Let $X = \mathrm{Spec}\ A$, and for each prime ideal $\mathfrak{p} \subseteq A$, let $A_{\mathfrak{p}}$ be the localization of $A$ at $\mathfrak{p}$. Now, for each open set $U \subseteq X$, define $\mathcal{O}(U)$ to be the set of functions
$$s: U \rightarrow \coprod_{\mathfrak{p} \in U} A_{\mathfrak{p}}$$
such that $s(\mathfrak{p}) \in A_{\mathfrak{p}}$ for each $\mathfrak{p}$, and such that $s$ is locally a quotient of elements of $A$, that is, for each $\mathfrak{p} \in U$, there is some neighborhood $V_{\mathfrak{p}}$ of $\mathfrak{p}$ with $V_{\mathfrak{p}} \subseteq U$ and elements $a,f \in A$ such that for each $\mathfrak{q} \in V$ and $f \notin \mathfrak{q}$, $s(\mathfrak{q} )= a/f$ in $A_{\mathfrak{q}}$. My question is
Why is the sheaf of rings $\mathcal{O}$ defined above a sheaf of rings?
While it seems clear that $\mathcal{O}$ is a presheaf, I fail to see why it satisfies the extra axioms that make it a sheaf. Hartshorne says "it is clear from the local nature of the definition that $\mathcal{O}$ is a sheaf", but while trying to prove it formally, I couldn't show that it is actually a sheaf.
Also, I've just been fiddling around with the definitions for the moment trying to gain some insight about why this "provides a systematic way of keeping track of local algebraic data on $\mathrm{Spec}\ A$", but for now, it's just as mysterious as it was at the beginning. So if anyone could provide me some insight about why this is what we want to be our definition of structure sheaf, I'd be very grateful for that. Thank you in advance!
Best Answer
As someone who has experienced nearly constant frustration in verifying "obvious" claims in algebraic geometry textbooks, I didn't find this one to be too bad.
Let $U_i$ be an open cover of $U$, let $s \in \mathcal O_X(U)$, and suppose $s|U_i = 0$ for all $i$. Now $s$ is a function from $U$ into the disjoint union of the rings $A_{\mathfrak p} : \mathfrak p \in U$. To say that $s$ is the zero element is to say that $s(\mathfrak p)$ is zero in $A_{\mathfrak p}$ for all $\mathfrak p$. So it is clear that $s = 0$.
Now suppose $s_i \in \mathcal O_X(U_i)$, and $s_i$ and $s_j$ agree on $U_i \cap U_j$ for all $i, j$. Then it is clear that there is a unique function $s: U \rightarrow \coprod\limits_{\mathfrak p \in U} A_{\mathfrak p}$ whose restriction to $U_i$ is $s_i$ for all $i$. Obviously, $s(\mathfrak p) \in A_{\mathfrak p}$ for all $\mathfrak p$, and the question now is whether $s$ is locally a quotient of elements from $A$.
Let $\mathfrak q \in U$. Then $\mathfrak q $ lies in some $U_i$, and since $s_i = s|U_i$ is in $\mathcal O_X(U_i)$, there exists an open neighborhood $V$ of $\mathfrak q$ and elements $a, f \in A$, with $f \not\in \mathfrak p$ for any $\mathfrak p \in V$, such that $s_i(\mathfrak p) = \frac{a}{f} \in A_{\mathfrak p}$ for all $\mathfrak p \in V$. But $s_i(\mathfrak p ) = s(\mathfrak p)$, done.