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Let E be an elliptic curve over $\mathbf{Q}$, and p be a prime of good reduction for E such that the Galois representation $\bar\rho_p$ of $\mathbf{Q}$ on the p-torsion of E surjects onto Aut(E[p]). Let now K be the number field cut out by the kernel of $\bar\rho_p$. What is known about the primes q that are completely split in K? Have they been characterized?

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Shimura looked at $E:y^2+y=x^3-x$ and showed that a prime $l\neq11,p$ splits in ${\bf Q}(E[p])$ if and only if $$\rho(Frob_l)=\pmatrix{1&0\cr0&1\cr}$$ where $\rho$ is the mod-$p$ representation associated with $$ q\prod_{k=1}^{+\infty}(1-q^k)^2(1-q^{11k})^2. $$ See my arxiv.org/abs/1007.4426 for an elementary introduction. –  Chandan Singh Dalawat Oct 10 '11 at 3:35
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Have you looked at Adelmann's LNM (vol. 1761) "The Decomposition of Primes in Torsion Point Fields"? –  Nicolas B. Oct 10 '11 at 7:37
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By the way, Shimura did not give any examples of primes which split in say ${\bf Q}(E[7])$. The first such examples $$ 4831, 22051, 78583, 125441, 129641, 147617, 153287, 173573, 195581, ... $$ were computed by Tim Dokchitser. –  Chandan Singh Dalawat Oct 11 '11 at 3:16
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@Chandan Singh Dalawat: I think that there is a mistake in the equation of the elliptic curve in your first comment. The one displayed is the curve labeled 37a1 in Cremona's table, but if I compute the $a_{\ell}$'s of $E$ for $\ell=4831, 22051,\ldots$, I don't find $2\pmod{7}$. Am I mistaken? –  Nicolas B. Oct 18 '11 at 20:50
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@Nicolas B, you are absolutely right. I was relying on my faltering memory; the equation should have been $$y^2+y=x^3-x^2,$$ which indeed has conductor $11$. Whoever sees this comment, please "upvote" it so that it is always visible. –  Chandan Singh Dalawat Oct 21 '11 at 3:26
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2 Answers

It is a difficult problem if you want really convenient criteria. The condition, for primes $\ell$ distinct from $p$ and not dividing the conductor, is that the Frobenius at $\ell$ acts trivially on the $p$-torsion points. However, most theoretical knowledge about this action are related to its characteristic polynomial $X^2-a(p)X+p$ (modulo $\ell$). So you get "for free" some simple congruence condition on $a(p)$ modulo $\ell$, that are necessary for $\ell$ to be totally split in the field generated by $p$-torsion. These are not sufficient -- in your situation, with surjective Galois action, the probability that a prime be totally split is about $1/p^4$, while the congruence condition only restricts prime to a set of density about $1/p^2$.

On the other hand, algorithmically, these primes can be investigated: $\ell$ is totally split if and only if the $p$-torsion elements of the curve modulo $\ell$ are defined over $F_{\ell}$, and there are fast algorithms to compute the group structure of $E(\mathbf{F}_{\ell})$ (in the unramified case), from which one can extract the set of primes $p$ such that $\ell$ is totally split on $p$-torsion points. So one can find, e.g., that the curve $y^2=x^3+6x-2$ has all its $140$-torsion points defined over the field with $196561$ elements (a much smaller prime than the Chebotarev-based heuristic would suggest.)

Another property worth keeping in mind, at least when making heuristics, is that if one has a "family" of elliptic curves, one can often say more. What happens (at least in many cases) is that after averaging over a suitable family, the problem boils down again to congruence conditions, though typically of a non-linear kind. This is related to the fact that the structure of all elliptic curves (up to isomorphism) modulo a fixed prime can often be investigated using Deuring's result on the distribution of their endomorphism rings, or the trace formula on various modular curves.

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I think I found a criterion which enables one to decide whether $Frob_\ell$ acts semisimply on E[p] or not. I wanted to how much has been already done on this. Thanks. –  Tommaso Centeleghe Oct 10 '11 at 7:21
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I think you want Proposition 5.6.3 of the book "The Decomposition of Primes in Torsion Point Fields" by Clemens Adelmann. If $p\geq 5$, then his result is that a prime of good reduction $q$ will split completely in $E(\mathbb{Q}[p])$ if and only if (1) $q\equiv 1\bmod p$, (2) the modular polynomial $\Phi_p(X,j(E))$ splits into distinct linear factors in $\mathbb{F}_q[X]$, and (3) $a_q \equiv 2 \bmod p$.

When $p=2,3$, he has similar theorems. In fact, there are results for splitting in $E(\mathbb{Q}[p^m])$ for all $p$ and all $m$.

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thanks! After Nicolas' comment above, I in fact had a chance to look at this proposition. –  Tommaso Centeleghe Nov 14 '11 at 14:46
    
Ha! I hadn't even seen that comment. Sorry for doubling up! –  user12750 Nov 14 '11 at 22:19
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