A q-th root of unity close to 1 is trivial near V(p)

Let $p$ be a prime natural number, $q$ a nonzero natural number, and $T$ a commutative ring with the property that multiplication by $p$ is injective on $T$, stated as: for every $x \in T$, $(p : T) \cdot x = 0$ implies $x = 0$. Suppose $u \in T$ satisfies $u^q = 1$ and $u - 1$ lies in the principal ideal of $T$ generated by $(p : T)^{v_p(q)+1}$, where $v_p(q)$ is padicValNat p q. Then there exists $a \in T$ with $a - 1$ in the principal ideal generated by $(p : T)$ and $a \cdot (u - 1) = 0$. Thus $u$ becomes equal to $1$ after inverting an element congruent to $1$ modulo $p$, i.e. on a distinguished open subset of $\operatorname{Spec} T$ containing the closed subscheme $V(p)$. No Noetherian or finiteness hypothesis on $T$ is imposed; the only hypothesis on $T$ beyond commutativity is the absence of $p$-torsion.

This is an elementary commutative-algebra statement to the effect that a $q$-th root of unity which is $p$-adically congruent to $1$ modulo $p^{v_p(q)+1}$ in a $p$-torsion-free ring is trivial in a neighbourhood of the fibre at $p$; the exponent $v_p(q)+1$ cannot be lowered, as $u=-1$ in $T=\mathbf Z$ with $p=q=2$ shows. It is used in the analysis of the punctured group scheme $\mu_q$ and its cohomology, being cited by AlgebraicGeometry.exists_shortExact_natCard_fppfCohomology_zero_dvd_of_injective_of_range_iff.