Erdős 699: conductor controlled by the cubic discriminant

P699 research · The elliptic curves · The integer cubics

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The monic elliptic curve admits a useful normalization: its conductor is bounded by an absolute constant times the discriminant of the coefficient cubic. Consequently a fixed upper bound on that cubic discriminant gives an effective finite set of possible bad triples. No uniform bound on the cubic discriminant is proved here, so global finiteness remains open.

1. Removing the odd twist

For a hypothetical bad triple, use the established notation

$$n=2^h u,\quad c=n-j,\quad d=n-2j,\quad t=v_2(j)=v_2(c),\quad m=h-t\ge2,$$

where $u$ is odd and $4\le j<n/2$. Put

$$w=u2^t,\qquad g=\gcd(u,3),\qquad v=u/g,\qquad D_0=\operatorname{Disc}P>0.$$

Here $D_0$ is an integer. The preceding notes give odd positive integers

$$k=\frac{3jc}{2^{2t}(n-1)},\qquad q=\frac{jcd}{2^{3t}(n-1)(n-2)},$$

and the integer moment $3j/u$. The latter implies $v\mid j$, hence $v\mid c,d$. If $p^a\Vert v$, both $n-1$ and $n-2$ are $p$-adic units: $p$ is odd and divides $n$. The displayed formulas therefore give

$$v_p(k)\ge2a,\qquad v_p(q)\ge3a.$$

Thus $k_0=k/v^2$ and $q_0=q/v^3$ are odd integers. Consider the integral curve

$$E_0:\quad y^2=x^3-k_0x+2q_0.$$

The earlier curve $E_H$ is its quadratic twist by $v$. The discriminant identity from the preceding notes becomes

$$\Delta(E_0)=64(k_0^3-27q_0^2) =2^{2m+4}g^6D_0.\tag{1}$$

In (1), $\Delta(E_0)$ is the discriminant of the displayed equation, which need not be minimal.

2. Conductor and a minimal-discriminant lower bound

Write $N_0$ for the conductor and $\Delta_{\min}$ for the minimal discriminant. The conductor exponent at any prime is at most the minimal-discriminant valuation, by Ogg's formula. Minimalization can only decrease the discriminant valuation of an integral equation. Hence the odd part of $N_0$ divides $g^6$ times the odd part of $D_0$. The universal bound $v_2(N_0)\le8$ then gives

$$N_0\le C D_0,\qquad C=2^8\cdot3^6.\tag{2}$$

The local conductor bound is recorded in von Känel, §2.2 and the proof of Theorem 7.1.

Since $c_4(E_0)=48k_0$ has valuation four at two, (1) gives the exact, model-independent identity

$$v_2(\mathfrak j(E_0))=8-2m-v_2(D_0).\tag{3}$$

On a minimal integral model $v_2(c_{4,\min})\ge0$. Applying $\mathfrak j=c_{4,\min}^3/\Delta_{\min}$ therefore yields

$$v_2(\Delta_{\min})\ge2m+v_2(D_0)-8, \qquad |\Delta_{\min}|\ge2^{2m-8}.\tag{4}$$

This remains valid when $t=0$ and $D_0$ has a nontrivial two-part.

3. Effective finiteness when the cubic discriminant is bounded

The exact formula for $D_0$ is

$$D_0=\frac{108j^2c^2(j-1)(c-1)} {w^4(n-1)^3(n-2)^2} \le\frac{384j^3}{w^4n^2} \le\frac{48n}{w^4}=\frac{48\cdot2^m}{w^3}.\tag{5}$$

For the first inequality, use $j-1<j$, $c-1<c<n$, $n-1\ge7n/8$, and $n-2\ge3n/4$; these hold since $n>2j\ge8$. The second uses $j<n/2$, and the last equality uses $n=2^m w$.

Proposition. For every fixed $B\ge1$, the hypothetical bad triples with $D_0\le B$ lie in an effectively bounded finite set.

Proof. The unconditional bound

$$\log|\Delta_{\min}(E)|\le 3N(E)(\log N(E))^2+124$$

holds for every elliptic curve over $\mathbb Q$ by von Känel, Corollary 6.2. Combining it with (2) and (4) bounds

$$m\le M_B:=4+ \frac{3CB(\log(CB))^2+124}{2\log2}.$$

Since $D_0\ge1$, (5) gives $w^3\le48\cdot2^m$. Consequently $n=2^m w\le48^{1/3}2^{4M_B/3}$. This is an explicit finite range for $n$, and each row has finitely many admissible columns. $\square$

Equivalently, Shafarevich finiteness applied to (2) leaves finitely many $\mathfrak j$-invariants. Equation (3) then bounds $m$, even if $v_2(D_0)$ varies. The displayed proof supplies an explicit bound directly.

4. What Szpiro would add

Assume Szpiro's conjecture $|\Delta_{\min}(E)|\ll_\epsilon N(E)^{6+\epsilon}$; this is the conjecture stated in Masser's publisher abstract. Equations (2), (4), and (5) then imply

$$j^{18+3\epsilon}\gg_\epsilon n^{14+2\epsilon}w^{22+4\epsilon}.\tag{6}$$

Indeed, $(n/w)^2=2^{2m}\ll_\epsilon D_0^{6+\epsilon}$, and substituting (5) gives (6). For every $\delta>0$, this entails

$$j\gg_\delta n^{7/9-\delta},\qquad u2^t=w\ll_\delta n^{2/11+\delta}.$$

These are conditional restrictions, not conditional global finiteness. When $u,t$ are fixed and $j/n$ stays inside $(0,1/2)$, one has $D_0\asymp n$; conductor and discriminant bounds of the sizes above are compatible with arbitrarily large $n$. Proving a uniform upper bound for $D_0$, or obtaining another restriction that excludes this regime, remains necessary. No enumeration was used in this note.