# $\operatorname{Hom}(Z/nZ, μ_n)$ is isomorphic to $μ_n$ as etale sheaf?

About etale sheaves, I saw $$\operatorname{Hom}(Z/n, μ_n)\cong μ_n$$ as isomorphism of etale sheaves here (https://mathoverflow.net/questions/52404/locally-constant-sheaves-for-the-%C3%A9tale-topology-lack-of-intuition-about-%C3%A9tale ) but I had an observation which seems to contradict this as below.

Let $$n=3$$, $$U_1 = \operatorname{Spec} Q[x]/(x^2+x+1)$$. Let $$Z/3Z$$, $$μ_3$$, and $$F=\operatorname{Hom}(Z/3Z,μ_3)$$ be etale sheaves over $$U_1$$. I think $$F$$ should be interpreted as "sheaf hom", so the section $$\Gamma(U_1,F)$$ is sheaf morphisms between partial sheaves $$Z/3Z|_{U_1} \to \mu_3|_{U_1}$$, that is for every etale $$U→U_1$$, there given group morphism $$Γ(U,Z/nZ)→Γ(U,μ_n)$$, compatible with restriction maps.

Here I try to construct sections of $$F$$ over $$U_1$$ by giving those group morphisms.

(1) I need to give morphism $$\Gamma(U_1,Z/nZ) \to \Gamma(U_1,μ_n)$$. Here I fix it to send $$1 \bmod 3$$ to $$x$$. According to $$\operatorname{Hom}(Z/n, μ_n)\cong μ_n$$, this choice will determine $$\Gamma(U_1,F)$$.

(2) Consider $$U_2 = Q[x,y]/(x^2+x+1,y^2+y+1) → U_1$$, correspond to the ring morphism sending $$x$$ to $$x$$. $$U_2$$ has two closed points. Restriction map $$Γ(U_1,F)→Γ(U_2,F)$$ sends $$a \bmod 3$$ to $$(a,a) \bmod 3$$, and $$Γ(U_1,G)→Γ(U_2,G)$$ sends $$x$$ to $$x$$. So to be compatible with (1), the group morphism $$Γ(U_2,Z/nZ)→Γ(U_2,μ_n)$$ must send $$(1,1) \bmod 3$$ to $$x$$.

But it seems that for example I can send $$(1,2) \bmod 3$$ to either $$y$$ or $$y^2$$ and thus the choice of (1) does not determine $$\Gamma(U_1,F)$$, which contradict $$\operatorname{Hom}(Z/n, μ_n)\cong μ_n$$. Am I missing something?

• MathJax works in the title section, don't you know? Sep 6 '20 at 13:02
• I replaced every occurence of \operatorname{mod} with \bmod and did some other MathJax copy-editing. Sep 6 '20 at 13:21
• From a more abstract non-sense point of view, you have a general fact that $\underline{Hom}(\mathbb{Z},\mathcal{F})=\mathcal{F}$. This is true for any sheaf $\mathcal{F}$ on any site. And similarly, $\underline{Hom}(\mathbb{Z}/n\mathbb{Z},\mathcal{F})$ is the subsheaf of $\mathcal{F}$ whose sections satisfy $ns=0$. From this, since all sections of $\mu_n$ are of $n$-torsion, $\underline{Hom}(\mathbb{Z}/n\mathbb{Z},\mu_n)=\mu_n$. Sep 6 '20 at 15:55

$$\newcommand{\Spec}{\mathrm{Spec}}\newcommand{\Hom}{\mathrm{Hom}}\newcommand{\Q}{\mathbb{Q}}\newcommand{\Z}{\mathbb{Z}}$$

You are trying to specify a sheaf homomorphism $$\underline{\mathbb{Z}/3\mathbb{Z}}\to \mu_3$$ over $$U_1:=\mathrm{Spec}(\Q[x]/(x^2+x+1))$$ for the choice of the element $$x\in \mu_3(U_1)$$.

By definition this means that for every etale map $$U\to U_1$$ we have a map of abelian groups $$\underline{\mathbb{Z}/3\mathbb{Z}}(U)\to \mu_3(U)$$ such that for any etale map $$V\to U$$ we have that the diagram

$$\begin{matrix}\underline{\Z/3\Z}(U) & \to & \mu_3(U)\\ \downarrow & & \downarrow\\ \underline{\Z/3\Z}(V) & \to & \mu_3(V)\end{matrix}$$

commutes.

You were then confused because it seemed like if we set $$U_2:=\Spec(\Q[x,y]/(x^2+x+1,y^2+y+1))$$ that there is ambiguity in the map

$$(\Z/3\Z)^2=\underline{\Z/3\Z}(U)\to \mu_3(U)$$

But, note that

$$U_2=V_1\sqcup V_2\cong U_1\sqcup U_1$$

essentially because

$$\Q[x,y]/(x^2+x+1,y^2+y+1)\cong (\Q[x]/(x^2+x+1))[y]/(y-x)\times (\Q[x]/(x^2+x+1))[y]/(y-x^2)$$

and where we set

$$V_1:=\Spec((\Q[x]/(x^2+x+1))[y]/(y-x)),\qquad V_2:=\Spec((\Q[x]/(x^2+x+1))[y]/(y-x^2))$$

So, from our compatability conditions we see that the map $$\underline{\Z/3\Z}(U_2)\to \mu_3(U_2)$$ is actually determined by the two maps

$$\underline{\Z/3\Z}(V_1)\to \mu_3(V_1),\qquad \underline{\Z/3\Z}(V_2)\to \mu_3(V_2)$$

But, since we have the commutativity of the diagrams

$$\begin{matrix}\underline{\Z/3\Z}(U_1) & \to & \mu_3(U_1)\\ \downarrow & & \downarrow\\ \underline{\Z/3\Z}(V_i) & \to & \mu_3(V_i)\end{matrix}$$

and the vertical maps are isomorphisms, we see that the map

$$\underline{\Z/3\Z}(V_1)\to \mu_3(V_1)$$

sends $$1$$ to $$x=y$$ and the map

$$\underline{\Z/3\Z}(V_2)\to \mu_3(V_2)$$

sends $$1$$ to $$x=y^2$$.

So, from this we see that the sheaf condition dictates that the map

$$(\Z/3\Z)^2=\underline{\Z/3\Z}(V_1)\times\underline{\Z/3\Z}(V_2)=\underline{\Z/3\Z}(U_2)\to \mu_3(U_2)$$

is given by

$$(a,b)\mapsto x^a y^{2b}$$

unless I've made a clerical error.

TL;DR: You didn't use the full presheaf compatability condition.

• Thank you for your clear answer. I understand that I can only send (1,2) to y if I declare that the first argument is corresponding to y=x and the second to y=x^2, or I can only send it to y^2 if the arguments are declared as the other order. But I thought (1,1) is sent to x and (1,2) is sent to y, then (a,b) is sent to x^(2a-b)*y^(b-a). Sep 6 '20 at 14:16