Localization of $\mathfrak p$ is $\mathfrak p R_{\mathfrak p}$ In the setting of this question it is mentioned that

Since localization preserves exact sequences we have that the exact sequence
$$0 \to \mathfrak p \to R \to R/\mathfrak p \to 0$$
becomes
$$0 \to \mathfrak p R_{\mathfrak p} \to R_{\mathfrak p} \to (R/\mathfrak p)_{\mathfrak p} \to 0.$$

I think this implies that the localization of $\mathfrak p$ at $\mathfrak p$ is $\mathfrak p R_{\mathfrak p}$. I am not sure how to get this result.
Any hints are appreciated!
 A: I'll prove that $\mathfrak p_\mathfrak p$ and $\mathfrak p R_\mathfrak p$ are isomorphic as $R$-modules. First, we have that for every $R$-module $M$ and multiplicative set $S\subset R$ $$S^{-1}M\simeq M\otimes_R S^{-1}R\tag{*}$$
I expect this to be known, if not you can find the proof here. In particular we have $\mathfrak p_\mathfrak p\simeq\mathfrak p\otimes_RR_\mathfrak p$, so we then apply that for a flat module $M$ and ideal $I$ $$IM\simeq I\otimes_RM\tag{**}$$
hence the conclusion that $\mathfrak p_\mathfrak p\simeq \mathfrak pR_\mathfrak p$ as $R_\mathfrak p$-modules. A proof of $R_\mathfrak p$ being flat ($-\otimes_RR_\mathfrak p:R\text{Mod}\rightarrow R_\mathfrak p\text{Mod}$ preserves injective morphisms) is immediate from (*), otherwise you can take the Stacks Project word for it. Finally, (**) can be prove as follows: Since $M$ is flat and $I\hookrightarrow R$ is injective $$I\otimes_RM\rightarrow R\otimes_RM\simeq M,\qquad r\otimes m\mapsto rm$$
is injective and clearly has image $IM$, so it restricts to an isomorphism $I\otimes_RM\simeq IM$.
A: This is an instance of much more general correspondence between ideals of $R$ and ideals of $S^{-1}R$ for a multiplicative subset $S\subset R$. This correspondence is between the sets of prime ideals
$$
\{P\le R\,|\,P\cap S=\emptyset\}\leftrightarrow\{\mathcal Q\le S^{-1}R\}
$$
and induced by the map $P\mapsto S^{-1}P:=(S^{-1}R)P$. Alternatively, this is just the extension of ideals along the canonical map $\epsilon\colon R\to S^{-1}R$ (with inverse the contraction of ideals).
I will not prove this (one can find this in a lot of books on commutative algebra) but the proof is of elementary nature. Indeed, the main idea is that localizations only give you new elements which you can multiply by but all these elements are units by construction, hence irrelevent for the generators of ideals.
A: The other answers have shown how to see this fact as a special case of more abstract results. But this special case is almost trivial to see directly, so I wanted to complement the other answers with a very concrete explanation.
First, let's be clear that for an $R$-module $M$, the $R_{\mathfrak{p}}$-module $M_{\mathfrak{p}}$ is (by definition) the set of equivalence classes of fractions $\frac{a}{b}$, where $a\in M$ and $b\in R\setminus \mathfrak{p}$.
So the $R_{\mathfrak{p}}$-module $\mathfrak{p}_{\mathfrak{p}}$ consists of equivalence classes of fractions $\frac{a}{b}$, where $a\in \mathfrak{p}$ and $b\in R\setminus \mathfrak{p}$.  The equivalence relation on fractions and the operations of addition and multiplication by elements of $R_{\mathfrak{p}}$ are defined exactly as they are in the ring $R_{\mathfrak{p}}$. So $\mathfrak{p}_{\mathfrak{p}}$ is naturally identified with a submodule of $R_{\mathfrak{p}}$, and it remains to show that $\mathfrak{p}_{\mathfrak{p}}$ is equal to $\mathfrak{p}R_{\mathfrak{p}}$ as subsets of $R_{\mathfrak{p}}$.
In one direction, if $\frac{a}{b}\in \mathfrak{p}_{\mathfrak{p}}$, then $a\in \mathfrak{p}$ and $b\in R\setminus \mathfrak{p}$, so $\frac{a}{b} = \frac{a}{1}\cdot \frac{1}{b}\in \mathfrak{p}R_{\mathfrak{p}}$. In the other direction, an element of $\mathfrak{p}R_{\mathfrak{p}}$ can be written as $\frac{a}{1}\cdot \frac{c}{b}$, with $a\in \mathfrak{p}$, $b\in R\setminus \mathfrak{p}$, and $c\in R$. Then $\frac{a}{1}\cdot \frac{c}{b} = \frac{ac}{b}\in \mathfrak{p}_{\mathfrak{p}}$, since $ac\in \mathfrak{p}$.
