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

• How to write the elements of $I$ ideal of $A_{\mathfrak{p}}$, in général ? Thanks. Feb 11, 2013 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! Feb 11, 2013 at 19:45
• 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 $i ( a ) = \frac{a}{1}$. Feb 11, 2013 at 19:53
• @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 $\mathfrak{p}A_{\mathfrak{p}} = \{\frac{p}{1} \cdot \frac{a}{b} \mid p\in\mathfrak{p}, a\in R, b\in A\setminus\mathfrak{p}\} = \{\frac{pa}{b} \mid p\in\mathfrak{p}, a\in R, b\in A\setminus\mathfrak{p}\}= \{\frac{p}{b} \mid p\in\mathfrak{p}, b\in A\setminus\mathfrak{p}\}$. Apr 21, 2017 at 20:02
• @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. May 23 at 15:49

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.

• How can we prove your claim in the beginning? Thanks. Dec 11, 2017 at 11:47

It is perhaps noteworthy 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 $$a+b\in S_\mathrm{sat}\implies a\in S_\mathrm{sat}\text{ or }b\in S_\mathrm{sat}.$$