2
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ MathJax works in the title section, don't you know? $\endgroup$
    – Shaun
    Sep 6 '20 at 13:02
  • 1
    $\begingroup$ I replaced every occurence of \operatorname{mod} with \bmod and did some other MathJax copy-editing. $\endgroup$ Sep 6 '20 at 13:21
  • 1
    $\begingroup$ 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, $\underline{Hom}(\mathbb{Z}/n\mathbb{Z},\mu_n)=\mu_n$. $\endgroup$
    – Roland
    Sep 6 '20 at 15:55
2
$\begingroup$

$\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.

$\endgroup$
1
  • $\begingroup$ 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). $\endgroup$
    – aerile
    Sep 6 '20 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.