2
$\begingroup$

I am working on theory of category and I found this exercise. I tried a lot but I didn't know how I could do. Let $A$ a discrete valuation ring. Show that the category of sheaves of abelian groups on $Spec(A)$ is equivalent to the category which objects are defined as below:

$\{ f: S \rightarrow L \ \quad S,L\in Ab \}$ and if we take two morphisms $f: S \rightarrow L$ and $f': S' \rightarrow L'$ then $g\circ f = f' \circ g'$ with $g: L \rightarrow L'$ and $g': S \rightarrow L$.

I would really appreciate your answers. Thanks!

$\endgroup$
  • $\begingroup$ This is almost immediate from the definitions. Have you tried writing down the definitions of everything involved? Do you know what $\operatorname{Spec}(A)$ is in this case? $\endgroup$ – Eric Wofsey Oct 8 '18 at 18:38
  • $\begingroup$ OK, so, what are the open sets of $\operatorname{Spec}(A)$? What data does a sheaf consist of? $\endgroup$ – Eric Wofsey Oct 8 '18 at 18:46
2
$\begingroup$

Let us write $X=\operatorname{Spec} A$. It seems that your main confusion is about what the topology on $X$ looks like in this case. If $A$ is a discrete valuation ring, it has two prime ideals $P=\{0\}$ and $Q$, the maximal ideal. So $X=\{P,Q\}$.

Now we need to determine the topology on $X$. By definition, a subset of $X$ is open iff it is a union of sets of the form $D(f)=\{x\in X:f\not\in x\}$ for elements $f\in A$. So, we must determine these sets $D(f)$. If $f=0$, then $D(f)=\emptyset$. If $f\not\in Q$, then $f$ is a unit, so $D(f)=X$. Finally, if $f\in Q$ is nonzero, then $f\in Q$ but $f\not\in P$, so $D(f)=\{P\}$. Since any union of these sets will again just give another one of these sets, we conclude that there are three open subsets of $X$: $\emptyset,\{P\}$, and $X$.

So a presheaf $F$ on $X$ consists of three abelian groups $F(X)$, $F(\{P\})$, and $F(\emptyset)$ together with restriction homomorphisms $F(X)\to F(\{P\})$ and $F(\{P\})\to F(\emptyset)$. For $F$ to be a sheaf, it needs to satisfy the gluing axiom, but in this case it is rather trivial, since no open subset of $X$ can be written as a union of other open subsets in a nontrivial way. The only restriction imposed is that $F(\emptyset)$ must be a trivial group, since $\emptyset$ is covered by the union of no open sets (this is true for a sheaf on any space).

So, since $F(\emptyset)$ must be trivial, we lose no information by ignoring it, and the data we are left with is two abelian groups $F(X)$ and $F(\{P\})$ together with a homomorphism $F(X)\to F(\{P\})$. This is exactly the data of the category you are asked to show is equivalent. I will leave it to you to write out all the details and verify that this really does give an equivalence of categories.

$\endgroup$
  • $\begingroup$ Yes, you want to show the functor taking $F$ to the restriction map $F(X)\to F(\{P\})$ is an equivalence of categories. (I had a typo before where I wrote $F(\{Q\})$ instead of $F(\{P\})$, now fixed.) I don't know what diagram you are talking about being commutative. $\endgroup$ – Eric Wofsey Oct 8 '18 at 19:42
  • $\begingroup$ I don't know how you think that result is relevant. $\endgroup$ – Eric Wofsey Oct 8 '18 at 21:54
  • $\begingroup$ Our sheaf is a sheaf on $X$, not a sheaf on a one point space. They're totally different things, even if our space happens to have a subset that has one pont... $\endgroup$ – Eric Wofsey Oct 9 '18 at 0:18

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.