Olympiad Proof Problem - Day 5

Prove that for all for primes pp the polynomial

xp1+xp2+xp3++1x^{p-1} + x^{p-2} + x^{p-3} + \ldots + 1

is irreducible over rational numbers.


Try more proof problems at Olympiad Proof Problems.

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Note by Surya Prakash
5 years, 7 months ago

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Comments

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

Let

Φp(x)=xp1+xp2++x2+x+1=xp1x1\Phi_p (x) = x^{p-1} + x^{p-2} + \ldots + x^2 + x + 1 = \dfrac {x^p - 1}{x-1}

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

f(x)=Φp(x+1)=(x+1)p1(x+1)1=k=0p1(pk)xpkxf(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}

=xp1+(p1)xp2+(p2)xp3++(pp2)x+(pp1)=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 pp, and the constant coefficient is exactly pp, so it is not divisible by p2p^2. Thus, Eisenstein's criterion applies, and ff is irreducible. Certainly, if Φp(x)=g(x)h(x)\Phi_p (x) = g(x) h(x) then f(x)=Φp(x+1)=g(x+1)h(x+1)f(x) = \Phi_p (x+1) = g(x+1) h(x+1) gives a factorisation of ff. Thus, Φp\Phi_p has no proper factorisation, i.e. it is irreducible.

Sharky Kesa - 5 years, 7 months ago

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

Surya Prakash - 5 years, 7 months ago

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I have written in the proof in the wiki.

Sharky Kesa - 5 years, 7 months ago

Well, is cyclotomic polynomials allowed?

Pi Han Goh - 5 years, 7 months ago

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@Pi Han Goh Yeah, sure. Post your solution using cyclotomic polynomials.

Sharky Kesa - 5 years, 7 months ago

When is Day 6 going to be posted?

Sharky Kesa - 5 years, 6 months ago
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