Two ways to localize a ring using a prime ideal 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 ?
 A: 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 $(3)$.
A: 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.
