I have started studying some étale cohomology and I am trying to build up some intuition about the concept of local for the étale topology. I can understand some nice examples (like Kummer exact sequence) but I am still quite confused by some "easy" notions such as locally constant sheaves.
I believe that an étale sheaf which is étale locally isomorphic to the same constant sheaf should be also globally isomorphic to that constant sheaf if the isomorphisms verify some cocycle condition, but here is a toy example which seems to contradict this:
Let $k$ be a field, $n$ an integer invertible in $k$ and assume that $k$ does not contain all $n$-th roots of unity. Now consider the two following étale sheaves on $X=Spec\; k$:
- The sheaf of n-th roots of unity $\mu_n$;
- The constant sheaf $\mathbb Z/n \mathbb Z$.
They are not isomorphic since their sections on $Spec\; k$ are different, but they become isomorphic after some finite separable extension of scalars so they are isomorphic étale locally. To be precise, $U=Spec(k[T]/(T^n-1))$ is an étale cover of $X$ such that the pullbacks of the two sheaves are isomorphic.
Why are this two sheaves locally isomorphic but not isomorphic?
Is it normal that this isomorphism doesn't "patch"? (which would imply that the sheaves over the small étale site on $Spec\; k$ don't form a prestack)
If I try to think to all this "stalkwise", changing to the point of view of topoi, (I'm not very familiar with the theory of topoi so please correct me if I am writing nonsense) I believe that:
the topos of sheaves over $Spec\;k$ with the small étale site has enough points, a family of conservative points consisting of just one element (the étale local ring is some separable closure $k^{sep}$ of $k$); and on this local ring the two sheaves above coincide.
It should follow that as soon as we have a morphism of sheaves inducing this isomorphism on the stalk the two sheaves should be isomorphic, which is not the case.
Is it just because we don't have such a morphism or am I missing something more fundamental here?
Best Answer
$Isom(F,G)$ is indeed an etale sheaf. If we take $F = \mathbb Z/n$ and $G = \mu_n$, then $G$ is a sheaf of $F$-modules, and so evaluation at the global section $1$ gives an isomorphism of sheaves $Hom(\mathbb Z/n,\mu_n) \cong \mu_n$, which identifies $Isom(\mathbb Z/n,\mu_n)$ with the subsheaf of $\mu_n$ whose sections are primitive $n$th roots of unity. Thus there is no global isomorphism precisely because (by assumption) there is no primitive $n$th root of $1$ in $k$.
Certainy if we take $l = k[X](X^n - 1)$ we can find a section of the $Isom$ sheaf over Spec $l$, but this section does not descend to a section over Spec $k$, because it does not satisfy the requisite gluing conditions on Spec $l \times$ Spec $l =$ Spec $l\otimes_k l$. (These gluing conditions amount to the Galois invariance that Tom Goodwillie refers to in his comment above.)
Perhaps the source of your confusion is that if $V$ is an open set of a topological space, then $V \cap V = V$, but in the etale site (in which generality intersection is replaced by fibre product), $V\times V$ is typically quite a bit larger than $V$.