The direct limit of $A_{f}$ for $f \notin P$ is $A_{P}$

Question

Let $A$ be a commutative ring with identity, and $P$ a prime ideal of $A$. I want to show that $\lim_{f \notin P} A_{f} = A_{P}$. $A_f$ is the localization of $A$ at $\{1\} \cup \{f^{n}: n \in \mathbb N\}$, and $A_P$ is the localization of $A$ at $A \setminus P$. There is a natural map from $A_{f} \to A_P$ if $f$ is not in $P$, so by the universal property of the direct limit, we get a map from $\Phi : \lim_{f \notin P} A_{f} \to A_{p}$. Then it remains to show that this canonical map is an isomorphism. Surjectivity can be proven just using the universal property without knowing what $\Phi$ looks like. However I cannot prove the injectivity using the universal property. It seems that one has to use the explicit form of $\Phi$ in order to achieve that.

Answer 1

It seems to me you're assuming the direct limit exists, but justifying that requires some sort of explicit construction anyway, which you're trying to avoid.

I think what you really what you want to do is to show that $A_P$ (with the system of natural maps $\phi_f \colon A_f \to A_P$) satisfies the universal property in question, using only the universal property of localization.

For that, suppose we have $Y$ with natural maps $\psi_f \colon A_f \to Y$. We need a unique map $u \colon A_p \to Y$ where $\psi_f = u \circ \phi_f$. Fix $f \not\in P$. For all $g \not\in P$, the map $\psi_f \colon A_f \to Y$ factors through the map $A_f \to A_{fg}$, and $g$ is a unit in $A_{fg}$, so $\psi_f(g)$ is a unit in $Y$. Hence there exists a unique map $u_f \colon A_P \to Y$ such that $\psi_f = u_f \circ \phi_f$.

All that's left is to show $u_f$ is independent of $f$, which is essentially formal. Write $\alpha_{f,g} \colon A_f \to A_g$ (when this makes sense). For all $f, g \not\in P$, $$\psi_f = \psi_{fg} \circ \alpha_{f, fg} = u_{fg} \circ \phi_{fg} \circ \alpha_{f, fg} = u_{fg} \circ \phi_f.$$ By the uniqueness of $u_f$, we have $u_{fg} = u_f$. Hence $u_f = u_{fg} = u_{gf} = u_g$.