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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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3 Answers 3

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

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    $\begingroup$ 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! $\endgroup$
    – azimut
    Commented Feb 11, 2013 at 19:45
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    $\begingroup$ @PrinceM: As usual, $R$ is seen as a subset of $A_{\mathfrak{p}}$ by identifying an element $a\in R$ with the fraction $\frac{a}{1}$ (which is $(a,1)$ in the pair notation). Thus the ideal $\mathfrak{p}$ is identified with the set $\frac{p}{1}$ with $p\in\mathfrak{p}$. Therefore, indeed $\mathfrak{p}A_{\mathfrak{p}} = \{\frac{p}{1} \cdot \frac{a}{b} \mid p\in\mathfrak{p}, a\in R, b\in A\setminus\mathfrak{p}\} = \{\frac{pa}{b} \mid p\in\mathfrak{p}, a\in R, b\in A\setminus\mathfrak{p}\}= \{\frac{p}{b} \mid p\in\mathfrak{p}, b\in A\setminus\mathfrak{p}\}$. $\endgroup$
    – azimut
    Commented Apr 21, 2017 at 20:02
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    $\begingroup$ @azimut What's $R$? Is it the ring we are working with and $A$ is a subring of $R$? $\endgroup$
    – kubo
    Commented May 22, 2022 at 21:16
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    $\begingroup$ @kubo Wow. 9 years and 75 upvotes later you are the first one spotting (or at least reporting) that I'm using a symbol $R$ without definition. It should be $R = A$ of course. I will correct my answer accordingly. $\endgroup$
    – azimut
    Commented May 23, 2022 at 15:49
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    $\begingroup$ @Avenger If $a\in \mathfrak{p}$, then $a/b\in \mathfrak{p}A_{\mathfrak{p}}$. So by the assumption $I\not\subseteq\mathfrak{p}A_\mathfrak{p}$, we must have an $a/b\in I$ such that $a\notin \mathfrak{p}$. $\endgroup$
    – azimut
    Commented Jun 5, 2022 at 7:44
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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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  • $\begingroup$ How can we prove your claim in the beginning? Thanks. $\endgroup$
    – Ninja
    Commented Dec 11, 2017 at 11:47
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It is perhaps worthwhile 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 $$a+b\in S_\mathrm{sat}\implies a\in S_\mathrm{sat}\text{ or }b\in S_\mathrm{sat}.$$

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