Why is the localization at a prime ideal a local ring? I would like to know, why $ \mathfrak{p} A_{\mathfrak{p}} $ is the maximal ideal of the local ring $ A_{\mathfrak{p}} $, where $ \mathfrak{p} $ is a prime ideal of $ A $ and $ A_{\mathfrak{p}} $ is the localization of the ring $ A $ with respect to the multiplicative set $ S = A -\mathfrak{p} $ ?
Thanks a lot.
N.B. : I have to tell you that I'm not very good at Algebra, so please, be more kind and generous in your explanation, and give me a lot of details about this subject please.
Thank you.
 A: The localization $A_\mathfrak{p}$ is given by all fractions $\frac{a}{b}$ with $a\in A$ and $b\in A\setminus\mathfrak{p}$.
So $\mathfrak{p}A_\mathfrak{p}$ consists of all fractions $\frac{a}{b}$ with $a\in\mathfrak{p}$ and $b\in A\setminus\mathfrak{p}$.
To show that $\mathfrak{p}A_\mathfrak{p}$ is the unique maximal ideal in $A_\mathfrak{p}$, let $I$ be an ideal in $A_\mathfrak{p}$ with $I\not\subseteq\mathfrak{p}A_\mathfrak{p}$.
Then there is an element $\frac{a}{b}\in I$ with $a,b\in A\setminus\mathfrak{p}$.
So $\frac{b}{a}$ is an element of $A_\mathfrak{p}$, and from $\frac{a}{b}\cdot\frac{b}{a} = 1$ we get that $I$ contains the invertible element $\frac{a}{b}$. Therefore, $I = A_\mathfrak{p}$.
A: Basically what you need to know is how the units in $A_\mathfrak{p}$ look like. More precisely, an element in the localization, say $\dfrac{a}{b} \in A_\mathfrak{p}$ is a unit if and only if $a \in A \setminus \mathfrak{p}$. Then what this tells you is that the set of non-units of $A_\mathfrak{p}$ is $\mathfrak{p}A_\mathfrak{p}$. 
Therefore now if you want to see why this shows that $\mathfrak{p}A_\mathfrak{p}$ is a maximal ideal, suppose that $I$ is an ideal in $A_\mathfrak{p}$ with $\mathfrak{p}A_\mathfrak{p} \subsetneq I$. Then $I$ must contain a unit, and therefore $I = A_\mathfrak{p}$, so $\mathfrak{p}A_\mathfrak{p}$ is indeed a maximal ideal.
Finally, you need to make sure that you understand why the characterization of $\mathfrak{p}A_\mathfrak{p}$ as the set of non-units in $A_\mathfrak{p}$ implies that it is the unique maximal ideal in $A_\mathfrak{p}$. Well, any proper ideal $\mathfrak{m} \subsetneq A_\mathfrak{p}$ would have to be contained in the set of non-units (since otherwise it would contain a unit and that would imply that the ideal is the whole ring), i.e. $\mathfrak{m} \subseteq \mathfrak{p}A_\mathfrak{p}$, so if $\mathfrak{m}$ is maximal, this implies that $\mathfrak{m} = \mathfrak{p}A_\mathfrak{p}$. 
Thus $\mathfrak{p}A_\mathfrak{p}$ is the unique maximal ideal, and hence $A_\mathfrak{p}$ is a local ring.
A: It is perhaps noteworthy to characterize when a localization at an arbitrary submonoid $S\subset A$ is a local ring. To this end recall a ring  $R$ is local iff $$a+b\in R^\times\implies a\in R^\times \text{ or }b\in R^\times.$$
Write $S_\mathrm{sat}\subset A$ for the saturation of $S\subset A$, i.e for the set of its divisors. It is a theorem that $A[S^{-1}]^\times=\lbrace\tfrac as :a\in S_\mathrm{sat},s\in S\rbrace$.
Exercise. Fix a submonoid submonoid $S\subset A$. Prove $A[S^{-1}]$ is local iff $$a+b\in S_\mathrm{sat}\implies a\in S_\mathrm{sat}\text{ or }b\in S_\mathrm{sat}.$$
