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On page 70 of Hartshrone's Algebraic Geometry he defined a sheaf $O$ of rings on $\operatorname{Spec} A$ for some ring $A$. In particular, $O(U)$ is the set of functions $s: U \rightarrow \coprod_{p \in U} A_{p}$ s.t. $s(p) \in A_p$ and $s$ is locally a quotient of elements of $A$: meaning for each $p \in U$, there is a nbhd $V$ of $p$, contained in $U$, and elements $a,f \in A$, s.t. for each $q \in V$, $f \not\in q$, and $s(q) = a/f$ in $A_q$.

My question is that if $A_q$ is the localisation of $A$ at $q$, then surely if $a/f \in A_q$ we must have $f \in q$, which is the opposite of what is written above. I feel like I am missing something crucial and would appreciate any help.

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    $\begingroup$ The localization of a ring $A$ at a prime ideal $q$ admits denominators in the complement of $q$; that is, elements of $A_{q}$ are equivalence classes of fractions of the form $a/f$, where $f \in A \setminus q$. $\endgroup$
    – Alex Wertheim
    Oct 9, 2019 at 8:15

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For a prime ideal $p$, $A_p := (A\setminus p)^{-1} A$ : you invert elements not in $p$ to make "everything be about $p$".

(Recall that $A_p$ is local with maximal ideal $pA_p$, so if $p$ was inverted, that wouldn't make sense)

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