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Oct 7, 2011 at 18:43 comment added Tommaso Centeleghe That's a very appropriate reference, thanks. I was aware of it, but I somehow forgot about.
Oct 7, 2011 at 18:02 comment added François Brunault The following paper by Koo, Stein and Wiese addresses the question of irreducibility of Hecke polynomials : wstein.org/papers/heckepoly
Oct 7, 2011 at 17:34 comment added Tommaso Centeleghe @Jo\"el: I don't know what to think. It just seems that in level one a single Hecke operator in enough to distinguish eigenforms!
Oct 7, 2011 at 17:31 comment added Tommaso Centeleghe @JSE: wait, do you have a proof of your last assertion, or this is just something with empirical evidence?
Oct 7, 2011 at 16:22 comment added Joël Nice answer, but does it mean that you guys think that the observation that in level 1 it never happens that two eigenforms have the same T_l eigenvalues for any l is false? But then it would be nice to have a counter-example.
Oct 7, 2011 at 16:12 comment added JSE Try it for a case where the space has dimension higher than 2 and the forms are all Galois conjugate!
Oct 7, 2011 at 15:30 comment added Tommaso Centeleghe Ok, thanks. I guess your argument can be adapted to the case where there is no form with rational coefficients in $S_2(\gamma_0(N))$. At least one can observe a similar phenomenon as that above even for values of $N$ (e.s $23$) for which there is not elliptic curve of that conductor.
Oct 7, 2011 at 15:21 comment added JSE By the way, this is somewhat badly written: I should have written "E_1 and E_2" since I'm really using the fact that the coefficients of f and g are rational integers.
Oct 7, 2011 at 15:11 vote accept Tommaso Centeleghe
Oct 7, 2011 at 14:54 history answered JSE CC BY-SA 3.0