# Direct limit of localizations of a ring at elements not in a prime ideal

For a prime ideal $P$ of a commutative ring $A$, consider the direct limit of the family of localizations $A_f$ indexed by the set $A \setminus P$ with partial order $\le$ such that $f \le g$ iff $V(f) \subseteq V(g)$. (We have for such $f \le g$ a natural homomorphism $A_f \to A_g$.) I want to show that this direct limit, $\varinjlim_{f \not\in P} A_f$, is isomorphic to the localization $A_P$ of $A$ at $P$. For this I consider the homomorphism $\phi$ that maps an equivalence class $[(f, a/f^n)] \mapsto a/f^n$. (I denote elements of the disjoint union $\sqcup_{f \not\in P} A_f$ by tuples $(f, a/f^n)$.) Surjectivity is clear, because for any $a/s \in A_P$ with $s \not\in P$, we have the class $[(s, a/s)] \in \varinjlim_{f \not\in P} A_f$ whose image is $a/s$. For injectivity, suppose we have a class $[(f, a/f^n)]$ whose image $a/f^n = 0/1 \in A_P$. Then there exists $t \notin P$ such that $ta = 0$. We want to show that $[(f, a/f^n)]] = [(f, 0/1)]$, which I believe is equivalent to finding a $g \notin P$ such that $V(f) \subseteq V(g)$ and $g^ka = 0$ for some $k \in \mathbb{N}$. Well, $t$ seems to almost work, but I couldn’t prove that $V(f) \subseteq V(t)$, so maybe we need a different $g$? Or am I using the wrong map entirely?

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clarify what $t$ is? – Ehsan M. Kermani Jun 25 '12 at 15:09
@ehsanmo, thanks! My $t$ was your $g$, your $gf$ was the $g$ I was looking for. (I couldn’t get $V(f) \subseteq V(g)$, but $V(f) \subseteq V(fg)$ is obvious.) – Adeel Jun 25 '12 at 16:30
@Adeel what is $V$, and why isn't this a consequence of the exactness of localization? – Exterior Oct 29 at 12:17
@Exterior, V(f) is the complement of the principal open subset defined by $f$. – Adeel Oct 29 at 13:34

If $a/f^n \in A_f$ is mapped to $0$ in $A_p,$ then there is a $g \not \in p,$ s.t. $ga=0,$ therefore, $a/f^n=0 \in A_{gf}.$ Hence the injectivity.