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?
Best Answer
$\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.