$k[x]/(x^n-1)$ is isomorphic to...?

Question

I am wondering if I can relate the quotient of $k[x]$ by the relation $x^n-1=0$, where $k$ is a field of characteristic zero, to roots of unity.

Answer 1

More generally, if the characteristic of $k$ is prime to $n$, the polynomial $x^n-1$ is separable, thus it factors as $f_1 \cdot \dotsc \cdot f_d$ with distinct irreducible polynomials $f_1,\dotsc,f_d$ and the Chinese Remainder Theorem implies $k[x]/(x^n-1) \cong k[x]/(f_1) \times \dotsc \times k[x]/(f_d)$, which is a product of fields. Note that $f_1,\dotsc,f_d$ are the irreducible factors of all the cyclotomic polynomials $\Phi_m$ with $m|n$ (since $x^n-1=\prod_{m|n} \Phi_m$). If $f$ is an irreducible factor of $\Phi_m$, then it is a minimal polynomial of $\zeta_m \in \overline{k}$, hence $k[x]/(f) = k(\zeta_m)$ is a cyclotomic field. For general $k$ we cannot make this more explicit. But of course one can work this out in special cases:

1. When the cyclotomic polynomials are irreducible (for example when $k=\mathbb{Q}$), we see that $k[x]/(x^n-1) \cong \prod_{m|n} k(\zeta_m)$.

2. For finite fields there are algorithms which compute the irreducible factors. For example, over $k=\mathbb{F}_5$ we have $\Phi_1=x-1$, $\Phi_2=x+1$, $\Phi_4=x^2+1=(x-2)(x+2)$, hence $k[x]/(x^4-1) \cong k^4$. Over $k=\mathbb{F}_{19}$ we have $\Phi_1=x-1$ and $\Phi_5=x^4+x^3+x^2+x+1=(x^2+5x+1)(x^2+15x+1)$, hence $k[x]/(x^5-1) \cong k \times k(\zeta_5)^2$.