I'm working through an exercise in Pinter's algebra book about different ways to determine irreducibility of polynomials over $\mathbb{Q}$.
It asks to show that $x^5+1$ is irreducible by sending it to $\mathbb{Z}_n[x]$ and showing it is irreducible there.
Namely, if $h : \mathbb{Z} \rightarrow \mathbb{Z}_n$ is the natural homomorphism, let $\bar{h} : \mathbb{Z}[x] \rightarrow \mathbb{Z}_n[x]$ be the homomorphism defined by $\bar{h}(a_0 + a_1 x + \cdots + a_n x^n) = h(a_0) + h(a_1)x + \cdots + h(a_n)x^n$. I showed that if $\bar{h}(a(x))$ is irreducible in $\mathbb{Z}_n[x]$ and $a(x)$ is monic, then $a(x)$ is irreducible in $\mathbb{Z}[x]$.
I know how to use this for polynomials like $x^4-10x^2+1$ (just send it to $\mathbb{Z}_{5}[x]$, and a previous exercise showed that $x^4+1$ is irreducible there), but I'm puzzled about how this technique helps with $x^5+1$, which doesn't have a term that would disappear under some mapping. I keep thinking there must be some use of Fermat's little theorem here, and am tempted to send $x^5+1$ to $\mathbb{Z_5}[x]$. Over $\mathbb{Z_5}[x]$, I know that $x^5+1$, viewed as a polynomial function, is equivalent to $x+1$ (which is irreducible). But I don't think this implies that $x^5+1$ is irreducible because as formal polynomial expressions $x^5+1$ and $x+1$ are different (i.e. coefficients of $x^k$ do not match).
Any tips?