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Sep 8, 2016 at 10:01 comment added Fedor Petrov @Mr.Kho total degree (not a number! above it is $m_1+2m_2$, not $m_1+m_2$) of linear and quadratic factors of a polynomial $g(x)\in \mathbb{F}_p[x]$ is a total number of roots of $g$ in $\mathbb{F}_{p^2}$. For $g(x)=x^t-1$ it equals gcd$(t,q^2-1)$ as may be seen by considering a primitive root $\alpha$ in $\mathbb{F}_{p^2}$ and checking which powers of $\alpha$, $\beta=\alpha^s$, satisfy $\beta^t=1$.
Sep 8, 2016 at 8:11 comment added Mr. Kho in one of your answers above you say that "this is standard ...total degree of linear and quadratic factors is gcd(t,$q^2−1$)". Is there a book which explains how to arrive at this?
Oct 28, 2015 at 11:11 comment added Fedor Petrov If characteristic polynomial is $X^2+X+1$, then determinant is 1.
Oct 28, 2015 at 9:20 comment added Mr. Kho @ Fedor Petrov, checking the determinants of the elements (matrices) of order 3 in GL(2,GF(32)), I find that all elements have determinant equal to 1. Is this a mere coincidence?
Oct 23, 2015 at 11:35 comment added Fedor Petrov For $t=11$ we see that $x^{11}-1$ has 1 linear multiple $x-1$ and 5 quadratic multiples, that is, $m_1=1$, $m_2=5$. Thus we should have $1+5\cdot 32\cdot 31=4961$ matrices for which $X^{11}=1$, out of them there is 1 matrix of order 1 and 4960 matrices of order 11.
Oct 23, 2015 at 7:43 comment added Mr. Kho Also, is there a paper/book/any reference material where I can find this?
Oct 23, 2015 at 7:41 comment added Mr. Kho @ Fedor Petrov, I have tried this and it works for t=3, however, over the same field, for t=11 and t=31, my computer results are different from those I compute by hand. Anything I am missing?
Oct 21, 2015 at 7:40 comment added Mr. Kho many thanks for your answer it solves the problem.
Oct 21, 2015 at 7:27 vote accept Mr. Kho
Oct 21, 2015 at 7:38
Oct 20, 2015 at 13:49 history edited Fedor Petrov CC BY-SA 3.0
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Oct 20, 2015 at 10:57 history edited Fedor Petrov CC BY-SA 3.0
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Oct 20, 2015 at 10:03 history answered Fedor Petrov CC BY-SA 3.0