In Harstshorne p.70, the definition of a sheaf of ring $\mathcal{O}$ on a the set of prime ideal Spec $A$ is:
For each prime ideal $\mathfrak{p} \subseteq A$, let $A_\mathfrak{p}$ be the localization of $A$ at $\mathfrak{p}$. For an open set $U \subseteq \textrm{Spec} A$, we define $\mathcal{O}(U)$ to be the set of functions $s: U \rightarrow \coprod_{p \in U} A_\mathfrak{p}$. […] We require that for each $\mathfrak{p} \in U$, there is a neighborhood $V$ of $\mathfrak{p}$ contained in $U$, and elements $a,f \in A$, such that for each $\mathfrak{q} \in V$, $f \notin \mathfrak{q}$, and $s(\mathfrak{q}) = \frac{a}{f}$ in $A_q$.
According to http://mathworld.wolfram.com/Localization.html, the localization of $R$ at $S$ is composed of the formal fractions:
\begin{equation}
\left\{\frac{a}{s}|a\in R, s \in S \right\}
\end{equation}
My question is, since $f \notin \mathfrak{q}$, how can $\frac{a}{f}$ be a member of $A_q$ ?
Best Answer
When $A$ is a commutative ring and $\mathfrak{q}\subset A$ is a prime ideal, then the localization $A_{\mathfrak{q}}$ is defined as the localization of $A$ with respect to the set $A\setminus\mathfrak{q}$. That is, in your notation, the set $S$ is $A\setminus\mathfrak{q}$, not $\mathfrak{q}$ itself.