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I was reading the part about localization of the Introduction to Commutative Algebra of Atiyah-MacDonald and I have a question I was not able to solve.

Let $R$ be a commutative ring with unit $1$ and $\mathfrak{p}$ a prime ideal of $R$. It seems to me that there are, a priori, two ways to use $\mathfrak{p}$ to localize $R$.

"Classical" way :

I consider the local ring $R_{\mathfrak{p}} := (R - \mathfrak{p})^{-1} R$. The prime ideals of $R_{\mathfrak{p}}$ are the prime ideals of $R$ which do not meet $(R - \mathfrak{p})$ i.e. those contained in $\mathfrak{p}$.

"Other" way :

Put $S := 1+ \mathfrak{p}$ the subset of $R$ of all the elements on the form $1+ x$ where $1$ is the unit and $x \in \mathfrak{p}$. The set $S$ is closed under multiplication and then I can localize $R$ with respect to $S$ i.e. I consider $S^{-1}R := (1 + \mathfrak{p})^{-1}R$. But this time if I'm not wrong the primes ideals of $S^{-1}R$ are the prime ideals of $R$ which contain $\mathfrak{p}$.

Here's my question : what is the use of the second localization ?

I mean if I want to look at the prime ideal which contain $\mathfrak{p}$ I can look at the quotient $R/\mathfrak{p}$. Is there more information in $S^{-1}R := (1 + \mathfrak{p})^{-1}R$ than in $R/\mathfrak{p}$ ? Or am I completly wrong and these are the same localization ?

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  • $\begingroup$ Look at exercise 7 on page 44. That might help you a bit along the way. $\endgroup$ – Arthur Nov 29 '14 at 23:51
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    $\begingroup$ I'm not sure if you can call the second one localization. How can you prove that you'll end up with a local ring? $\endgroup$ – brick Nov 30 '14 at 0:33
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    $\begingroup$ @brick Localisation is not defined by the local rings it makes. It's defined from the process of introducing inverses of a subset of the elements in a ring. In practice you often get a local ring as a result, but as long as $S\subset R$ contains $1$ and is multiplicatively closed, $S^{-1}R$ makes sense and is called a localisation regardless of the properties of the resulting ring. $\endgroup$ – Arthur Nov 30 '14 at 1:15
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This is a different kind of localization. I have never seen it before, to be honest. And it makes sense for arbitrary ideals, not just prime ideals $\mathfrak{p}$. The prime ideals of $(1+\mathfrak{p})^{-1} R$ correspond 1:1 to the prime ideals of $R$ which are disjoint to $1+\mathfrak{p}$. Prime ideals which contain $\mathfrak{p}$ (which correspond to prime ideals of $R/\mathfrak{p}$) have not much to do with these prime ideals. For example, let $R=\mathbb{Z}$ and $\mathfrak{p}=(3)$. This is a maximal ideal, so there is only one prime ideal containing $\mathfrak{p}$. The prime ideals disjoint from $1+(3)$ are $(0)$ and $(3)$.

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I like to look at the quotient ring a lot.

An element of $R$ is inverted in $R_\mathfrak{p}$ if and only if its image in $R / \mathfrak{p}$ is nonzero. Also, $R_\mathfrak{p} / \mathfrak{p}_\mathfrak{p} \cong (R / \mathfrak{p})_\mathfrak{p} = (R/\mathfrak{p})_{(0)}$, so localizing at a prime makes the quotient ring into a field.

An element of $R$ is inverted in $(1 + \mathfrak{p})^{-1} R$ if and only if its image in $R / \mathfrak{p}$ is a unit. In fact, this is the largest localization with the property that the canonical map $R/\mathfrak{p} \to (S^{-1} R) / (S^{-1} \mathfrak{p})$ is an isomorphism.

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