# 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.

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

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Thank you very much. :) –  Bryan Feb 11 '13 at 19:40
How to write the elements of $I$ ideal of $A_{\mathfrak{p}}$, in général ? Thanks. –  Bryan Feb 11 '13 at 19:42
An ideal $I$ of $A_\mathfrak{p}$ is a subset of $A_\mathfrak{p}$. So the elements of $I$ consist of a selection of elements of $A_\mathfrak{p}$, which have the form $\frac{a}{b}$ with $a\in R$ and $b\in R\setminus\mathfrak{p}$. BTW, if my above answer was helpful, feel free to upvote! –  azimut Feb 11 '13 at 19:45
I don't know how to upvote. sorry. –  Bryan Feb 11 '13 at 19:49
Thanks for this answear, but if we have an ideal in $R$, How to write his image in $A_{\mathfrak{p}}$. I'm not certain to be clear. What is $J$ ideal in $A_{\mathfrak{p}}$ such that $i^{-1} ( J )$ is an ideal in $R$ ? $i$ is the map : $i : R \to A_{\mathfrak{p}}$ such that $i ( a ) = \frac{a}{1}$. –  Bryan Feb 11 '13 at 19:53

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.

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Thank you very much. :) –  Bryan Feb 11 '13 at 19:40
@Bryan I'm glad to help =) –  Adrián Barquero Feb 11 '13 at 19:53