Vtgyjfy

$mathfrak p$ is prime ideal of commtative ring $R$.
localization $R_{mathfrak p}:=(R-{mathfrak p})^{-1}R$.

We know $mathfrak pR_{mathfrak p}=(R-{mathfrak p})^{-1}mathfrak p$ is prime ideal of $R_{mathfrak p}$. Define $mathfrak q ^c:={rin R | frac r1in mathfrak q}$.

Suppose ${mathfrak q}$ is prime ideal of $R_{mathfrak p}$, if we have $color{blue}{mathfrak q ^c cap (R-{mathfrak p})=varnothing}$, then $mathfrak q^cin mathfrak p, mathfrak qin mathfrak pR_{mathfrak p}$.

So $mathfrak pR_{mathfrak p}$ is the only maximal ideal of $R_{mathfrak p}$.

So how to prove $mathfrak q ^c cap (R-{mathfrak p})=varnothing$ ? Thanks in advance.

As @jgon said, if not, then $mathfrak q$ contains a unit in $R_{mathfrak p}$.

For commutative ring $R$ and multiplicative set $S$ on it, there's a bijection

{prime ideal in $R$ disjoint to $S$} $to$ {prime ideal in $S^{-1}R$}, $quad$ $mathfrak p mapsto S^{-1}mathfrak p$

The reverse map is $mathfrak q mapsto mathfrak q^c$.

And as @reuns said, if commutative ring $R$ has only one maximal ideal $mathfrak m$, then $mathfrak m =R-R^times.$

proof: if $a not in $maximal ideal $mathfrak m$, then $(a)=R$. So $exists b in R $ s.t. $ab=1, a in R^times$.

The exercise asks you to prove that $mathfrak{p}R_{mathfrak{p}}$ is the only maximal ideal of $R_{mathfrak{p}}$, so starting with a prime ideal $mathfrak{q}$ is not the best way (the ring $R_{mathfrak{p}}$ can actually have other prime ideals).

Let $r/s$ be an element of $R_{mathfrak{p}}$ that's not in $mathfrak{p}R_{mathfrak{p}}$. Then $rnotinmathfrak{p}$, otherwise
$$
frac{r}{s}=frac{r}{1}frac{1}{s}inmathfrak{p}R_{mathfrak{p}}
$$
by definition. Then $s/rin R_{mathfrak{p}}$ and so $r/s$ is invertible. Therefore an ideal of $R_{mathfrak{p}}$ properly containing $mathfrak{p}R_{mathfrak{p}}$ contains a unit and so it is the whole ring.

There's however another fact to verify, namely that $mathfrak{p}R_{mathfrak{p}}ne R_{mathfrak{p}}$. This is equivalent to showing that $1/1notinmathfrak{p}R_{mathfrak{p}}$. Suppose the contrary; then
$$
frac{1}{1}=frac{r}{s}
$$
for some $rinmathfrak{p}$ and $sin R-mathfrak{p}$. This amounts to saying that there exists $tin R-mathfrak{p}$ such that $(s-r)t=0$, that is, $rt=st$ which is impossible because $rtinmathfrak{p}$ and $stnotinmathfrak{p}$, the latter due to $mathfrak{p}$ being a prime ideal and $s,tnotinmathfrak{p}$.

$R_P$ is a local ring, i.e. it has a unique maximal ideal of the form
$$ m_P={ frac {p}{s} ; pin P, sin R-P } $$
to see that $m_p$is maximal ideal, every element of $R_P$ not in $m_P$ is the form $frac{s'}{s}$ with $s'in R-P$, and so has an inverse $frac{s}{s'}$ and is hence a unit.