# Given a prime $p\in\mathbb{N}$, is $A=\frac{x^{p^{2}}-1}{x^{p}-1}$ irreducible in $\mathbb{Q}[x]$?

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.

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.

-
 Qiaochu (note spelling) gave you a link. Did you check there to see whether it says anything about showing the polynomials are irreducible? – Gerry Myerson Nov 17 '11 at 12:15 Yes I checked it, it says one can apply the Eisenstein criterion. But the cyclotomic polynomial is of the form $z^{p^{2}}-1$. – Tashi Nov 17 '11 at 15:23 The Wikipedia link says the proof is not trivial, and gives a link to Lang's Algebra textbook. There is a proof at planetmath.org/encyclopedia/… but it uses Galois Theory. Maybe math.columbia.edu/~pugin/Teaching/USemBlog_files/CycloRed.pdf is easier to read. Or just type cyclotomic polynomial irreducible into Google and see what comes up. – Gerry Myerson Nov 17 '11 at 23:20

It is the minimal polynomial for the roots of unity of order $p^2$, and it is irreducible.