I am looking at the ring $R = \mathbb{Z}[X]/(\Phi_m(X), p)$ where $\Phi_m$ is the $m$th cyclotomic polynomial and $p$ is a prime. What I want to show is that $R$ can be split into $l$ copies of $\mathbb{F}_{p^d}$ where $d$ is the smallest integer such that $p^d \equiv 1 \mod m$ and $l\cdot d = \phi(m)$, the degree of the polynomial.
The condition in some sources I have looked at says that $p$ cannot divide $m$ and that $p$ cannot be ramified in $\mathbb{Q}[\zeta_m]$ where $\zeta_m$ is the $m$th root of unity. I cannot make the connection of why this is necessary. I don't even know if the polynomial have to be cyclotomic, or that it is sufficient that they are irreducible in $\mathbb{Z}[X]$. Any tips would be much appreciated.
Best Answer
So $R$ is actually $\mathbb{F}_p[X]/(\Phi_m(X))$.
Assume that $R$ is a product of fields. In particular, $R$ is reduced (that is the problem for general $m$); in other words, $\Phi_m$ mod $p$ is separable. But if $p|m$, using complex roots, we can see that $\Phi_m(X)=\Phi_{m/p}(X^p)$ so is not separable mod $p$ (it’s equal to $\Phi_{m/p}(X)^p)$.
Now, assume that $p$ does not divide $m$. Then $\Phi_m$ divides $X^m-1$ which is separable mod $p$, so $\Phi_m$ is separable mod $p$, and coprime mod $p$ to any $\Phi_d$ with $d|m$ and $d < m$.
In particular, if $F$ is a field of characteristics $p$ and $\omega \in F$ is a root of $\Phi_m$, then $\omega$ has multiplicative order $m$ – and conversely.
Let $\Phi_m=P_1\ldots P_l$ be the decomposition in products of irreducible polynomials mod $p$, with the $P_i$ being monic, irreducible mod $p$ and pairwise distinct. Then $R$ is the product of the $\mathbb{F}_p[X]/(P_i)$ (by the CRT) which are actually fields generated by an element which is a root of $\Phi_m$, so is an element of multiplicative order $m$.
To conclude, we show that if $F$ is a field of degree $d$ over $\mathbb{F}_p$ generated by an element $\omega$ of multiplicative order $m$, then $d$ is the multiplicative order $u$ of $p$ mod $m$. This shows that all the $P_i$ have the same degree.
Indeed, $1=\omega^{|F^{\times}|}=\omega^{p^d-1}$ so that $m|p^d-1$ and $u|d$. But $F’=\{z \in F,\,z^{p^u}=z\}$ is a subfield of $F$ containing $\mathbb{F}_p$ and $\omega$ so is $F$. But $F’$ is the set of roots of the polynomial $X^{p^u}-X$ so $|F|=|F’| \leq p^u$ so $d \leq u$ and $d=u$.