Question about an isomorphism of cyclic groups

Question

Let's say that we have a $p$-modular system $(K,R,k)$ where $K$ is a field of characteristic $0$ with a discrete valuation whose ring of integers $R$ and prime ideal $\mathfrak{m}$ satisfy $R/\mathfrak{m} \cong k$ (so that $k$ is a perfect field) with positive characteristic $p$ dividing $|G|$. Let $G$ be a finite group and set $g=p^rm, (p,m)=1$. Assume that the field $k$ is sufficiently large to contain the group of $m^{th}$ roots of unity. Did I miss something ?

Then the residue class map $R \longrightarrow k$ induces an isomorphism between the group of $m^{th}$ roots of unity in $K$ and $k$ itself.

I don't know how to prove this. Is it equivalent to state that the sets of roots of polynomial $X^m-1$ in $K[X]$ and in $k[X]$ have the same cardinal ? Can I prove this by factoring $X^g-1=(X^m-1)^{p^r}$?

I thank you for any suggestions

Answer 1

Let $R$ be a complete DVR, then every root of unity of $Frac(R)$ is in $R$,

every root of unity in $k=R/\mathfrak{m}$ is the reduction of a root of unity in $R$,

and the kernel of $R^\times_{tors}\to k^\times_{tors}$ is trivial if $char(k)=0$, otherwise it is $R^\times[p^\infty]$ where $p=char(k)$.

Proof : Hensel lemma, it constructs from $\zeta_n\in k$ a root $\omega_n$ of $X^n-1\in R[X]$ which satisfies $\omega_n\equiv \zeta_n\bmod \mathfrak{m}$.

Hensel lemma is just the gradient descent : if $f(a)=0\bmod \pi^m$ and $f'(a)\not \equiv 0 \bmod \pi$ then $f(a+b\pi^m) = f(a)+bf'(a)\pi^m + O(\pi^{m+1})$, thus we choose $b= -f(a)/(f'(a)\pi^m)$ to obtain $f(a+b\pi^m) =0\bmod \pi^{m+1}$, repeating, we obtain a sequence which converges in $R[[\pi]]$ (which is the completion of $R$) to a root of $f$.