So the problem is this
Let $R$ be a commutative ring and let $P$ be a prime ideal.
Prove that the set of non-units in $R_{P}$ is the ideal $P_{P}$ , moreover $R_{P}$ is a quasi local ring.
I searched and found this solution.
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 $(a/s)^{-1}=s/a$.
I don't get why we have two direction prove here , isn't it enough to show the first direction only?
Let $I \nsubseteq P_p$. Then there exists some $\frac{a}{b} \in I$ with neither $a,b$ in $P$. But them, $a$ appear among denominators. Conclude that $1 \in I$