeness proves

$$\sum_{i=1}^4S_{r_i}=2Q.\tag{10}$$

They are not an integral basis: the full Mordell–Weil lattice has determinant $1/12$, from the fiber lattice determinant $3\cdot2\cdot2$ and the unimodularity of the rational surface. Hence their sublattice has index two. Adding $Q$ fills this index, with relation (10). Independently of this lattice description, (8) proves that $Q$ cannot be a nontrivial integer multiple of another section.

5. Galois action on the four sections

The reciprocal polynomial of (6), up to a nonzero scalar, is

$$g_M(s)=s^4+2s^3-3Ms^2-2Ms-M.\tag{11}$$

First, it has no rational root for any integer $n\ge4$. A rational root $b$ would be an integer because (11) is monic, and would give

$$n=\frac{2(b^2+b+1)^2}{3b^2+2b+1}.$$

Put $A_b=b^2+b+1$, $B_b=3b^2+2b+1$. Then $\gcd(A_b,B_b)=\gcd(A_b,b+2)$ divides three. If three divides $A_b$, its valuation there is exactly one. Since $B_b\mid2A_b^2$, necessarily $B_b\mid18$. The inequality $B_b\le18$ leaves $b=-2,-1,0,1,2$, whose integral values of $n$ are only $1,2,3$.

Now suppose $M$ is odd and not a cube. Choose an odd prime $p$ with $3\nmid e=v_p(M)$. The Newton polygon of (11) has segments from $(0,e)$ to $(3,0)$ and from $(3,0)$ to $(4,0)$. The first segment gives an irreducible local factor of degree three, with ramification index three; the second gives a linear factor. Any proper global factorization would therefore have a linear factor, which was excluded above. Thus $g_M$ is irreducible. Its local ramification also supplies a three-cycle in the global Galois group.

Finally,

$$\operatorname{Disc}g_M =-432M^2(M+1)^2(2M+1)<0.\tag{12}$$

The discriminant is not a rational square. A transitive subgroup of $S_4$ containing a three-cycle is $A_4$ or $S_4$; (12) excludes $A_4$. Therefore (6) has Galois group $S_4$.

Because (7) is a rational expression in $r$, this group permutes the four independent sections exactly as it permutes the four roots. Its invariant subspace is the line spanned by their sum, hence by $Q$ using (10). The rational Mordell–Weil group therefore has rank one. It has no torsion, and (8) makes $Q$ primitive. This proves (3).

6. Applicability and the remaining gap

Every hypothetical bad row satisfies $n=2^hu$, $u$ odd, and $3u^3<2^h$. Such a row has $M=(n-2)/2$ noncubical. Indeed, if $M=v^3$ with odd $v\ge3$, then

$$n=2(v+1)(v^2-v+1),\qquad u\ge v^2-v+1,\qquad 2^h\le2(v+1),$$

contradicting $3u^3<2^h$. The remaining value $v=1$ gives $n=4$, which has no admissible column. Thus the theorem applies to every possible bad row, including all powers of two with $n\ge8$.

The theorem classifies points whose coordinates are rational functions of $J$ on this fixed-row family. It does not classify all rational points after setting $J=j$: specialized curves can acquire additional points and higher rank. Moreover, the known point $Q$ exists for every column before the endpoint divisibility conditions are imposed. Therefore (3) supplies neither a contradiction to badness nor a finite exceptional set. It shows that an additional rational section on the same family cannot provide a new descent independent of multiples of $Q$; a successful argument still has to use the arithmetic restrictions on the specialization.