Let $k$ be an algebraically closed field of characteristic $0$, let $C_{/k}$ be a nice (smooth, projective, geometrically integral curve), let $K = k(C)$, and let $\overline{K}$ be an algebraic closure of $K$. Let $E_{/K}$ be an elliptic curve with $j(E) \notin k$. Let $P \in E(\overline{K}) \setminus E(K)$ be a point of infinite order. Let $K(P)$ be the field of definition of $P$ (equivalently, the field obtained by adjoining to $K$ the coordinates of $P$ in a Weierstrass equation for $E$). Then

$\langle P, E(K) \rangle \subset E(K(P))$.

Must we have equality?


1) For my application, I may assume that $E_{/K}$ is semistable, so please feel free to address the question under that additional hypothesis if it helps. (But I don't see how it does...)

2) I am not able to assume anything about $C$.

3) For my application, I have already dealt with the case in which $P$ has finite order, and in that case I could assume that $C = X_m(n)$ -- the elliptic modular curve parameterizing $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ torsion structures -- and $E$ is the universal elliptic curve over $C$ (the four pairs $(m,n) \in \{(1,1), (1,2), (1,3), (2,2)\}$ in which there is no universal elliptic curve are excluded). In this case the result follows from work of Shioda and Cox-Parry.

4) If this is true, it seems to be closely related to Shioda's landmark 1972 paper on elliptic modular surfaces. I confess that I am asking this question before I have fully absorbed this important paper: I have several collaborators who would be happy if my plate were cleaner.

Added: If I may, I'll try one variant of the question: what if $[n]P \in E(K)$ for some $n \in \mathbb{Z}^+$?

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    $\begingroup$ I fear I agree with Remke's answer below that equality is a bit much to ask for in general. Also if for instance $K(P)/K$ happened to be Galois of odd prime degree, then you should certainly add some conjugates of $P$, too. $\endgroup$ – Chris Wuthrich Jul 26 '13 at 12:57

I do not believe you have equality in general. I sketch a counterexample below, which is a geometric version of the fact that if $E/K$ is an elliptic curves such that the quadratic twist $E^{(d)}/K$ has Mordell-Weil rank at least 2 then the Mordell-Weil of $E/K(\sqrt{d})$ is at least two bigger than the Mordell-Weil rank of $E/K$.

Take an elliptic K3 surface $X_0$ with two $I_{\nu}^*$-fibers, Mordell-Weil rank at least two and such that all further singular fibers are semistable. (Such surfaces exist by the surjectivity of the period map for K3 surfaces). Take now the quadratic twist $X$ of $X_0$ by the two *-fibers. Then $X$ is a rational elliptic surface and all singular fibers are semistable.

The base curve $C$ of each of the elliptic fibrations is $\mathbb{P}^1$. Let now $C'=\mathbb{P}^1$ and let $C'\to C$ be the double cover ramified over the two points where the $I_{\nu}^*$ fibers are.

Then the relatively minimal smooth models of $X\times_{C}C'$ and $X_0\times_{C} C'$ coincide and call this model $Y$. The Mordell-Weil rank of $Y$ equals the sum of the Mordell-Weil ranks of $X$ and $X_0$ and hence the elliptic fibration $X\to C$ is a counterexample.

[If you assume that $X$ has at least 6 singular fibers then $X_0$ is also a counterexample, but $X_0$ is not semistable.]

  • $\begingroup$ More generally, I think you can just start with $E$ of rank bigger than one, find a quadratic extension in which the rank doesn't increase and twist $E$ by that quadratic extension. The twisted curve has rank zero but extending to the quadratic extension will make the rank jump by more than one. $\endgroup$ – Felipe Voloch Jul 26 '13 at 15:00
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    $\begingroup$ Pete wants his elliptic curve to be semistable. Therefore you need to require that "the E you start with" has an even number of `unstable' fibers and each unstable fiber is of type $I^*_{\nu}$. Such surfaces exist, but it requires a little work to show the existence by given a Weierstrass equation for exmaple. The K3 case is easier since you can just refer to the surjectivity of the period map. $\endgroup$ – Remke Kloosterman Jul 26 '13 at 15:17
  • $\begingroup$ As an additional comment to see why we have such surjectivity of period map for $K3$ surface .If $K_X$ the canonical bundle of $X$, is trivial, $X$ Kähler, then the monodromy on $H^2$ is finite if and only if the degeneration is trivial (i.e. $X_0$ is smooth) This implies that if the period map can be extended, then the degenerate fibre is smooth, as a result we have the surjectivity of the period map in the case of Kähler $K3 $ surfaces. $\endgroup$ – user21574 Jul 25 '17 at 20:38

I will sketch a counterexample for the modified question. The idea behind the construction is similar to the counterexample for the original question. Only the geometric details of this construction are trickier and does not integrate so well in my previous answer. For this reason I will post this a new answer.

For this example we take $k=\mathbb{C}$ and $C=\mathbb{P}^1$. Take two polynomials $a(t), b(t)$ of degree 2 and 4 and consider the elliptic curve $E_{a,b}: y^2=x(x^2+a(t)x+b(t))$. This is a rational elliptic surface, with Mordell-Weil group $\mathbb{Z}^4 \times \mathbb{Z}/2\mathbb{Z}$, provided that $a$ and $b$ are sufficiently general, e.g., $a$ and $b$ have no common zero and both $b$ and $a^2-4b$ are squarefree.

Take a point $Q\in E_{a,b}(K(t))$ of infinite order. Let $T$ be a point of order 2, different from $(0,0)$ and take for $P$ the point $Q+T$. Then $2P\in E_{a,b}(\mathbb{C}(t))$ and if $a^2-4b$ is not a square then $P$ is not in $E_{a,b}(\mathbb{C}(t))$.

As in my previous answer, it suffices to find a $(a,b)$ such that the twist of $E_{a,b}$ over $K(P)=K(\sqrt{a^2-4b})$ has positive rank. I will sketch two arguments why such an $(a,b)$ exists.

Let $U \subset \mathbb{C}[t]_2\times \mathbb{C}[t]_4$ be the locus of polynomials such that $E_{a,b}$ defines a rational elliptic surface with positive Mordell-Weil rank and precisely one point of order $2$. (The complement of $U$ is the union of the locus where $a$ is a multiple of $b^2$, together with the locus where $a^2-4b$ is a square, and finitely many codimension 4 components parametrizing the Mordell-Weil rank $0$ surfaces.)

For any integer $m$ consider the locus $L_m$ of $(a,b)$ such that the twisted elliptic curve $(a^2-4b)y^2=x(x^2+ax+b)$ has a section of infinite order which intersects the zero section in $m$ points. Using Hodge theory, Lefschetz $(1,1)$ and the fact that the corresponding elliptic surface has $h^{2,0}=2$ it follows that this locus is locally given by two equations. This implies that $L_m$ is either contained in the complement of $U$ or $L_m$ contains a component of codimenion at most two in $U$. Now the complement of $U$ has only one component of codimension two. Since for most $m,m'$ we have that $L_m\neq L_{m'}$ we find at least one component of some $L_m$ which is not contained in $U$. Actually, there is a standard argument in Noether-Lefschetz theory to check whether the union of $L_m$ is dense in the analytic topology on $U$ and I expect that this argument applies also here.

A second way to construct an explicit example. It is rather easy to find equations for the loci $L_m$. For $L_0$ you have $19$ equations in $25$ variables. Since the equations are of small degree I expect that with some help from a computer you can find an explicit example. Moreover, modern computer algebra software should be able to check whether $L_0$ is in the complement of $U$ or not.

Edit: If you want that $E_{a,b}$ is semistable then you need to do a bit more work. In this case you need to exclude all $(a,b)$ with a common zero from $U$. This defines a codimension one locus. Hence, depending on the strategy you need to show that $\cup L_m$ is dense or that $L_0$ is not contained in the complement of $U$.


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