Prove $R_{\mathfrak p}$ has only one maximal ideal $\mathfrak pR_{\mathfrak p}$

Question

$\mathfrak p$ 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 $\mathfrak q ^c:=\{r\in R\ |\ \frac r1\in \mathfrak q\}$.

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

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.

Answer 1

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 $r\notin\mathfrak{p}$, otherwise $$ \frac{r}{s}=\frac{r}{1}\frac{1}{s}\in\mathfrak{p}R_{\mathfrak{p}} $$ by definition. Then $s/r\in R_{\mathfrak{p}}$ 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 $1/1\notin\mathfrak{p}R_{\mathfrak{p}}$. Suppose the contrary; then $$ \frac{1}{1}=\frac{r}{s} $$ for some $r\in\mathfrak{p}$ and $s\in R-\mathfrak{p}$. This amounts to saying that there exists $t\in R-\mathfrak{p}$ such that $(s-r)t=0$, that is, $rt=st$ which is impossible because $rt\in\mathfrak{p}$ and $st\notin\mathfrak{p}$, the latter due to $\mathfrak{p}$ being a prime ideal and $s,t\notin\mathfrak{p}$.

Answer 2

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 $\mathfrak q \mapsto \mathfrak q^c$.

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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. $ab=1, a \in R^\times$.

Answer 3

$R_P$ is a local ring, i.e. it has a unique maximal ideal of the form $$ m_P=\{ \frac {p}{s} ; p\in P, s\in R-P \} $$ 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.