Timeline for If Ramanujan's tau function has a prime power zero then $\ldots$
Current License: CC BY-SA 2.5
17 events
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| Jan 17, 2011 at 17:15 | comment | added | GH from MO | @ndkrempel: Thanks for the clarification, I misunderstood indeed. @Luis: I apologize for misunderstanding you! | |
| Jan 17, 2011 at 16:06 | comment | added | ndkrempel | @GH: I think you misunderstood - Luis was concerned that the proof would only work for N being the smallest prime power with $\tau(N) = 0$, as opposed to larger prime powers for which $\tau(N) = 0$. | |
| Jan 17, 2011 at 14:52 | comment | added | GH from MO | @Luis: Sure. I just said something easy (that you did not understand before) hoping that you would be grateful. | |
| Jan 17, 2011 at 14:48 | comment | added | Luis H Gallardo | @GH:The important things about this post appears in Laurent's answer and in Francois's comment. | |
| Jan 17, 2011 at 14:34 | comment | added | GH from MO | @Luis: Yes, but you seemed not to understand that the smallest $N$ such that $\tau(N)=0$ is trivially a prime power, which follows from the multiplicativity of $\tau$ (as I explained). Earlier you stated this as a nontrivial assumption, read back your second comment in this thread. | |
| Jan 17, 2011 at 11:40 | comment | added | Luis H Gallardo | @GH: Right, but $\tau(p_1^n)=0$ does not (generally) implies $\tau(p^h)=0$ for $p^h < p_1^n.$ | |
| Jan 17, 2011 at 10:54 | comment | added | GH from MO | @Luis, $\tau$ is a multiplicative function, hence if $\tau(N)=0$ then $\tau(p^n)=0$ for some prime power $p^n$ exactly dividing $N$. This is simply because nonzero complex numbers have a nonzero product. | |
| Jan 17, 2011 at 10:23 | comment | added | Luis H Gallardo | @Fancois: This is very nice ! | |
| Jan 17, 2011 at 10:16 | comment | added | François Brunault | @Luis, condition (c) also implies $\tau(p)=0$ : let $1-\tau(p)X+p^{11} X^2 = (1-\alpha X)(1-\beta X)$ with $|\alpha|=|\beta| = p^{11/2}$ (by Deligne). Condition (c) says that $\sum_{i=0}^{2r} \alpha^i \beta^{2r-i} = (-1)^r p^{11r}$ which implies that $\alpha/\beta$ is a root of unity, so some $\tau(p^k)$ is zero. | |
| Jan 17, 2011 at 9:41 | comment | added | Luis H Gallardo | Assuming only (c) (instead of (b)) I do not know what happens. | |
| Jan 17, 2011 at 9:36 | vote | accept | Luis H Gallardo | ||
| Jan 17, 2011 at 9:36 | comment | added | Luis H Gallardo | @Laurent: You are right ! Lehmer's result uses formula (5), in page 667, line -3. I wondered (a little quickly) to get the result only from formulas (3) and (4), as in my question here... | |
| Jan 17, 2011 at 9:05 | comment | added | Laurent Berger | @Luis: Where in the proof of "if $\tau(p^n)=0$ then $\tau(p)=0$" (bottom of page 430 and top of page 431) is it assumed? | |
| Jan 17, 2011 at 8:59 | comment | added | Luis H Gallardo | @Laurent: Sure, but assuming that $N=p^n$ is the "smallest" $N$ such that $\tau(N)=0.$ If we do not assume this the proof of Lehmer vanishes ! | |
| Jan 17, 2011 at 8:51 | comment | added | Laurent Berger | @Luis: if I'm not mistaken, the proof of theorem 2 consists in showing that if $\tau(p^n)=0$ then $\tau(p)=0$ which does answer your question. | |
| Jan 17, 2011 at 8:46 | comment | added | Luis H Gallardo | @Laurent: I know the paper. Using it I putted in the question that we may assume that the power of $p$ (assumed to be a possible zero of $\tau$) is not the smallest one. Indeed, Lehmer proved these result by playing with the known congruences for $\tau$ together with some computations. (He succedded since being the smallest possible zero implies reasonable computations). | |
| Jan 17, 2011 at 8:29 | history | answered | Laurent Berger | CC BY-SA 2.5 |