If $p \in \mathbb{N}$ is a prime, is $\displaystyle A=\frac{x^{p^{2}}-1}{x^{p}-1}$ irreducible in $\mathbb{Q}[x]$?
I don't think it is. If somebody sees a contradiction, I would be glad to see it.
The link of Qiaochu links to the Eisenstein criterion.
There it is written that:
cyclotomic polynomials can be obtained by dividing the polynomial $x^{p}-1$ (in this case $x^{p^{2}}-1$) by $x-1$ (in this case $x^{p}-1$, which is it's obvious root.
Then the article makes a substitution so the criterion can be applied: by substituing x+1 for x this gives : $((x+1)^{p}-1)/x = x^{p-1} + (p nCR p-1)x^{p-2} + .... + (p nCR 1) $. The coefficients are divisible by p because of the properties of binomial coefficients, and therefore not divisible by $p^{2}$.
How to find the substitution so Eisenstein criterion can be applied?
I am very thankful for any insights.
It is the minimal polynomial for the roots of unity of order $p^2$, and it is irreducible.