How to prove that the polynomial $x^n+x^{n-1} +⋯+x^2+x−1$ is irreducible over $ \mathbb Q$?

How to prove that $x^n+x^{n-1} +⋯+x^2+x−1$ is irreducible over $\mathbb Q$?

Below I show you my (unsuccessful) attempt to prove it using Eisenstein's criterion when $n+1$ is a prime. First, observe

$$\frac{x^{n+1} - 2x + 1 }{ x-1} = x^n + x^{n-1} + \cdots + x^2 + x - 1.$$

Next, substitute $x+1$ for $x$, and obtain

$$ {\frac {(x+1)^{n+1}-2x - 1}{x}}=x^n + {\binom {n+1}{n}}x^{n-1}+\cdots +{\binom {n+1}{2}}x+ (n-1).$$

We need that all coefficients, except the coefficient of $x_n$, be divisible by $n+1$ and $(n+1)^2$ must not divide the constant coefficient $n-1$. But $n+1$ must divide the constant coefficient! And this is not the case...

How to proceed further? What to do in general case?

Solution 1:

You can also use the following criterion by Brauer on the (negative) reciprocal $$-x^nf(1/x)=x^n-x^{n-1}-\dots-1.$$

Let $a_1 \geq a_2 \geq \dots \geq a_n$ be positive integers and $n \geq 2$. Then the polynomial $p(x)=x^n-a_1x^{n-1}-a_2x^{n-2}-\dots-a_n$ is irreducible over $\mathbb{Z}$.

For the proof see for example Prasolov's Polynomials (Theorem 2.2.6).

Solution 2:

This is result of Perron and Selmer, after your $x+1$ trick. See Theorem 2 here:

Selmer, Ernst S., On the irreducibility of certain trinomials, Math. Scand. 4, 287-302 (1956). ZBL0077.24602.