Prove that a given family of polynomials is reducible over ZZ

Prove that \psi_n(x) = \sum_{i=0}^n x^i is irreducible over \mathbb{Z} if and only if n+1 is prime.

Note that if n+1 is prime, then \psi_n is cyclotomic of prime index. We have already seen that \psi_n \circ (x+1) is Eisenstein at n+1 and hence irreducible. (Note that \psi_n(x) = (x^n - 1)/(x-1) and use the Binomial theorem.)

Now suppose n+1 = ab is composite. Using the division algorithm, every number in [0,n] can be written uniquely in the form ia+j where i \in [0,b-1] and j \in [0,a-1]. Thus we have \psi_n(x) = \sum_{j=0}^{b-1} \sum_{i=0}^{a-1} x^{ia+j} = \left( \sum_{i=0}^{b-1} x^{ia} \right) \left( \sum_{j=0}^{a-1} x^j \right), so that \psi_n is reducible.

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  • Kreizhn  On July 15, 2011 at 11:28 am

    In the problem statement, should the polynomial be exponentiated in “i” rather than merely indexed? Namely, x^i rather than x_i

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