3
$\begingroup$

I have a question about common divisors of some expressions involving Gaussian coefficients, in particular in the case ${n \brack 1}_{q} = \frac{q^{n}-1}{q-1}$ where $q$ is a prime power.

It is well known that the greatest common divisor $$\gcd{\left( {a \brack 1}_{q}, {b \brack 1}_{q}\right)} = {d \brack 1}_{q}$$ where $d = \gcd{(a,b)}$.

If we assume that $a,b \geq 1$, $a+b < n$, is there an easy way to determine the following? $$\gcd{\left({n \brack 1}_{q},\ q^{a}{b \brack 1}_{q} \pm 1\right)}$$

This has come up in some research I am doing. In particular I am hoping to find a nice way to determine when this $\gcd$ is 1.

I do know that if $n = a+kb+s$ for some $0 \leq s < b$ then $${n \brack 1}_{q} = q^{a}{b \brack 1}_{q}q^{s}\frac{q^{kb}-1}{q^{b}-1} + {s \brack 1}_{q}$$ which I'm hoping can help.

$\endgroup$
5
  • $\begingroup$ In the ring $\Bbb Q[q]$? $\endgroup$
    – anon
    Commented Apr 19, 2022 at 6:00
  • $\begingroup$ @runway44 I was thinking of $q$ being a fixed (prime power) integer, and just looking at divisibility in $\mathbb{Z}$. But certainly thinking of $q$ as an indeterminate, and looking at factoring in $\mathbb{Q}[q]$ is close to equivalent. $\endgroup$
    – xxxxxxxxx
    Commented Apr 19, 2022 at 6:55
  • $\begingroup$ @runway44 Your deleted answer seems useful, with the modification that I think we need to consider that $\zeta$ could be a $d$th root of unity for some $d \mid n$. $\endgroup$
    – xxxxxxxxx
    Commented Apr 19, 2022 at 9:28
  • $\begingroup$ That's not a modification - if $d\mid n$ then any $d$th root of unity is an $n$th root of unity. I assume you're thinking of primitive roots of unity. $\endgroup$
    – anon
    Commented Apr 19, 2022 at 9:35
  • $\begingroup$ @runway44 I'm not saying that in this case $\zeta$ wouldn't also be an $n$th root of unity, but I'm thinking of your conclusion at the end, that if $\zeta^{b} = \zeta^{-1}$ then $b = n-1$; if $\zeta$ is also a $d$th root of unity for some $d < n$, there are other possibilities for $b$. $\endgroup$
    – xxxxxxxxx
    Commented Apr 19, 2022 at 10:15

0

You must log in to answer this question.