Prove that if $p$ is prime then $\frac{X^p-1}{X-1}=X^{p-1}+\cdots+X+1$ is irreducbile in $\mathbb{Q}[X]$.

Question

I have no idea how to approach this.

I'm supposed to use $f\in \mathbb{Q}[X]$ is irreducible $\iff \exists{a}\in\mathbb{Q}$ such that $f(X+a)$ is irreducible.

I tried to use $a = 1 \in \mathbb{Q}$ so $\frac{(X + 1)^p-1}{X}=(X+1)^{p-1}+\cdots+ (X+1) + 1$ but I don't see how I can get from there to being able to use any other criterion.

See http://math.stackexchange.com/questions/215042/irreducibility-of-xp-1-ldots-x1 and several linked questions. — Bart Michels, Jun 03 '14 at 18:21
But I don't see how I can easily find the divisor $d$ to apply Eisenstein. — jjjx, Jun 03 '14 at 18:24
In a case like this you should always look at a few examples. The cases $p=2,3,5$ would show you immediately what the story is. — Lubin, Jun 03 '14 at 18:34

score 0 · Answer 1 · answered Jun 03 '14 at 18:47

From where you left off:

$\displaystyle \sum_{i=0}^{p-1} (X+1)^i = \displaystyle \sum_{i=0}^{p-1} \displaystyle \sum_{k=0}^i \binom{i}{k} X^k = 1 + (1 + X) + \left(1 + \binom{2}{1}X + \binom{2}{2}X^2\right) + \left(1 + \binom{3}{1}X + \binom{3}{2}X^2 + X^3\right) + ... + \left(1 + \binom{k}{1}X + \binom{k}{2}X^2 +...+ \binom{k}{j}X^j +..+ X^k\right) + ...+ \left(1 + \binom{p-1}{1}X + \binom{p-1}{2}X^2 +...+ \binom{p-1}{j}X^j +...+ X^{p-1}\right) = p + \displaystyle \sum_{k=1}^{p-1} \binom{k}{1}X + \displaystyle \sum_{k=2}^{p-1} \binom{k}{2}X^2 + ...+ \displaystyle \sum_{k=q}^{p-1} \binom{k}{q}X^q + ...+ X^{p-1}$. You can use the Eisenstein's criterion for this polynomial.

Prove that if $p$ is prime then $\frac{X^p-1}{X-1}=X^{p-1}+\cdots+X+1$ is irreducbile in $\mathbb{Q}[X]$.

1 Answers1