Lemma 59.28.3. For any $n \in \mathbf{N}$ the sequence

$0 \to \mu _{n, S} \to \mathbf{G}_{m, S} \xrightarrow {(\cdot )^ n} \mathbf{G}_{m, S} \to 0$

is a short exact sequence of sheaves on the site $(\mathit{Sch}/S)_{fppf}$ and $(\mathit{Sch}/S)_{syntomic}$.

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. Since the syntomic topology is weaker than the fppf topology, see Topologies, Lemma 34.7.2, it suffices to prove this for the syntomic topology. 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 a syntomic cover of $U$ over the members of which the restriction of $f$ is an $n$th power. Set

$U' = \underline{\mathop{\mathrm{Spec}}}_ U(\mathcal{O}_ U[T]/(T^ n-f)) \xrightarrow {\pi } U.$

(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, $B$ is a global relative complete intersection over $A$, so the ring map $A \to B$ is standard syntomic, in particular syntomic. Since this holds for every affine open of $U$ we conclude that $\pi$ is syntomic. Hence $\mathcal{U} = \{ \pi : U' \to U\}$ is a syntomic 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. $\square$

Comment #2048 by Yuzhou Gu on

Should "stronger than" be "weaker than"?

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