Is polynomial $1+x+x^2+\cdots+x^{p-1}$ irreducible?

Question

Let $p$ a prime number, is the polynomial

$$1+x+x^2+\cdots+x^{p-1}$$ irreducible in $\mathbb{Z}[x]$ ? Thanks in advance.

Answer 1

Mr. Eisenstein certainly thinks so!

The polynomial can be rewritten as $\frac{x^p-1}{x-1}$. Setting $x=(y+1)$ has no effect on (ir)reducibility of the polynomial; doing so and applying the binomial theorem yields: $$\frac{(y+1)^p-1}{(y+1)-1}=\frac{1}{y}\sum_{k=1}^{p}\binom{p}{k}y^{k}=y^{p-1}+\binom{p}{p-1}y^{p-2}+\ldots+\binom{p}{2}y^1+\binom{p}{1}y^0$$

Since all the coefficients apart from the leading one are divisible by $p$ and the constant term is too small to be divisible by $p^2$, we can apply the criterion and conclude that the polynomial is irreducible over rationals (and thus also over integers).

Answer 2

$1+x+x^2+\ldots +x^{p-1}=\frac{x^p-1}{x-1}$. So if we set $x=y+1$, then our polynomial becomes $\frac{(y+1)^p-1}{y}=y^{p-1}+py^{p-2}+\frac{p(p-1)}{2}y^{p-3}+\ldots+p$, which is irreducible by Eisenstein. But since $x \mapsto y+1$ is an isomomorphism from $\mathbb{Z}[x]$ to $\mathbb{Z}[y]$, our original polynomial is irreducible as well.