Prove that for all for primes \(p\) the polynomial

\[x^{p-1} + x^{p-2} + x^{p-3} + \ldots + 1\]

is irreducible over rational numbers.

Try more proof problems at Olympiad Proof Problems.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestI used Eisenstein's Irreducibility Criterion to solve this question.

Let

\[\Phi_p (x) = x^{p-1} + x^{p-2} + \ldots + x^2 + x + 1 = \dfrac {x^p - 1}{x-1}\]

We claim that \(\Phi_p (x)\) is irreducible over the rational numbers. Although \(\Phi_p (x)\) itself does not directly admit application of Eisenstein's criterion, a minor variant of it does. That is, consider

\[f(x) = \Phi_p (x+1) = \dfrac {(x+1)^p - 1}{(x+1)-1} = \dfrac {\displaystyle \sum_{k=0}^{p-1} {p \choose k} x^{p-k}}{x}\]

\[=x^{p-1}+{p \choose 1} x^{p-2} + {p \choose 2} x^{p-3} + \ldots + {p \choose {p-2}} x + {p \choose {p-1}}\]

All the lower coefficients are divisible by \(p\), and the constant coefficient is exactly \(p\), so it is not divisible by \(p^2\). Thus, Eisenstein's criterion applies, and \(f\) is irreducible. Certainly, if \(\Phi_p (x) = g(x) h(x)\) then \(f(x) = \Phi_p (x+1) = g(x+1) h(x+1)\) gives a factorisation of \(f\). Thus, \(\Phi_p\) has no proper factorisation, i.e. it is irreducible.

Log in to reply

Nice solution. But why don't you do it another way or rather prove Eisenstein's criterion?

Log in to reply

Well, is cyclotomic polynomials allowed?

Log in to reply

Log in to reply

I

havewritten in the proof in the wiki.Log in to reply

When is Day 6 going to be posted?

Log in to reply