If x = [$X$ mod $\Phi_m(X)$], why $\Phi(x) = 0$?

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On https://eprint.iacr.org/2020/1481.pdf (page 3), it says:

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where $\Phi_m(X)$ is the m-th cyclotomic polynomial. First of all, when we do the quotient $\mathbb{Z}[X]/(x^{4096}+1)$ for example, won't my result be the polynomials with degree maximum 4096? Why $A=\mathbb{Z}[X]$? Shouldn't it be just the polynomials with coefficient 4096 maximum?

Anyways, why $\Phi_m(x) = 0$? As far as I understood, $x$ is a polynomial, it doesn't even make sense to do $\Phi_m(x)$ technically.