basic question regarding the definition of sheaf of rings

Question

I was wondering and got confused about something. Say we have a sheaf of rings $F$ on a topological space $X$. Let $U$ be an open set of $X$, then by the definition $F(U)$ is a ring. I was wondering, when defining a sheaf $F$ is it necessary to be able to say what ring $F(U)$ is exactly (ie $F(U) = R$) or is it ok if I define $F(U)$ up to isomorphism (ie just have $F(U) \cong R $)?

I guess I was wondering if we have the latter then it gives me a well defined sheaf or not?

Answer 1

A sheaf so defined would be well-defined up to a natural isomorphism of sheaves.

Exercise: let $F$ be a sheaf of rings, $U$ an open set, $S$ a ring, and $f : F(U) \to S$ a ring isomorphism. Define a sheaf $G$ with the property that

$$ G(V) = \begin{cases} F(V) & V \neq U \\ S & V = U \end{cases} $$

and a natural isomorphism of sheaves of rings $\eta : F \to G$.

Exercise: Let $F$ be a sheaf of rings, and for each $U$ let $S_U$ be a ring, and $f_U : F(U) \to S_U$ be a ring isomorphism. Define a sheaf $G$ with the property that $G(V) = S_V$, and a natural isomorphism of sheaves of rings $\eta : F \to G$.

Hint:

In both exercises, you could define $\eta$ first then solve for $G$

Answer 2

I believe that this shouldn't matter. Recall that a sheaf of rings is not just a collection of rings $F(U)$ for each open set $U$, but also the restriction maps $F(U) \to F(V)$ for each inclusion $V \subset U$. So this gives you a lot of compatibility data.

If you were then to replace $F(U)$ by some isomorphic ring, all this would do is require you to modify the restriction maps by composition with said isomorphism. In particular, there would be a natural isomorphism between the two sheaves.

Answer 3

In the definition of a sheaf of rings $\mathscr{F}$, $\mathscr{F}(U)$ is a ring and not an isomorphism class of rings. However, usually you will only need to know the existence of a sheaf such that for each open subset $U$, $\mathscr{F}(U)$ is isomorphic to some ring in a famlily already known. For example when you define the sheaf of rings in the prime spectrum $\mathop{Spec}(A)$ of a commutative ring $A$, you state that there is a sheaf $\mathcal{O}_A$ such that for each $f\in A$, $\mathcal{O}_A(D(f))\cong A_f$ and such that for each inclusion $D(fg)\subset D(f)$, the restriction map $\mathcal{O}_A(D(f))\rightarrow \mathcal{O}_A(D(fg))$ corresponds to the canonical map $A_f\rightarrow A_{fg}$