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.
Best Answer
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 $P_P$.