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.
Best Answer
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, $I = A_\mathfrak{p}$.