Prove that for all for primes ppp the polynomial
xp−1+xp−2+xp−3+…+1x^{p-1} + x^{p-2} + x^{p-3} + \ldots + 1xp−1+xp−2+xp−3+…+1
is irreducible over rational numbers.
Try more proof problems at Olympiad Proof Problems.
Note by Surya Prakash 5 years, 7 months ago
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I used Eisenstein's Irreducibility Criterion to solve this question.
Let
Φp(x)=xp−1+xp−2+…+x2+x+1=xp−1x−1\Phi_p (x) = x^{p-1} + x^{p-2} + \ldots + x^2 + x + 1 = \dfrac {x^p - 1}{x-1}Φp(x)=xp−1+xp−2+…+x2+x+1=x−1xp−1
We claim that Φp(x)\Phi_p (x)Φp(x) is irreducible over the rational numbers. Although Φp(x)\Phi_p (x)Φp(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)p−1(x+1)−1=∑k=0p−1(pk)xp−kxf(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}f(x)=Φp(x+1)=(x+1)−1(x+1)p−1=xk=0∑p−1(kp)xp−k
=xp−1+(p1)xp−2+(p2)xp−3+…+(pp−2)x+(pp−1)=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}}=xp−1+(1p)xp−2+(2p)xp−3+…+(p−2p)x+(p−1p)
All the lower coefficients are divisible by ppp, and the constant coefficient is exactly ppp, so it is not divisible by p2p^2p2. Thus, Eisenstein's criterion applies, and fff is irreducible. Certainly, if Φp(x)=g(x)h(x)\Phi_p (x) = g(x) h(x)Φp(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)f(x)=Φp(x+1)=g(x+1)h(x+1) gives a factorisation of fff. Thus, Φp\Phi_pΦp has no proper factorisation, i.e. it is irreducible.
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Nice solution. But why don't you do it another way or rather prove Eisenstein's criterion?
I have written in the proof in the wiki.
Well, is cyclotomic polynomials allowed?
@Pi Han Goh – Yeah, sure. Post your solution using cyclotomic polynomials.
When is Day 6 going to be posted?
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\sin \theta
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Comments
I used Eisenstein's Irreducibility Criterion to solve this question.
Let
Φp(x)=xp−1+xp−2+…+x2+x+1=x−1xp−1
We claim that Φp(x) is irreducible over the rational numbers. Although Φp(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)−1(x+1)p−1=xk=0∑p−1(kp)xp−k
=xp−1+(1p)xp−2+(2p)xp−3+…+(p−2p)x+(p−1p)
All the lower coefficients are divisible by p, and the constant coefficient is exactly p, so it is not divisible by p2. Thus, Eisenstein's criterion applies, and f is irreducible. Certainly, if Φp(x)=g(x)h(x) then f(x)=Φp(x+1)=g(x+1)h(x+1) gives a factorisation of f. Thus, Φp has no proper factorisation, i.e. it is irreducible.
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Nice solution. But why don't you do it another way or rather prove Eisenstein's criterion?
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I have written in the proof in the wiki.
Well, is cyclotomic polynomials allowed?
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When is Day 6 going to be posted?