# Irreducibility of $\frac{X^{p^n}-1}{X^{p^{n-1}}-1}$.

Exercise: Let $p$ be a prime number. Then, the polynomial $$\frac{X^{p^n}-1}{X^{p^{n-1}}-1}$$ is irreducible over $\mathbb Z[X]$, for any integer $n \geq 1$.

I'm able to work out the case $n=1$ (substitution $Y=X+1$ and Eisenstein's criterion); how can I prove the general case?

• The same argument that works for $n=1$ can work for any $n$, with a bit more work. – Thomas Andrews Mar 17 '14 at 16:04
• Could you give me some more hints? – Francesco Genovese Mar 17 '14 at 16:36
• How did you prove it for $n=1$? – Thomas Andrews Mar 17 '14 at 16:56

## 1 Answer

Interesting fact: This is the $p^n$th cyclotomic polynomial $\Phi_{p^n}(x)$. Proving this for any $n$ is similar to the case $n=1$.Assume we are working over a field $\Bbb F$. We have $(x+1)^{p^{n-1}}=x^{p^{n-1}}+1\mod p$ . This allows us to write in $\Bbb F_p[x]$ \begin{align}\frac{(x+1)^{p^{n}}-1}{(x+1)^{p^{n-1}}-1}\end{align}=\\((x+1)^{p^{n-1}})^{p-1}+((x+1)^{p^{n-1}})^{p-2}+\dots+((x+1)^{p^{n-1}})+1=\\(x^{p^{n-1}}+1)^{p-1}+(x^{p^{n-1}}+1)^{p-2}+\dots+(x^{p^{n-1}}+1)+1=\\ \frac{(x^{p^{n-1}}+1)^{p}-1}{(x^{p^{n-1}}+1)-1}=x^{(p-1)p^{n-1}}

Which shows that $p$ divides all the coefficients but that of the highest degree. Note that the constant term is $p$(Prove it) which is not divisible by $p^2$, and you get your result by the Eisentstein criterion.