Lemma 59.28.5. Let $S$ be a scheme. There is a canonical identification

if $n$ is invertible on $S$. In general we have

The same result holds with fppf replaced by syntomic.

Lemma 59.28.5. Let $S$ be a scheme. There is a canonical identification

\[ H_{\acute{e}tale}^1(S, \mu _ n) = \text{group of pairs }(\mathcal{L}, \alpha )\text{ up to isomorphism as above} \]

if $n$ is invertible on $S$. In general we have

\[ H_{fppf}^1(S, \mu _ n) = \text{group of pairs }(\mathcal{L}, \alpha )\text{ up to isomorphism as above}. \]

The same result holds with fppf replaced by syntomic.

**Proof.**
We first prove the second isomorphism. Let $(\mathcal{L}, \alpha )$ be a pair as above. Choose an affine open covering $S = \bigcup U_ i$ such that $\mathcal{L}|_{U_ i} \cong \mathcal{O}_{U_ i}$. Say $s_ i \in \mathcal{L}(U_ i)$ is a generator. Then $\alpha (s_ i^{\otimes n}) = f_ i \in \mathcal{O}_ S^*(U_ i)$. Writing $U_ i = \mathop{\mathrm{Spec}}(A_ i)$ we see there exists a global relative complete intersection $A_ i \to B_ i = A_ i[T]/(T^ n - f_ i)$ such that $f_ i$ maps to an $n$th power in $B_ i$. In other words, setting $V_ i = \mathop{\mathrm{Spec}}(B_ i)$ we obtain a syntomic covering $\mathcal{V} = \{ V_ i \to S\} _{i \in I}$ and trivializations $\varphi _ i : (\mathcal{L}, \alpha )|_{V_ i} \to (\mathcal{O}_{V_ i}, 1)$.

We will use this result (the existence of the covering $\mathcal{V}$) to associate to this pair a cohomology class in $H^1_{syntomic}(S, \mu _{n, S})$. We give two (equivalent) constructions.

First construction: using Čech cohomology. Over the double overlaps $V_ i \times _ S V_ j$ we have the isomorphism

\[ (\mathcal{O}_{V_ i \times _ S V_ j}, 1) \xrightarrow {\text{pr}_0^*\varphi _ i^{-1}} (\mathcal{L}|_{V_ i \times _ S V_ j}, \alpha |_{V_ i \times _ S V_ j}) \xrightarrow {\text{pr}_1^*\varphi _ j} (\mathcal{O}_{V_ i \times _ S V_ j}, 1) \]

of pairs. By (59.28.4.1) this is given by an element $\zeta _{ij} \in \mu _ n(V_ i \times _ S V_ j)$. We omit the verification that these $\zeta _{ij}$'s give a $1$-cocycle, i.e., give an element $(\zeta _{i_0i_1}) \in \check C(\mathcal{V}, \mu _ n)$ with $d(\zeta _{i_0i_1}) = 0$. Thus its class is an element in $\check H^1(\mathcal{V}, \mu _ n)$ and by Theorem 59.19.2 it maps to a cohomology class in $H^1_{syntomic}(S, \mu _{n, S})$.

Second construction: Using torsors. Consider the presheaf

\[ \mu _ n(\mathcal{L}, \alpha ) : U \longmapsto \mathit{Isom}_ U((\mathcal{O}_ U, 1), (\mathcal{L}, \alpha )|_ U) \]

on $(\mathit{Sch}/S)_{syntomic}$. We may view this as a subpresheaf of $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}, \mathcal{L})$ (internal hom sheaf, see Modules on Sites, Section 18.27). Since the conditions defining this subpresheaf are local, we see that it is a sheaf. By (59.28.4.1) this sheaf has a free action of the sheaf $\mu _{n, S}$. Hence the only thing we have to check is that it locally has sections. This is true because of the existence of the trivializing cover $\mathcal{V}$. Hence $\mu _ n(\mathcal{L}, \alpha )$ is a $\mu _{n, S}$-torsor and by Cohomology on Sites, Lemma 21.4.3 we obtain a corresponding element of $H^1_{syntomic}(S, \mu _{n, S})$.

Ok, now we have to still show the following

The two constructions give the same cohomology class.

Isomorphic pairs give rise to the same cohomology class.

The cohomology class of $(\mathcal{L}, \alpha ) \otimes (\mathcal{L}', \alpha ')$ is the sum of the cohomology classes of $(\mathcal{L}, \alpha )$ and $(\mathcal{L}', \alpha ')$.

If the cohomology class is trivial, then the pair is trivial.

Any element of $H^1_{syntomic}(S, \mu _{n, S})$ is the cohomology class of a pair.

We omit the proof of (1). Part (2) is clear from the second construction, since isomorphic torsors give the same cohomology classes. Part (3) is clear from the first construction, since the resulting Čech classes add up. Part (4) is clear from the second construction since a torsor is trivial if and only if it has a global section, see Cohomology on Sites, Lemma 21.4.2.

Part (5) can be seen as follows (although a direct proof would be preferable). Suppose $\xi \in H^1_{syntomic}(S, \mu _{n, S})$. Then $\xi $ maps to an element $\overline{\xi } \in H^1_{syntomic}(S, \mathbf{G}_{m, S})$ with $n \overline{\xi } = 0$. By Theorem 59.24.1 we see that $\overline{\xi }$ corresponds to an invertible sheaf $\mathcal{L}$ whose $n$th tensor power is isomorphic to $\mathcal{O}_ S$. Hence there exists a pair $(\mathcal{L}, \alpha ')$ whose cohomology class $\xi '$ has the same image $\overline{\xi '}$ in $H^1_{syntomic}(S, \mathbf{G}_{m, S})$. Thus it suffices to show that $\xi - \xi '$ is the class of a pair. By construction, and the long exact cohomology sequence above, we see that $\xi - \xi ' = \partial (f)$ for some $f \in H^0(S, \mathcal{O}_ S^*)$. Consider the pair $(\mathcal{O}_ S, f)$. We omit the verification that the cohomology class of this pair is $\partial (f)$, which finishes the proof of the first identification (with fppf replaced with syntomic).

To see the first, note that if $n$ is invertible on $S$, then the covering $\mathcal{V}$ constructed in the first part of the proof is actually an étale covering (compare with the proof of Lemma 59.28.1). The rest of the proof is independent of the topology, apart from the very last argument which uses that the Kummer sequence is exact, i.e., uses Lemma 59.28.1. $\square$

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## Comments (1)

Comment #1234 by jojo on

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