The Stacks project

Proof. By definition the sheaf $\mu _{n, S}$ is the kernel of the map $(\cdot )^ n$. Hence it suffices to show that the last map is surjective. Let $U$ be a scheme over $S$. Let $f \in \mathbf{G}_ m(U) = \Gamma (U, \mathcal{O}_ U^*)$. We need to show that we can find an étale cover of $U$ over the members of which the restriction of $f$ is an $n$th power. Set

(See Constructions, Section 27.3 or 27.4 for a discussion of the relative spectrum.) Let $\mathop{\mathrm{Spec}}(A) \subset U$ be an affine open, and say $f|_{\mathop{\mathrm{Spec}}(A)}$ corresponds to the unit $a \in A^*$. Then $\pi ^{-1}(\mathop{\mathrm{Spec}}(A)) = \mathop{\mathrm{Spec}}(B)$ with $B = A[T]/(T^ n - a)$. The ring map $A \to B$ is finite free of rank $n$, hence it is faithfully flat, and hence we conclude that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is surjective. Since this holds for every affine open in $U$ we conclude that $\pi $ is surjective. In addition, $n$ and $T^{n - 1}$ are invertible in $B$, so $nT^{n-1} \in B^*$ and the ring map $A \to B$ is standard étale, in particular étale. Since this holds for every affine open of $U$ we conclude that $\pi $ is étale. Hence $\mathcal{U} = \{ \pi : U' \to U\} $ is an étale covering. Moreover, $f|_{U'} = (f')^ n$ where $f'$ is the class of $T$ in $\Gamma (U', \mathcal{O}_{U'}^*)$, so $\mathcal{U}$ has the desired property. $\square$