# 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.

The localization $$A_\mathfrak{p}A_\mathfrak{p}$$ is given by all fractions $$\frac{a}{b}\frac{a}{b}$$ with $$a\in Ra\in R$$ and $$b\in R\setminus\mathfrak{p}b\in R\setminus\mathfrak{p}$$.
So $$\mathfrak{p}A_\mathfrak{p}\mathfrak{p}A_\mathfrak{p}$$ consists of all fractions $$\frac{a}{b}\frac{a}{b}$$ with $$a\in\mathfrak{p}a\in\mathfrak{p}$$ and $$b\in R\setminus\mathfrak{p}b\in R\setminus\mathfrak{p}$$.
To show that $$\mathfrak{p}A_\mathfrak{p}\mathfrak{p}A_\mathfrak{p}$$ is the unique maximal ideal in $$A_\mathfrak{p}A_\mathfrak{p}$$, let $$II$$ be an ideal in $$A_\mathfrak{p}A_\mathfrak{p}$$ with $$I\not\subseteq\mathfrak{p}A_\mathfrak{p}I\not\subseteq\mathfrak{p}A_\mathfrak{p}$$.
Then there is an element $$\frac{a}{b}\in I\frac{a}{b}\in I$$ with $$a,b\in R\setminus\mathfrak{p}a,b\in R\setminus\mathfrak{p}$$.
So $$\frac{b}{a}\frac{b}{a}$$ is an element of $$A_\mathfrak{p}A_\mathfrak{p}$$, and from $$\frac{a}{b}\cdot\frac{b}{a} = 1\frac{a}{b}\cdot\frac{b}{a} = 1$$ we get that $$II$$ contains the invertible element $$\frac{a}{b}\frac{a}{b}$$. Therefore, $$I = A_\mathfrak{p}I = A_\mathfrak{p}$$.