Alternating irreducibility of polynomial $t^n + t^{n-1} + \ldots + t + 1$?

Question

I was doing a problem on Stewart's Galois Theory. In section 3, we were asked to decide the irreducibility or otherwise of the following polynomials:

$t^4 + t^3 + t^2 + t + 1$

$t^5 + t^4 + t^3 + t^2 + t + 1$

$t^6 + t^5 + t^4 + t^3 + t^2 + t + 1$

After proving polynomials in this form with degree 4, 5, 6, 7, 8, I realize that, for $p(t) = t^n + t^{n-1} + \ldots + t + 1$, we have two cases:

1. If $n$ is odd, the polynomial is reducible with the factorization $t^n + t^{n-1} + \ldots + t + 1 = (t + 1)(t^{n-1} + t^{n-3} + \ldots + t^2 + 1)$.
2. If $n$ is even, then the polynomial $p(t) = t^n + t^{n-1} + \ldots + t + 1$ is irreducible if and only if $p(t+1)$ is irreducible. Then, substituting $t$ with $t+1$, $p(t + 1) = a_n t^n + a_{n-1} t^{n-1} + \ldots + a_{1}$, where the coefficient $a_i$ is the $(n+ 1 - i)$-th item in the $(n+1)$-th row of Pascal's Triangle. For example: when $n = 4$, $p(t+1) = t^4 + 5t^3 + 10t^2 + 10t + 5$. Thus we could use Eisenstein's Criterion and let the prime $q = n+1$ or in this example, $q = 5$, such that the irreducibility of $p(t)$ immediately follows.

This alternating pattern of irreducibility looks very fascinating to me but I have no idea why it is this. What is, if there is any, the deep idea behind this? Why the (ir)reducibility depends on the parity of the degree?

Answer 1

C. Falcon’s response was just on the mark, but there’s something even deeper going on here.

In fact, it makes more sense to look at the polynomials $X^n-1$, which always have a factor $X-1$, and no others if $n$ is prime, but the general fact is this: $$ X^n-1=\prod_{d|n}\Phi_d(X)\,, $$ the $\Bbb Z$-polynomials $\Phi_d$ being the cyclotomic polynomials mentioned in Falcon’s answer. These start out with $\Phi_1(X)=X-1$, $\Phi_2(X)=X+1$, and $\Phi_3(X)=X^2+X+1$. Using the Möbius Inversion Formula, you may prove that $$ \Phi_m(X)=\prod_{d|m}\bigl(X^d-1\bigr)^{\mu(m/d)}\,, $$ where $\mu$ is the Möbius function, which I’ll give you the pleasure of finding out about.

Answer 2

One has: $$t^8+t^7+t^6+t^5+t^4+t^3+t^2+t+1=\left(t^2+t+1\right)\left(t^6+t^3+1\right).$$ Hence, $t^8+t^7+t^6+t^5+t^4+t^3+t^2+t+1$ is reducible over $\mathbb{Q}$ and your conjecture is false.

However, your argument is perfectly fine when $n+1$ is a prime number (notice that it's not always the case) and in this very case you are dealing with cyclotomic polynomials of prime order.