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.

**Answer**

The localization A_\mathfrak{p} is given by all fractions \frac{a}{b} with a\in R and b\in R\setminus\mathfrak{p}.

So \mathfrak{p}A_\mathfrak{p} consists of all fractions \frac{a}{b} with a\in\mathfrak{p} and b\in R\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 R\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}.

**Attribution***Source : Link , Question Author : Bryan , Answer Author : azimut*