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In Propriétés galoisiennes des points d'ordre fini des courbes elliptiques, Invent. Math. 15, 259--331 (1972), Serre proved the following (Theorem 6 ′′, p. 325):

Let $K$ be a number field and let $K^{cycl}$ be the cyclotomic extension of $K$ generated by all roots of unity. Let $E$ and $E'$ be two elliptic curves such that, over $\bar{K}$,

(i) $E$ and $E'$ have no complex multiplication;

(ii) The $l$-adic representations $(\rho_l)$, $(\rho'_l)$ attached to $E$ and $E'$ don't become isomorphic over any finite extension of $K$.

Then $K(E_{tors}) \cap K(E'_{tors})$ is finite over $K^{cycl}$.

My question is whether this holds for $E$ and $E'$ defined over a function field? If this hasn't already been considered somewhere with an argument specific to the function field case, then maybe a specialization argument might work? Could anyone please provide a reference where there are similar specialization arguments used, or a standard reference for the basic theory of these specialization theorems?

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  • $\begingroup$ If $j,j'$ are non-constant functions on the same algebraic curve, then it's not hard to show that they would have the same poles. Beyond that, I have nothing else to add, except that making trivial edits to keep bumping your question to the front page is not cool. $\endgroup$
    – Felipe Voloch
    Commented Sep 30, 2011 at 17:21
  • $\begingroup$ I mean if they don't have the same poles the answer is yes. $\endgroup$
    – Felipe Voloch
    Commented Sep 30, 2011 at 17:24
  • $\begingroup$ I think Serre's proof may go through for a finite extension of $\mathbb{Q}(T)$, but I'll have a look at the more geometric approach you suggest also. Thanks for the help - sorry about the edits, won't be doing that again! $\endgroup$
    – Adam Harris
    Commented Oct 2, 2011 at 8:59

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If $K$ is a function field over an algebraically closed field and one of your elliptic curves is constant (which does not necessarily violate your hypotheses unless the constant field is the algebraic closure of a finite field) then the answer is no. What kind of constant field are you interested in? You might want to add some non-isotriviality condition. The person to ask is probably Chris Hall, but I don't think he reads MO.

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  • $\begingroup$ Thanks Felipe - I should have given more information: I specifically am thinking of a situation where I have two non-CM, non-isogenous curves $E$ and $E'$ with $j$-invariants $j(\tau)$ and $j(\tau')$ which are transcendental over $\mathbb{Q}$ but $j(\tau')$ is algebraic over $\mathbb{Q}(j(\tau))$. Also $E$ is defined over $\mathbb{Q}(j(\tau))$ and $E'$ over $\mathbb{Q}(j(\tau'))$. $\endgroup$
    – Adam Harris
    Commented Sep 28, 2011 at 14:47

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