# 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.

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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.

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More generally, the cyclotomic polynomials (en.wikipedia.org/wiki/Cyclotomic_polynomial) are all irreducible. – Qiaochu Yuan Nov 17 '11 at 1:36

Here is a proof based on the Eisenstein criterion, without using the general result about the irreducibility of cyclotomic polynomials.

For the $p^2$-th cyclotomic polynomial $A = \frac{x^{p^2} - 1}{x^p - 1}$, we are going to show that $A(x+1)$ is an Eisenstein polynomial for the prime $p$.

Substituting $x+1$ into the equation $(x^p - 1)A(x) = x^{p^2} - 1$ yields (after the application of the binomial theorem and division by $x$) $$\left(\sum_{i=0}^{p-1}\binom{p}{i+1}x^i\right)\cdot A(x+1) = \sum_{i=0}^{p^2 - 1}\binom{p^2}{i+1}x^i.$$ Comparison of the constant coefficients shows that $pa = p^2$, where $a$ denotes the constant coefficient of $A(x+1)$. So $a = p$.

Furthermore, reading the above equation modulo $p$ we get $$x^{p-1} \cdot A(x+1) \equiv x^{p^2 - 1}\mod p$$ and therefore $$A(x+1) \equiv x^{p^2 - p}\mod p.$$ So $A(x+1)$ is indeed an Eisenstein polynomial.

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