Prime-to-p roots lift uniquely in a complete DVR

Normalize the valuation by $v(\pi )=1$. An element is a unit exactly when its value is zero: if $v(x)=0$, then $v(x^{-1})=0$; if both $x,x^{-1}$ are integral, their nonnegative values sum to zero. Thus the maximal ideal is $(\pi )$. For ${\lambda }^m=1$ choose one representative $x_0$; it is a unit, as is $f^{\mathrm{\prime }}(x_0)=m{x}_0^{m-1}$ for $f(X)=X^m-1$.

Define deterministically $x_{n+1}=x_n-f(x_n)\mathrm{/}{f}^{\mathrm{\prime }}(x_n)$. Taylor expansion $f(x+h)=f(x)+f^{\mathrm{\prime }}(x)h+h^{2}R$ with $R\in \mathcal{O}$ shows $v(f(x_{n+1}))\ge 2v(f(x_n))$. All $x_n$ retain residue $\lambda$, so every derivative is a unit. Starting with $v(f(x_0))\ge 1$, we get $v(x_{n+1}-x_n)\ge 2^n$ (a zero error stays zero). The sequence converges by completeness to $x\equiv \lambda$; continuity of the finite polynomial operations and separation give $f(x)=0$. This is the simple-root Newton construction used in the Stacks complete-local-ring lifting proof specialized to a DVR.

If $x^m=y^m=1$ and $\bar{x}=\bar{y}=\lambda$, then $0=(x-y){\sum }_{j=0}^{m-1}{x}^{m-1-j}{y}^j$. The sum has residue $m{\lambda }^{m-1}\mathrm{\ne }0$, hence is a unit by step 1.1. Thus $x=y$, the Stacks simple-root uniqueness argument.

Products and inverses of lifted roots are roots lifting the corresponding products and inverses. Uniqueness therefore gives $\widehat{\lambda \nu }=\widehat{\lambda }\widehat{\nu }$, $\widehat{1}=1$, and $\widehat{{\lambda }^{-1}}=(\widehat{\lambda }{)}^{-1}$. If two exponents occur, both lifts are roots for their least common multiple, still prime to $p$, so uniqueness there identifies them. When $m=1$ both groups are $\{1\}$.