Show that product of integer polynomial values for arguments being complex roots of unity is integer.

P(x)P(x) is polynomial with integer coefficients and ww is complex (primitive) root of unity of a degree nn. Show that P(1)∗P(w)∗P(w2)∗…∗P(wn−1)P(1)*P(w)*P(w^2)*…*P(w^{n-1}) is integer number for every natural number nn.

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1 Answer
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Let wk=wkw_k=w^k. Then Q(w0,w1,…,wn−1)=P(w0)⋅P(w1)⋅P(w2)⋯P(wn−1)Q(w_0, w_1, \dots,w_{n-1})=P(w_0) \cdot P(w_1) \cdot P(w_2) \cdots P(w_{n-1}) is a polynomial symmetric in w0,w1,…,wn−1w_0, w_1, \dots,w_{n-1}. Therefore QQ can be expressed as a polynomial in the elementary symmetric polynomials in wkw_k with integer coefficients. But, since ww is a primitive nthn^{th} root of unity, wk=wk|k=0…n−1w_k=w^k \;|\; k=0 \dots n-1 are the nn roots of xn−1=0x^n – 1=0 which is a monic polynomial with integer coefficients, so by Vieta’s formulas all the elementary symmetric polynomials in wkw_k are integers (in fact, all of them 00 except the full product which is 11), thus QQ is an integer as well.