Given a closed immersion of integral (and maybe normal) schemes $i^*:Z\hookrightarrow X$. Is it true that $i^*\mu_n$ is isomorphic to $\mu_n|_{Z}$? Here by pullback we consider the sheaf of $n$-th root of unity as a sheaf on the small etale site. By the restriction to $Z$, I mean the sheaf of $n$-th root of unity naturally defined on $Z$.

  • $\begingroup$ Thanks for your comment. Is the reason for this being true because $\mu_n$ is locally constant? It becomes constant on some etale cover. $\endgroup$
    – user127776
    Commented Mar 20, 2021 at 15:43
  • $\begingroup$ So by that logic it is true for $\mathbb{G}_m$ too. But let's assume $Z$ is a closed point corresponding to some maximal ideal. I think $i^*\mathbb{G}_m$ is the stalk of $\mathbb{G}_m$, which is given by $\mathbb{G}_m$ of the etale local ring (strict henselization of the loacl ring at the maximal ideal.). This seems to be different than $\mathbb{G}_m$ on a point ($\text{Spec}(k)$ for a field $k$, which is the residue field of the maximal ideal.) they have different global sections, as the local ring has more invertible elements. $\endgroup$
    – user127776
    Commented Mar 20, 2021 at 16:17
  • $\begingroup$ Yes you are right and I was too quick. This is definitively true if $n$ is invertible on the base scheme. I will give a thought if $n$ is not invertible. I'll remove my comments. $\endgroup$
    – Roland
    Commented Mar 20, 2021 at 16:47

1 Answer 1


This isn't true in general. Let me give a counterexample first, and then discuss two cases in which it works.

Counterexample. Consider $X=\operatorname{Spec} \mathbb Z_p[\zeta_p]$, where $\zeta_p$ is a $p$-th root of unity, and let $Z=\operatorname{Spec}\mathbb F_p$, with $i\colon Z\hookrightarrow X$ being the inclusion of the special point. All étale $\mathbb F_p$-algebras are finite products $\prod_i\mathbb F_{p^i}$, hence they don't contain any non-trivial $p$-th root of unity. Thus, the stalk of $\mu_{n,Z}$ at any geometric point $\overline{z}$ of $Z$ is $\{1\}$. However, any étale $\mathbb Z_p[\zeta_p]$-algebra $B$ contains the element $\zeta_p$, and $\zeta_p-1$ is a nonzerodivisor by flatness. Hence also the stalk of $\mu_{n,X}$ at $i(\overline{z})$ contains the non-trivial root of unity $\zeta_p$. Since $(i^*\mu_{n,X})_{\overline{z}}=(\mu_{n,X})_{i(\overline{z})}$ (pullback of étale sheaves preserves stalks), this shows that $i^*\mu_{n,X}$ cannot coincide with $\mu_{n,Z}$.

And now for the two cases in which it works. Neither case needs that $i$ is a closed immersion.

Case 1. Assume $n$ is invertible on $X$ and thus on $Z$ too (I believe you already discussed that case in the now deleted comments, but let me spell it out for completeness). Then the canonical map $i^*\mu_{n,X}\rightarrow \mu_{n,Z}$ is always an isomorphism, even for $i$ not a closed immersion and $X$, $Z$ not reduced.

This can be seen as follows: $\mu_{n,X}$ is always represented by the scheme $Y=\operatorname{\underline{Spec}}\mathcal O_X[T]/(T^n-1)$, and similarly $\mu_{n,Z}$ is represented by the base change $i^*Y$. If $n$ is invertible, then $Y$ is étale over $X$ and (thus) $i^*Y$ is étale over $Z$. In this case it is completely formal to show that the induced map $i^*h_Y\rightarrow h_{i^*Y}$ on representable sheaves is an isomorphism. Indeed, by Yoneda's lemma, it suffices to check that $$\operatorname{Hom}_{\mathrm{Sh}(Z_{\mathrm{\acute{e}t}})}(h_{i^*Y},\mathcal F)\rightarrow \operatorname{Hom}_{\mathrm{Sh}(Z_{\mathrm{\acute{e}t}})}(i^*h_Y,\mathcal F)$$is an isomorphism for all sheaves $\mathcal F\in \mathrm{Sh}(Z_{\mathrm{\acute{e}t}})$. This follows from the calculation $$ \operatorname{Hom}_{\mathrm{Sh}(Z_{\mathrm{\acute{e}t}})}(h_{i^*Y},\mathcal F)\cong\Gamma(i^*Y,\mathcal F)\cong\Gamma(Y,i_*\mathcal F)\cong\operatorname{Hom}_{\mathrm{Sh}(X_{\mathrm{\acute{e}t}})}(h_{Y},i_*\mathcal F)\cong \operatorname{Hom}_{\mathrm{Sh}(Z_{\mathrm{\acute{e}t}})}(i^*h_{Y},\mathcal F)\,.$$ The first and third isomorphism follow from Yoneda's lemma, the second holds by definition of $i_*\mathcal F$, and the fourth by the $(i^*,i_*)$-adjunction. In the first and the third isomorphism, we critically used that $i^*Y$ and $Y$ are étale over $Z$ and $X$, respectively. This explains why $i^*\mathbb G_{m,X}\rightarrow \mathbb G_{m,Z}$ may fail to be an isomorphism, as you note in the comments: $\mathbb G_{m,X}$ is representable by an $X$-scheme, but not by an étale one.

Case 2. $X$ and (thus) $Z$ have characteristic $p$. We'll only need that $X$ and $Z$ are reduced and $i$ can be any map. Moreover, the question is local, so we may assume $X=\operatorname{Spec} A$ and $Z$ are affine. Finally, $\mu_{n,X}$ is the direct sum of its subsheaves $\mu_{\ell^{m}}$, where $\ell$ is a prime divisor of $n$ and $\ell^m$ is the highest power of $\ell$ dividing $n$. For $\ell\neq p$, case 1 does it, hence we may assume $n=p^m$.

Note that every $\mathbb F_p$-algebra $B$ with a non-trivial $p^m$-th root of unity $\zeta$ is necessarily non-reduced, as $0=\zeta^{p^m}-1=(\zeta-1)^{p^m}$. But every étale algebra over the reduced ring $A$ is reduced again. For $A$ noetherian, this follows from the fact that a noetherian ring is reduced iff it satisfies Serre's conditions $(R_0)$ and $(S_1)$ (the little brother of Serre's normality criterion) and that the conditions $(R_k)$ and $(S_k)$ ascend along étale maps. In general, we can write any étale map $A\rightarrow B$ as a filtered colimit $\operatorname{colimit} (A_\alpha\rightarrow B_\alpha)$ of étale maps between finite type $\mathbb Z$-algebras. Replacing all $A_\alpha$ by their reductions, the above argument shows that $B$ can be written as a filtered colimit of reduced rings $B_\alpha$, whence $B$ is reduced itself.

Therefore, the étale sheaves $\mu_{p^m,X}$ and $\mu_{p^m,Z}$ vanish if $X$ and $Z$ are reduced of characteristic $p$, so $i^*\mu_{p^m,X}\rightarrow \mu_{p^m,Z}$ is trivially an isomorphism.


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