TITLE: Irreducibility issue
QUESTION [7 upvotes]: This is a homework question. 
Given $f(x)=x^{p-1}+x^{p-2}+\cdots+x+1$, where $p$ is any prime. 
Prove that $f(x)$ is irreducible over $\mathbb{Z}[x]$?
Any idea, hint, etc? Hint given by my book was to use Eisenstein's Irreducibility Criterion. But I see that the coefficients of each term is 1 which is not divisible by any prime number, so how can the criterion be satisfied?

REPLY [3 votes]: As a matter of interest I think you can do this problem without using Eisenstein if you accept the uniqueness of factorization in $F[x],$ where $F$ is the field with $p$ elements.For, as others have already remarked, we have $x^p - 1 = (x-1)^p$ and $1+x \ldots + x^{p-1} = (x-1)^{p-1}$ in $F[x].$ This means that if we can write $1+x \ldots + x^{p-1} = f(x).g(x)$ with $f(x),g(x) \in \mathbb{Z}[x],$ each of degree at least $1,$ then we have $f(x) = (x-1)^j + p.h(x)$ and $g(x) = (x-1)^{p-1-j} + p.k(x)$ for some integer $j$ with $1\leq j \leq p-1$ and polynomials $h(x),k(x) \in \mathbb{Z}[x].$ But notice then that $f(1)$ is an integer multiple of $p$ and $g(1)$ is an integer multiple of $p.$ However, evaluating $1 + x + \ldots + x^{p-1}$ at $1$ gives $p$, which is certainly not an integer multiple of $p^2.$ I remark, since it has been mentioned in another problem about $p$-power cyclotomic polynomials on this site, that this method also works for the cyclotomic polynomial $\Phi_{p^k}(x)$ for any positive integer $k,$ since we have $\Phi_{p^k}(1) = p$ and the image of $\Phi_{p^k}(x)$ factors as $(x-1)^{\phi(p^k)}$ in $F[x].$