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For the Mordell equation $$ Y^2-X^3 = c, $$ Bachet gave a famous duplication formula which translates one rational solution $(x_1,y_1)$ into a second rational solution $(x_2,y_2)$. Réalis gave a second such formula.

My question is: How many others are there?

Bart Michels
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Kieren MacMillan
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  • What do you mean by a "duplication formula"? – Bruno Joyal Nov 01 '14 at 23:41
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    See http://www.ugrad.math.ubc.ca/~scarfy/Seminar/2013-09-05.pdf or http://math.stackexchange.com/questions/485079/on-bachets-duplication-formula-and-the-number-432 for more details. – Kieren MacMillan Nov 02 '14 at 00:56
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    Bachet's formula is multiplication by 2 on an elliptic curve, hence why it is called "duplication formula". There is a formula for multiplication by n for any n; I was wondering whether you would count these as "duplication" formulae, although they would not be duplication, but rather triplication, etc. I stand by my question: what do you mean by a "duplication formula"? – Bruno Joyal Nov 02 '14 at 01:13
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    For the scope of this discussion, I would definitely call those all "duplication" formulae, in that they take a rational solution $(x_1,y_1)$ and generate a second [potentially distinct] rational solution $(x_2,y_2)$. Perhaps "translation formulae" is a better term?

    So a better question, I suppose, would be: Is there an effective/useful way to order all of the "translation formulae", so that the solutions generated are ranked (e.g., by height, or some other computable quantity)?

     – Kieren MacMillan Nov 02 '14 at 02:09

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The projective curve $E$ given by the equation $Y^2Z = X^3 + cZ^3$ is a smooth curve of genus $1$ over $R = \mathbb Q[c, c^{-1}]$. You are essentially asking for a description of $$\text{End}_{R}({E}).$$

The curve $E$ has the rational point

$$\mathcal O := [0:1:0] \in E(R)$$

which we can use as a base-point, so that $E$ is an elliptic curve over $R$. I believe that the only nonconstant endomorphisms of the curve $E/R$ are the multiplication by $n$ maps $[n]: E \to E$.

Any nonconstant endomorphism $\sigma$ is of the form $\sigma = t \circ \sigma'$ where $t$ is a translation $P \mapsto P+\sigma(\mathcal O)$ and $\sigma'$ is an isogeny (an endomorphism preserving the point $\mathcal O$).

Therefore, to describe $\text{End}_{R}({E})$, we should describe:

  1. The Mordell-Weil group $E(R)$;
  2. The endomorphism ring of $E/R$ as an elliptic curve.

I claim that $E(R) = \{\mathcal O\}$, assuming that the following is true:

For infinitely many $c \in \mathbb Q^\times$, the specialization $E_c/\mathbb Q$ has trivial Mordell-Weil group.

In fact, one certainly expects that for a positive proportion of $c \in \mathbb Q^\times$, $c$ ordered by height, the specialization $E_c$ has rank $0$. This is likely true, and the data is convincing, but I'm not sure if this is known at the moment.

Let me explain how $E(R) = \{\mathcal O\}$ follows from this conjecture. Suppose that $P = [f : g : h] \in E(R)$, with $f,g,h \in R$. Suppose that $P \neq \mathcal O$, so that $h \neq 0$. Then for almost all $c_0$, $h(c_0) \neq 0 $. Choose such a $c_0$ for which $E_{c_0}$ has trivial Mordell-Weil group. Then $[f(c_0), g(c_0), h(c_0)] \in E_{c_0}(\mathbb Q)$ is a rational point with nonzero $Z$-coordinate, hence a nontrivial point on $E_{c_0}$; this contradicts the choice of $c_0$. Hence $E$ has trivial Mordell-Weil group.

I claim also that $\text{Hom}(E, E) = \mathbb Z$, and that this follows from the following fact:

For almost all $c \in \mathbb Q^\times$, the specialization $E_c/\mathbb Q$ has endomorphism ring $\mathbb Z.$

This is obviously true as finitely many $j$-invariants are excluded, and the $j$-invariant of $E_c$ is a polynomial in $c$. The proof of $\text{Hom}(E, E) = \mathbb Z$ using this fact is similar to the one above.

Thus, the only nonconstant endomorphisms of the curve $E/R$ are the multiplication by $n$ maps $[n]: E \to E$.

Bruno Joyal
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