Timeline for The sum of same powers of all matrices modulo p
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 11, 2013 at 8:19 | vote | accept | Anton Klyachko | ||
| Apr 11, 2013 at 8:19 | comment | added | Anton Klyachko | Matthieu, the solution goes as follows: Ilya showed that the sum in question vanishes for sufficiently small $k$; then ya-tayr showed that if the sum vanishes for all sufficiently large $k$, than it vanishes always; when these bounds have met, "we're done!". This is all right, but honestly I hoped for a simpler solution. ---- Thanks, ya-tayr, Ilya, Will, and everybody involved! | |
| Apr 10, 2013 at 12:50 | comment | added | Matthieu Romagny | If "we're done" then could someone sum things up, please? I'm a bit lost. | |
| Apr 9, 2013 at 17:30 | comment | added | Will Sawin | It's $\prod_{k=1}^p(x^{p^k}−x)$, I guess, whose degree is $(p^{p+1}−1)/(p−1)$. So I think we're done! | |
| Apr 9, 2013 at 17:24 | history | edited | ya-tayr | CC BY-SA 3.0 |
added 537 characters in body; added 9 characters in body
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| Apr 9, 2013 at 7:01 | comment | added | Anton Klyachko | ... and it is easy to find the LCM of all unitary polynomials of degree $p$. | |
| Apr 9, 2013 at 6:45 | comment | added | Anton Klyachko | ya-tayr, Ilya, I am not sure I understand. Should "determinant of $A$" read "determinant of $I-Ax$"? If so, there are exactly $p^{p-1}$ different denominators: they are just polynonomials reciprocal to characteristic polynomials of all matrices (= all unitary polynomials of degree $p$). | |
| Apr 9, 2013 at 5:42 | comment | added | ya-tayr | Oh! - | |
| Apr 9, 2013 at 4:50 | comment | added | Ilya Bogdanov | The approach is interesting. But it seems that you think there are $p^3$ matrices; in fact, there are $p^{p^2}$ of them. But you do not need so much, since there are less than $p^p$ distinct denominators (all costant terms are ones). This gives the estimate of about $2p^{p+1}$ coefficients to check. Unfortunately, now we have a half of this amount. | |
| Apr 8, 2013 at 23:50 | history | answered | ya-tayr | CC BY-SA 3.0 |