Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Everyone, I am now reading a paper named The Irreducible Factors of $(cx+d)x^{q^m}-(ax+b)$ over $GF(q)$, http://qjmath.oxfordjournals.org/content/14/1/61.extract. And I’m confused with one of its details, described in the following here. So can someone help me with it:

Theorem 5 of this paper says "… the number of monic irreducible factors of degree $rt$ which divide some $h_m(x)$ is $\frac{\phi(t)}{rt}\sum\mu(w)q^{r/w}$, ($w|r, (w,t)=1$)…". Here $\phi(t)$ is Euler’s function and $\mu(t)$ is Mobius’ function, and $h_m(x)$ is $(cx+d)x^{q^m}-(ax+b)$.

To get this conclusion, the author just says "we simply apply the usual counting process to the products $P_m(x)=\prod h_{m,u}(x)$, ($1\leq u\leq t, (u,t)=1,m\geq 1$)…", where $h_{m,u}(x)$ is $(c_ux+d_u)x^{q^m}-(a_ux+b_u)$ and he has proved that the number of irreducible polynomials of a given degree which divide $h_{m,u}(x)$ is the same for all $u$ prime to $t$. My question is: what is the "counting process"? Does this mean the way we use to compute the usual formula for the number of all monic irreducible polynomials of a given degree without any restrictions using Mobius inversion, as when $t=1$ this theorem gets the same conclusion for degree $r$? But how to deal with $rt$ and where does $(w,t)=1$ in the formula come from? What is the function like when its Mobius inversion is $\sum\mu(w)q^{r/w}$, ($w|r, (w,t)=1$)?

A useful conclusion may be Theorem 4 in the paper, saying that $f(x)$ of degree $rt$ divides $h_m(x)$ if and only if $m\equiv rs (\mod rt)$.

share|improve this question
    
I find this paper might be useful: link –  a guest Feb 25 at 2:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.