TITLE: Kummer sequence etale topology
QUESTION [9 upvotes]: Consider the category $C=Sch/S$ of schemes over $S$ and let $n \in \Gamma(S,\mathcal{O}_S)^{*}$. It is possible to show that $$0 \rightarrow \mu_{n,S} \rightarrow \mathbb{G}_m \rightarrow \mathbb{G}_m \rightarrow 0$$
is exact in the etale topology, where the last map is given by $x \to x^n$. 
As an exercise, I should show with a counterexample that this is not true anymore if now does not give the condition $n$ being invertible as a global section.
I tried to take $S=Spec(\mathbb{Z})$ and what I was looking for was a scheme $U$ , a section $s \in \mathcal{O}_U(U)$ such that for every etale cover $U_i \to U$ there is no $x \in \mathbb{G}_m(U_i)$ such that $x^m=s|_{U_i}$.
Let's take $U=Spec(A)$ and let's strike to standard etale cover(which should be enough to check). So what I 've got now is $U_i=Spec(A[t]_h/(f))$ for some monic $f$ with $(f,f')= (A[t]_h/(f))$. Now i tried to take $s=1$ and write down the condition explicitly but I couldn't come up with anything.

REPLY [7 votes]: I am writing up (after discussions with Alex Youcis) an argument by Piotr Achinger.
We show that for any prime $p$ the Kummer sequence \begin{align*} 1 \to \mu_{p} \to \mathbb{G}_{m} \stackrel{\times p}{\to} \mathbb{G}_{m} \to 1 \end{align*} is not exact on the etale site of $\mathbb{Z}$.
It suffices to show that there is an integer $q$ for which the unit $q \in \mathbb{G}_{m}(\mathbb{Z}[\frac{1}{q}])$ does not etale-locally have a $p$th root. Choose $q$ so that $\gcd(p,q) = 1$ and $q^{p-1} \not\equiv 1 \pmod{p^{2}}$; we can do this for any prime $p$ since $(\mathbb{Z}/(p^{2}))^{\times}$ is cyclic of order $p(p-1)$. (For example if $p = 2$ then $q = 3$ works.) 
Suppose there exists an etale cover $f : U \to \operatorname{Spec} \mathbb{Z}[\frac{1}{q}]$ such that $f^{\ast}q$ has a $p$th root in $\mathbb{G}_{m}(U)$. After replacing $U$ by the disjoint union finitely many affine open subschemes of $U$ whose images in $\operatorname{Spec} \mathbb{Z}[\frac{1}{q}]$ form a covering, we may assume that $U$ is affine. Since $\gcd(p,q) = 1$, the closed immersions $\operatorname{Spec} \mathbb{Z}/(p^{n}) \to \operatorname{Spec} \mathbb{Z}$ factor through $\operatorname{Spec} \mathbb{Z}[\frac{1}{q}]$. Set $U_{n} := U \times_{\operatorname{Spec} \mathbb{Z}[1/q]} \operatorname{Spec} \mathbb{Z}/(p^{n+1})$. Here each $U_{n}$ is nonempty, quasi-compact, etale over $\mathbb{Z}/(p^{n+1})$, hence is the disjoint union of finitely many $\operatorname{Spec}$ of finite etale $\mathbb{Z}/(p^{n+1})$-algebras (see the lemma below; here we use that etale maps are locally quasi-finite, so etale extensions of Artinian rings are Artinian). The underlying maps of $U_{n} \to U_{n+1}$ are homeomorphisms; choose a connected component $V_{0}$ of $U_{0}$, and let $V_{n}$ be the unique connected component of $U_{n}$ corresponding to $V_{0}$ under the map $U_{0} \to U_{n}$. Say $V_{n} := \operatorname{Spec} A_{n}$. We consider the $n=0,1$ cases; since $\mathbb{Z}/(p^{2}) \to A_{1}$ is flat, tensoring the sequence \begin{align*} 0 \to (p)/(p^{2}) \to \mathbb{Z}/(p^{2}) \to \mathbb{Z}/(p) \to 0 \end{align*} by $\mathbb{Z}/(p^{2}) \to A_{1}$ gives an exact sequence \begin{align*} 0 \to (p)/(p^{2}) \otimes_{\mathbb{Z}/(p^{2})} A_{1} \to A_{1} \to A_{1} \otimes_{\mathbb{Z}/(p^{2})} \mathbb{Z}/(p) \to 0 \end{align*} of $A_{1}$-modules. Here $A_{1} \otimes_{\mathbb{Z}/(p^{2})} \mathbb{Z}/(p) \simeq A_{0}$ (isomorphism as $A_{1}$-algebras) by definition of the $A_{n}$, and $pA_{1} \simeq (p)/(p^{2}) \otimes_{\mathbb{Z}/(p^{2})} A_{1} \simeq A_{0}$ (isomorphism as $A_{1}$-modules) since $(p)/(p^{2}) \simeq \mathbb{Z}/(p)$. Thus the above exact sequence is equivalent to \begin{align*} 0 \to pA_{1} \to A_{1} \to A_{0} \to 0 \end{align*} which induces an exact sequence \begin{align*} 1 \to 1+pA_{1} \to A_{1}^{\times} \to A_{0}^{\times} \to 1 \end{align*} involving the groups of units since $pA_{1}$ is a square-zero ideal of $A_{1}$. We have that $A_{0}$ (being a (connected) finite etale extension of $\mathbb{F}_{p}$) is isomorphic to $\mathbb{F}_{p^{k}}$ for some $k \ge 1$, hence $|A_{0}^{\times}| = p^{k}-1$ and $|pA_{1}| = |A_{0}| = p^{k}$; thus $|A_{1}^{\times}| = p^{k}(p^{k}-1)$; in fact every element of $A_{1}^{\times}$ is $p(p^{k}-1)$-torsion since $pA_{1}$ is $p$-torsion. 
By the assumption that $U$ has a $p$th root of $q$, we have that $A_{1}$ has a $p$th root of $q$, say $x^{p} = q$ for some $x \in A_{1}^{\times}$. Raising both sides to the power of $p^{k}-1$, we have $q^{p^{k}-1} = 1$ in $A_{1}$. Since the map $\mathbb{Z}/(p^{2}) \to A_{1}$ is injective (since it is faithfully flat), we have that $q^{p^{k}-1} = 1$ in $\mathbb{Z}/(p^{2})$. We have $q^{p(p-1)} \equiv 1 \pmod{p^{2}}$ since $p(p-1) = \varphi(p^{2})$; since $p^{k}-1 \equiv p-1 \pmod{p(p-1)}$ for any $k \ge 1$, we have $q^{p-1} \equiv 1 \pmod{p^{2}}$; this contradicts our choice of $q$.

Lemma: Let $\varphi : A \to B$ be a finite type ring map where $A,B$ are Artinian rings. Then $\varphi$ is finite.

Proof: We may assume that $A$ is local, say with maximal ideal $\mathfrak{m}$; since $B$ is the finite product of Artin local rings, we may assume that $B$ is local, say with maximal ideal $\mathfrak{n}$; there exists some $s \gg 0$ such that $\mathfrak{n}^{s} = 0$. Let $k := A/\mathfrak{m}$ and $\ell := B/\mathfrak{n}$ denote the residue fields of $A$ and $B$; then $\ell/k$ is a finite extension (this is the Nullstellensatz), and $B$ has a filtration $0 = \mathfrak{n}^{s} \subseteq \mathfrak{n}^{s-1} \subseteq \dotsb \subseteq \mathfrak{n} \subseteq B$ where each successive quotient $\mathfrak{n}^{i}/\mathfrak{n}^{i+1}$ are finite-dimensional as $\ell$-vector spaces, hence finite-dimensional as $k$-vector spaces, hence finitely generated as $A$-modules; hence $B$ itself is finitely generated as an $A$-module.