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Nov 15, 2010 at 4:05 vote accept schur
Oct 21, 2010 at 6:11 answer added user631 timeline score: 9
Oct 19, 2010 at 15:44 comment added Tony Scholl @Jared: good point. I had in mind trying to find directly a Hecke character with the right local factor at $p$ - which would give an AV over $\mathbb{Q}$ with real multiplication by some tot.real field containing $F^+$. But I haven't had time to give it serious thought and maybe there is a trivial obstruction.
Oct 19, 2010 at 14:29 comment added Jared Weinstein @Tony: I tried thinking CM forms. Let's say $a_p$ is totally real, let $\pi$ be a root of $X^2+a_pX+p$, let $F^+=\mathbf{Q}(a_p)$ and let $F=F^+(\pi)$. Then $F/F^+$ is CM. Now $\pi$ is a Weil $p$-integer: the proof of Honda-Tate supplies us with an abelian variety $A$ over $\mathbf{F}_p$ with CM by $F$ on which Frob is $\pi$ (up to a root of unity). The problem is lifting $A$ to an abelian variety over $\mathbf{Q}$. (It lifts to some undetermined number field; restricting scalars to $\mathbf{Q}$ results in something too big.) I don't see any way around this.
Oct 19, 2010 at 11:33 comment added François Brunault @Olivier: yes, I think so. In fact if $f$ is a newform of level $N$ and Nebentypus character $\psi$ modulo $N$ then $\overline{f} = f \otimes \overline{\psi}$ so that $\overline{a_n(f)} = a_n(f) \overline{\psi}(n)$ for any $n$ prime to $N$. In particular we have the condition that $x/\overline{x}$ should be a root of unity. I suggest first looking at the case of a totally real algebraic integer.
Oct 19, 2010 at 10:37 comment added Olivier And if it is not totally real, by the stability of the Hecke algebra under the Rosatti involution and its positivity, the field generated by the coefficients of $f$ is a CM field, isn't it?
Oct 19, 2010 at 10:25 comment added François Brunault There might also be some restriction on the number field generated by $x$. For example, if $f$ is a weight $k$ newform on $\Gamma_0(N)$, then the number field gen'd by the coeffs of $f$ is totally real (this is because of the Atkin-Lehner involution).
Oct 19, 2010 at 10:03 comment added François Brunault In the case $x$ is an integer, I think it is known that there exists an elliptic curve $E$ over $\mathbf{F}_p$ with $a_p(E)=x$. So in this case the answer to Ian's question is yes.
Oct 19, 2010 at 9:51 history edited Tony Scholl
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Oct 19, 2010 at 7:46 comment added Tony Scholl Have you thought about finding a CM form with $a_p=x$?
Oct 19, 2010 at 5:10 comment added Jared Weinstein I think you have to impose the condition that all of the Q-conjugates of $a_p$ also have to lie inside the Hasse-Weil bound. (In this context I'd call it the Ramanujan bound, but I know what you mean.) For if $f$ is a normalized newform, then so is $\sigma(f)$, where $\sigma$ is any automorphism of $\overline{\mathbf{Q}}$.
Oct 19, 2010 at 2:26 comment added schur Dear Ben and Alex, that's right, I was being imprecise. I meant normalized newform.
Oct 19, 2010 at 2:24 history edited schur CC BY-SA 2.5
Normalized newform is what I meant.
Oct 19, 2010 at 2:13 comment added user1073 Alex - Even if the modular form had to be normalized, the question would still follow more or less trivially from the theory of newforms (in fact, you could let x be any complex number and get the same result). Perhaps he requires the modular form to be a normalized newform?
Oct 19, 2010 at 1:34 comment added Alex B. I assume that Ian means a normalised modular form.
Oct 19, 2010 at 1:15 comment added user1073 Because if it is, I'm afraid that I don't really understand the question; why couldn't I just take any modular form whose p-th fourier coefficient is nonzero and multiply it by a suitable scalar?
Oct 19, 2010 at 1:10 comment added user1073 Is a_p supposed to be the p-th fourier coefficient of your modular form?
Oct 19, 2010 at 0:54 history asked schur CC BY-SA 2.5