In a previous post on a $q$-analog of number theory I present the fact that any $q$-number can be written as a unique product of cyclotomic polynomials, similar to the fundamental theorem of arithmetic. Namely, that if $n = p_1^{\nu_{p_1}(n)}p_2^{\nu_{p_2}(n)}…p_i^{\nu_{p_i}(n)}$ then I show that
$${[n]}_q = \prod_i\left(\prod_{j=0}^{\nu_{p_i}(n)-1}\Phi_{p_i}\left(q^{p_i^j\prod_{r \leq i}p_r^{k_r}}\right)\right)$$
Some notation quickly:
A $q$-number is denoted as ${[n]}_q$ and defined as
$${[n]}_q = \sum_{k=0}^{n-1}q^k$$
We have the following rules for multiplication and addition
$${[nm]}_q = {[n]}_q{[m]}_{q^n}$$
$${[n+m]}_q = {[n]}_q+q^n{[m]}_q$$
I was just playing around with this and was curious as to what the 'discriminant of a number' would look like through its extension as a $q$-analog. I tried two different attempts to come to a closed form solution. My first attempt was simply in seeing that $\{q = \zeta_n^k, \,\,\,k<n: {[n]}_q=0\}$ and hence the discriminant is
$$D{[n]}_{q} = \prod_{k<j}(\zeta_n^k-\zeta_n^j)^2$$
$$ = \prod_{k<j}(e^{\frac{2\pi i k}{mn}}-e^{\frac{2\pi i j}{mn}})^2$$
This was my first attempt. I stopped here mainly because I didn't immediately see where to go from here, and because I wanted to use the '$q$-analog of prime factorization' to find a closed form solution. So here it is,
Given that
$${[n]}_q = \prod_i\left(\prod_{j=0}^{\nu_{p_i}(n)-1}\Phi_{p_i}\left(q^{p_i^j\prod_{r \leq i}p_r^{k_r}}\right)\right)$$
We know that
$$D(\Phi_p(x)) = (-1)^{(p-1)/2} p^{p-2}$$
for prime $p$
We also have that for two monic polynomials $f, g$ with real coefficients
$$D(fg)
= D(f)D(g)\prod_{i,j}(a_i-b_j)^2$$
So, for example
$$D{[6]}_q = D{[2]}_qD{[3]}_q\prod_{i,j}(\zeta_2^i-\zeta_3^j)^2$$
Since ${[2]}_q$ and ${[3]}_q$ are 'prime' and cannot be factored as cyclotomic polynomials other than $\Phi_2(q)$ and $\Phi_3(q)$, we get the simplified expression
$$D{[6]}_q = -6i\prod_{k,j}(\zeta_2^k-\zeta_3^j)^2$$
I couldn't find a way to come to a 'nice' closed form as suggested by this post. All I seem to get is
$$D{[n]}_q = \prod_i(-1)^{(p_i-1)/2} p_i^{p_i-2}\prod_{k,j}…\prod_{l,m}(\zeta_{p_1}^k-\zeta_{p_2}^j)^2…(\zeta_{p_{i-1}}^l-\zeta_{p_i}^m)^2$$
without any way to simplify further. I know I must be missing something obvious. Any help is appreciated!
Best Answer
It is a standard calculation in number theory that $\text{disc}(x^n-1) = (-1)^{n(n+1)/2 + 1}n^n$, and this implies that $\text{disc}([n]_q) = (-1)^{n(n+1)/2 + 1}n^{n-2}$. Let me prove both these statements.
Let $f(x)=x^n-1$ and $g(x)=[n]_x$ in your notation. Then $f(x)=g(x)(x-1)$, which is what will allow us to prove statement $2$ from statement $1$. Obviously $f(x)=\prod_{0\leq i \leq n-1}(x-\zeta^i)$, so the discriminant of $f$ is $$\text{disc}(f) = \prod_{0\leq i<j\leq n-1}\left(\zeta^i-\zeta^j\right)^2 = (-1)^{\frac{n(n-1)}{2}}\prod_{i\neq j}\left(\zeta^i-\zeta^j\right).$$ The sign here is just for ordering reasons. We will have to be very careful with our indices, so keep that in mind - when writing $i\neq j$, both indices will always run from $0$ inclusive to $n-1$ inclusive. From the expression of $f$ above, it is clear that $$f'(x)=\sum_{i=1}^n\prod_{j\neq i} (x-\zeta^j),$$ and so $f'(\zeta^{i_0})=\prod_{j\neq i_0} (\zeta^{i_0}-\zeta^j)$. However, we also have $f'(x)=nx^{n-1}$, and so $f'(\zeta^{i_0})=n\zeta^{i_0(n-1)}$. Combining these two identities, can compute $$\prod_{i\neq j}\left(\zeta^i-\zeta^j\right) = \prod_{i=1}^n f'(\zeta^i) = \prod_{i=1}^n n \zeta^{i(n-1)} = n^n \zeta^{\frac{n(n-1)(n+1)}{2}}.$$ Now just check that $\zeta^{\frac{n(n-1)(n+1)}{2}}=(-1)^{n+1}$, and you are done.
For the second statement, it is clear that $$\text{disc}(g) = \prod_{1\leq i<j\leq n-1}\left(\zeta^i-\zeta^j\right)^2 = \prod_{0\leq i<j\leq n-1}\left(\zeta^i-\zeta^j\right)^2\cdot \left(\prod_{1\leq j\leq n-1}(1-\zeta^j)^2\right)^{-1}.$$
The first term on the RHS is exactly $\text{disc}(f)$, and the second term is computed by $$\prod_{1\leq j\leq n-1}(1-\zeta^j)^2 = \left(\prod_{1\leq j\leq n-1}(1-\zeta^j)\right)^2 = (g(1))^2 = n^2.$$
This completes the proof.