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Problem is:

Let $R$ be a commutative ring and let $P$ be a prime ideal.

(a) Prove that the set of non-units in $R_{P}$ is the ideal $P_{P}$.

(b) Deduce that $R_{P}$ has a unique maximal ideal.

I tried (a) as:

Let $u$ be a non-unit in $R_{p}$ and $p$ be any elements in $R_{p}.$ Then, since $u$ is not unit, $up\neq1$ and $pu\neq1.$ Suppose that $up$ or $pu$ is a unit in $R_{p}.$ Then there exists $q$ such that, say, $upq=1.$ Then $u(pq)=1.$ So, $u$ is a unit in $R_{p}.$ This is contradiction. So, the set of non-units in $R_{p}$ is an ideal in $R_{P}.$

So, I have to show that the set of non-units is equal to $P_P$ and (b)

But I am stuck at this point.

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3 Answers 3

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The ideal $P_P$ is $$ P_P=\left\{\frac{a}{s}: a\in P, s\in R\setminus P\right\} $$ Let's prove that no element of $P_P$ is a unit. Suppose $a\in P$, $s\in S=R\setminus P$, $x\in R$ and $t\in S$ be such that $$ \frac{a}{s}\frac{x}{t}=\frac{1}{1} $$ By definition there exists $u\in S$ with $u(ax-st)=0$ or $uax=ust$. This is a contradiction because $uax\in P$, but $ust\in S$.

On the other hand, every element of $R_P$ not in $P_P$ is a unit. Indeed, if $a/s\notin P_P$, we have $a\notin P$, so $(a/s)^{-1}=s/a$.

Therefore the set of nonunits is an ideal and the ring is local, by general results. The maximal ideal is the set of nonunits, which we proved to be $P_P$.

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  • $\begingroup$ Which is the same argument in the picture I gave one hour ago :) $\endgroup$
    – user 1
    Commented Jun 13, 2016 at 14:06
  • $\begingroup$ @user1 Basically so; but I redid it without looking in a book. By the way, I'm unsure adding a picture of a book is not a copyright violation. Of course, the general argument is folklore and not subject to copyright. $\endgroup$
    – egreg
    Commented Jun 13, 2016 at 14:16
  • $\begingroup$ it is not copyright violation. it is from google books. $\endgroup$
    – user 1
    Commented Jun 13, 2016 at 14:17
  • $\begingroup$ @user1 Do you think that all books on that site are free? You're wrong. $\endgroup$
    – egreg
    Commented Jun 13, 2016 at 14:34
  • $\begingroup$ not the book, but parts of it. BTW do you think an snapshot of (half of) a page is copyright violation? $\endgroup$
    – user 1
    Commented Jun 13, 2016 at 14:36
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Let $x\in R$, denote by $[x]$ the class of $x$ in $R_P$, let $[M]=\{[p],p\in P\}$,

$M$ is an ideal, if $[x],[y]\in \in M, [z]\in R_P$ $[x]+[y]=[x+y]\in M$, $[zx]\in M$ since $x+y, zx\in P$

$M$ is distinct of $R_P$, suppose $1\in M$, this implies $1=p\in P$ impossible since $P$ is an ideal.

$M$ is maximal. Let $N$ an ideal which contains strictly $M$, there exists $[s]\in N$, $[s]$ is not in $M$ this implies $s\in R-P$ thus $[s]$ is invertible and $1\in N$.

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Part (a) is Lemma 5.20 of the Sharp's book, "Steps in Commutative Algebra" (page 89):

enter image description here

It is easy to see that $I$ is an ideal of $R_P$. Sharp proves that $R_p\setminus I$ is the set of units. One direction is easy: for any $a/s\notin I,$ one has $a\in S$, so $\dfrac{a}{s}\dfrac{s}{a}=\dfrac{1}{1}.$
On the other hand, if $b/t$ is a unit, say $\dfrac{b}{t}\dfrac{c}{v}=\dfrac{1}{1},$ then there exists $w\in S$ such that $wbc=wtv\in R\setminus P.$ Hence $b\notin P,$ which means $b/t\notin I.$

For part (b), note that a ring is local if and only if the set of non-units of the ring is an ideal (this is lemma 3.13 of the book).

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