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Let $R$ be a commutative ring with $1 \neq 0$ and let $P \subset R$ be a prime ideal. Apparently we have $$\varinjlim\limits_{f \in R \setminus P} R_f \cong R_P$$ where $R_f$ the the localization of $R$ at the set $\{1,f,f^2,\ldots\}$. Why is this the case?

edit 1: I'd like to use the construction of the direct limit and avoid the universal property.

edit 2: In case it's unclear, $R_P$ is the localization at the prime $P$ i.e. $R_P = (R \setminus P)^{-1}R$.

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    $\begingroup$ Where do you encounter difficulties? Everything is straight forward. (Ok you will have to do some calculations when you want to ignore the (essential!) universal properties.) $\endgroup$ – Martin Brandenburg Nov 29 '14 at 15:30
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    $\begingroup$ Two useful posts: 1, 2. $\endgroup$ – André 3000 Oct 23 '17 at 3:48
  • $\begingroup$ See Proposition 2.14 in Ueno's Algebraic Geometry 1. It has a complete proof. Or Exercise 3.23 in Atiyah's Introduction to Commutative Algebra. $\endgroup$ – bfhaha Apr 21 at 16:53
  • $\begingroup$ Could you please tell me where did you see the proof by using universal property? $\endgroup$ – bfhaha Apr 21 at 16:54

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