Why is the sum of the ($\varphi(d)$-1)th coefficients of $\Phi_d$ equal to the ($\varphi$(n)-1)th coefficient of $\prod_{d\mid n} \Phi_d(X)$

Question

I grasp this answer, except for one identity. To quote: "$\sum_{d\mid n}\left[\Phi_d(X)\right]_{-1} = \left[\prod_{d\mid n} \Phi_d(X)\right]_{-1}$"

It isn't so simple i think because you don't take the coefficients corresponding to the same place/x-power. What is my misunderstanding?

Update: I simply had missed reading the updates, which made it all reasonable. Thanks to https://math.stackexchange.com/a/69548/218659 I understand that "the coefficient in question is the negative sum of the roots".

Though I feel somewhat silly, i honestly can't produce myself why "Even without considering roots, this follows from looking at how the product expands."

See the updates in that answer. Then apply induction to get this for product of any number of polynomials. — Wojowu, Dec 21 '15 at 14:30

score 3 · Accepted Answer · answered Dec 31 '15 at 13:55

Given two monic polynomials $f_1=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0,f_2=x^m+b_{m-1}x^{m-1}+...+b_1x+b_0$, after we multiply them we will get a polynomial of degree $n+m$. After grouping terms containing $x^{n+m-1}$ in the product, we notice we get $1\cdot b_{m-1}+a_{n-1}\cdot 1=a_{n-1}+b_{m-1}$. This is precisely the statement that next-to-leading coefficient of the product of two monic polynomials is the sum of their next-to-leading coefficients. Now, by induction, this is true for any number of polynomials in a product. After applying this to the cyclotomic polynomials and their product we get what was asked for.

Why is the sum of the ($\varphi(d)$-1)th coefficients of $\Phi_d$ equal to the ($\varphi$(n)-1)th coefficient of $\prod_{d\mid n} \Phi_d(X)$

1 Answers1