$1+x^p+x^{2p}+\dotsb+x^{p(p-1)}$ irreducible

Question

Let $p$ be a prime number. Is $$f(x)=1+x^p+x^{2p}+\dotsb+x^{p(p-1)}$$ an irreducible polynomial over $\Bbb Z$?

Can we use Eisenstein's criterion?

$f(x+1)=1+(x+1)^p+(x+1)^{2p}+\dotsb+(x+1)^{p(p-1)}$

I am stuck. Thanks a lot!

$t=x^p\iff f(x)=\dfrac{t^p-1}{t-1}$ – Lucian Apr 25 '14 at 03:43 — Lucian, Apr 25 '14 at 03:43

DeepSea · Answer 1 · 2014-04-25T02:18:36.873

2

The constant term of $f(x+1)$ is $p$. So you can use the Eisenstein's criterion for this question.

edited Apr 25 '14 at 02:18

answered Apr 24 '14 at 23:07

DeepSea

coefficients of $x^p, x^{2p}$ ? – ziang chen Apr 24 '14 at 23:13
@ziangchen: no, the constant term of the polynomial p(x+1) is 1 + 1 +... + 1 = p – DeepSea Apr 24 '14 at 23:23

1 Answers1