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Dear MO,

There is a classical theorem (cf. Theorem 4.1 on Husemoller "Elliptic Curve" book) that states conditions on the coordinates of a point P in an elliptic curve to be twice another point. The result is the following:

Theorem: Let $E$ be an elliptic curve defined over a field $K$ by the equation $$ E:y^2=(x-\alpha)(x-\beta)(x-\gamma),\quad\mbox{with $\alpha,\beta,\gamma\in K$}. $$ For $(x',y')\in E(K)$ there exists $(x,y)\in E(K)$ with $2(x,y)=(x',y')$ if and only if $x'-\alpha$, $x'-\beta$ and $x'-\gamma$ are squares.

I would like to know if there is something similar for 3. That is, condition on the coordinates $(x',y')\in E(K)$ such that there exists $(x,y)\in E(K)$ with $3(x,y)=(x',y')$.

Thanks for any answer or suggestions.


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The fact that you quote comes from the Kummer sequence. Let $G_K=\text{Gal}(\bar K/K)$. Then one gets $$ E(K)/2E(K) \hookrightarrow H^1(G_K,E[2]) = \text{Hom}(G_K,\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}) \cong K^*/(K^*)^2 \times K^*/(K^*)^2. $$ This works because you're taking a curve for which all of the $2$-torsion is rational. You can do something similar if all of the $m$ torsion is rational, yielding $$ E(K)/mE(K) \hookrightarrow H^1(G_K,E[m]) = \text{Hom}(G_K,\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z}) \cong K^*/(K^*)^m \times K^*/(K^*)^m. $$ (Note that the assumption that the $m$ torsion is rational implies that $K$ contains the $m$'th roots of unity, which is why the last isomorphism is okay.) From this one can in principle derive two functions $F$ and $G$ on $E$ with the property that $P$ is $m$ times a point if and only if $F(P)$ and $G(P)$ are $m$'th powers. But I don't know a reference offhand.

This may be in Husemoller's book, or you can find it in Chapter X of my Arithmetic of Elliptic Curves.

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[Clarifying remark, perhaps obvious anyway: in the original question, the $m=2$ case, it looks like you have three functions which you need to check are squares, but if any two of them are squares then the third is automatically because their product is a square. So Joe's argument really does generalise the $m=2$ case above.] – user30035 Mar 26 '13 at 22:46
@wccanard: That was my first thought, too. Except it's not quite right, because if one of the three quantities is 0, then you really do need to check that the other two are squares. So my answer isn't quite right, either. It will work as long as P is not itself an m-torsion point, but if F(P) or G(P) vanishes, then there will be a third condition H(P) that needs to be checked. I believe that the ambiguity comes from making a specific choice of an isomorphism $E[m]\cong\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z}$. – Joe Silverman Mar 26 '13 at 23:55
@Joe Silverman. Thanks for the answer. Although it seems that it is not very explicit. – Enrique Gonzalez-Jimenez Mar 28 '13 at 18:09
@Enrique: I didn't say it was explicit, I said it's where the result comes from. In principle, one should be able to make it explicit by writing down explicit functions F, G and H, but first you'd need to write the elliptic curve in a form where its $m$-torsion is rational. So there's a fair amount of work involved in doing this, and I don't recall seeing it. But at least for $m=3$, it's likely to be feasible. – Joe Silverman Mar 28 '13 at 21:06

A similar criterion, although more complicated, can be obtained using division polynomials. Let $P'=(x',y') \in E(K)$ be the given point. Then $P'=3P$, for some $P=(x,y)$ if and only if $$(x',y')=\left(\frac{\phi_3(x)}{\psi_3^2(x)}, \frac{\omega_3(x)}{\psi_3^3(x)} \right),$$ where $\phi_3, \psi_3$ and $\omega_3$ are defined as in

So the question can be rephrased to asking whether the polynomial $\phi_3(x)-x'\psi_3^2(x)$ (note that $\psi_3(x)=0$ will hold only if $P'=O$) has a $K$-rational root such that the corresponding $y$-coordiante is defined over $K$. This polynomial will be a degree 9 polynomial whose roots correspond to the $x$-coordinates of the 9 solutions (in $E(\overline K )$) of the equation $3P=P'$.

The upside of this criterion is that you do not need any $3$-torsion in $E(K)$.

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@Filip Najman. This sounds very explicit. I will try with division polynomials. – Enrique Gonzalez-Jimenez Mar 28 '13 at 18:15

In this paper by Ed Schaefer and myself, there is a reasonably explicit description of `3-descent' on an elliptic curve. There is a somewhat more detailed version of the paper here.

Assuming that your curve $E$ over $K$ is such that the Galois action on the points of order 3 is transitive, you look at the field $L = K(P_0)$, where $P_0$ is any point of order 3 on $E$. Then $P_0 = (\xi,\eta) \in E(K)$ and you can consider the tangent line $y = \lambda x + \mu$ at this point $P_0$. The map $$E(K) \longrightarrow L^\times/L^{\times 3}, \qquad P \longmapsto f(P) = y(P) - \lambda x(P) - \mu$$ (modulo $L^{\times 3}$, where $L^{\times 3}$ means the subgroup of cubes) has kernel $3 E(K)$. This means that $P$ is divisible by 3 in $E(K)$ if and only if ($P = O$ or) $f(P)$ is a cube in $L$.

This is in fact the correct generalization of Joe's reply (for $m=3$) to the case when you don't have full rational 3-torsion: $f$ is the correctly scaled function with divisor $3(P_0) - 3(O)$, and working over $L$ basically means that we look at all the $f$'s for the various choices of $P_0$ at the same time.

When the Galois action is not transitive, you may have to consider two such functions $f$ (such that the orbits of the corresponding points generate $E[3]$ and have total cardinality not a multiple of 3).

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@Michael: I had assumed the OP was primarily interested in the situation where one can reduce to questions about cubes in $K$, but you're quite right that this is the right generalization if one doesn't assume that the 3-torsion is rational. Of course, now the arithmetic of $L$ will come into play, especially the class group and unit group. I first saw this used (for 2-torsion) in a letter that Tate wrote computed $E(Q)$ for $E=X_1(11)$. For general $m$, Bob Wake in his thesis partially generalized the idea of looking at $L^*/{L^*}^m$, but I don't think that he ever published his thesis. – Joe Silverman Mar 29 '13 at 12:45
@Joe: For the OP's question (how to determine if a point is divisible by 3), we only need to be able to check if an element of $L$ is a cube, which is a purely algebraic question and does not require any arithmetic information. (Take the minimal polynomial $p$ of your element and check if $p(x^3)$ has a root in $L$.) – Michael Stoll Mar 29 '13 at 22:18
@Michael: Sorry, you're right. Given a specific point, its divisibility by 3 (or by 2, or by $m$) is a purely algebraic question. However, usually the reason to do this is to compute the Mordell-Weil group, so (for $m=2$) one looks at equations $x-\alpha_i=y_i^3$ for $i=1,2,3$. And then the arithmetic of $L$ will come into play. For $X_1(11)$ and $m=2$, my recollection is that $L$ has class number 1 and that one can completely characterize the units that are squares, which was crucial in showing that $X_1(11)(\mathbb{Q})$ has rank $0$. – Joe Silverman Mar 30 '13 at 17:40
@Joe: The OP didn't say what he was needing the answer for. You're right, of course, for the application to computing MW ranks. But then, the same will be true for the arithmetic of $K$ in the case when $E[m] \subset E(K)$. – Michael Stoll Mar 31 '13 at 14:15
@Michael: Thanks for the answer. @All: Yes, I didn't say nothing if I would like something explicit or not. Sorry. – Enrique Gonzalez-Jimenez Apr 4 '13 at 9:37

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