Timeline for Normal basis in cyclotomic number fields
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 9, 2020 at 3:50 | comment | added | Gerry Myerson | Link to the paper, please? | |
| Jan 9, 2020 at 3:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 29, 2015 at 12:04 | comment | added | Angel del Rio | The answer is positive. I got a solution by René Schoof. The solution is too long to include it here. I will include it in a paper that I am writing. | |
| Apr 14, 2015 at 16:27 | comment | added | Angel del Rio | Another equivalent form of the question: Let $k\mod p$ denote the representative of $k$ modulo $p$ in the interval $[-n,n]$ where $n=\frac{p−1}{2}$. Consider the $n\times n$ matrix having $1$ in the $(i,j)$-th entry if $ij \mod p$ is odd and $0$ otherwise. Is A invertible in the ring of rational matrices? | |
| Apr 8, 2015 at 12:53 | comment | added | Gerry Myerson | Also posted as mathoverflow.net/questions/202321/… | |
| Apr 7, 2015 at 7:18 | comment | added | Angel del Rio | Observe that the original basis is the normal basis generated by $b=\zeta+\zeta^{-1}$. Taking a generator $\sigma$ of the Galois group and reordering the origianal basis as $\{\sigma^i(b):i=0,1,\dots,\frac{p-1}{2}\}$ one can see see the coefficient matrix as a circulant matrix. This reduces to see if the polynomial $P$ with coefficients the first row of the circulant matrix has a root which is a $\frac{p-1}{2}$ root of unity, but I don't see how to prove the latter because I don't see regularity on the polynomial $P$. | |
| Apr 7, 2015 at 6:57 | comment | added | Angel del Rio | Dear Cam, thank you for the hint. I tried to do what you suggest but I don't see any regularity in the matrix of coefficients which can help to obtain . Well, writing $\alpha$ in a power basis (for example, $1$, $\zeta+\zeta^{-1}$, $(\zeta^2+\zeta^{-2}$, $\dots$, seems a bit messy. This is why I prefer to use that more "suitable" basis $b_1=\zeta+\zeta^{-1}, \dots, b_{\frac{p-1}{2}}=\zeta^{\frac{p-1}{2}}+\zeta^{-\frac{p-1}{2}}$. Then the coefficient matrix is formed by rows formed with 0 and -2, (0 or 1 more -2). | |
| Apr 7, 2015 at 1:48 | comment | added | Cam McLeman | I haven't given this too much thought, but I would've thought the natural approach would've been to write the conjugates in terms of the standard power basis, and then show that the matrix of coefficients was invertible. | |
| Apr 6, 2015 at 22:51 | comment | added | Angel del Rio | I encountered this question when studying the Zassenhaus Conjecture on torsion units of integral group rings of some finite groups. | |
| Apr 6, 2015 at 22:49 | comment | added | Angel del Rio | Yes! I also have verified that $\alpha$ generates a normal basis for all the odd primes up $p\le 1700$. But I don't have a clue why $\alpha$ always generates a normal basis. | |
| Apr 6, 2015 at 16:59 | comment | added | Cam McLeman | Have you tried writing down some of its conjugates? | |
| Apr 6, 2015 at 15:29 | history | asked | Angel del Rio | CC BY-SA 3.0 |