Understanding local rings

Question

I don't understand the last two sentence of the following text (from Adkins' Algebra):

$1-$ Why $\phi(P)$ is maximal ideal? My attempt: It is an idea because for $$\phi(P) = {\{ab^{-1} \in Q(R) \ : \ a\in P \ \text{and} \ b\notin P}\} $$ and for $a'b'^{-1} \in R_S$, $a'b'^{-1}\phi(P)=\phi(P)a'b'^{-1} \subseteq \phi(P)$. But How to prove that it is maximal? Especially I believe that $\phi(P) = R_S$ so it is not maximal!

$2-$ Why there is only one maximal ideal of $R_S$, i.e. $\phi(P)$? (i.e. text says "unique" maximal ideal).

Answer 1

1) $\phi(P)$ is maximal because every element not contained in $\phi(P)$ is a unit, and an ideal that contains a unit must be the whole ring. So suppose $\phi(P) \subsetneq I$ for some ideal $I$. Then there is some $x\in I$ with $x \notin \phi(P)$. Then $x$ is a unit, so, as mentioned above, $I = R_S$.

As for why $\phi(P) \neq R_S$: By definition, prime ideals are proper, so $1 \notin P$. Could $1 = \frac{1}{1}$ be in $\phi(P)$? If it were, then $1 = \frac{a}{b}$ with $a \in P$ and $b \notin P$. But then clearing denominators implies $a = b$, a contradiction.

2) To show this, try proving the following result: Let $A$ be a commutative ring with $1$. If the set $A \setminus A^\times$ of nonunits is an ideal, then it is the unique maximal ideal. We have already seen that $\phi(P) = R_S \setminus R_S^\times$, so this will answer your question.

Answer 2

There's a unique answer to both questions: for any multiplicative subset $S$ of a ring $R$, there is a bijection between the prime ideals of $S^{-1}R$ and the prime ideals of $R$ which do not meet $S$. If $i\colon R\to S^{-1}R$ is the canonical map, the bijections are \begin{align} \operatorname{Spec}(S^{-1}R)&\to\operatorname{Spec}R\\ \mathfrak q &\mapsto \mathfrak i^{-1}(\mathfrak q)\\ i(\mathfrak p)S^{-1}R&\leftarrow \mathfrak p \end{align} In particular, the maximal ideals of $S^{-1}R$ correspond to the prime ideals of $R$ which are maximal among those which do not meet $S$.

Now, if $S=R\setminus\mathfrak p$, the ideals which do not meet $S$ are the ideals contained in $\mathfrak p$, hence every prime ideal $\mathfrak q$ which does not meet $S$ is contained $\mathfrak p$, hence its image $\mathfrak q R_{\mathfrak p}$ is contained in $\mathfrak p R_{\mathfrak p}$, which is by definition the maximal ideal of $R_{\mathfrak p}$.

Proof of the bijection (sketch): We prove that composing maps in both orders yields the identity map:

• If $\mathfrak p$ is a prime ideal of $R$ which does not meet $S$, the ideal generated by $\mathfrak p$ in $S^{-1}R$ is the set of elements of the form $\dfrac ps$ $\;(p\in \mathfrak p,\, s\in S$), and if an element in $i(\mathfrak p)S^{-1}R$ lies in $i(R)$, we can write $$\frac x1=\frac ps\iff\exists t\in S,\;tsx=tp.$$ Thus $tsx\in \mathfrak p$. As $ts\notin \mathfrak p$, $x\in\mathfrak p$. This proves $\;i^{-1}(\mathfrak pS^{-1}R)\subset \mathfrak p$. The reverse inclusion is obvious.
• Conversely, let $\mathfrak q$ be an ideal of $S^{-1}R$, and set $\mathfrak p=i^{-1}(\mathfrak q)$. If $\dfrac qs\in \mathfrak q$, $\;\dfrac q1=\dfrac s1\cdot\dfrac qs\in \mathfrak q\cap i(R)$, so $\;q\in i^{-1}(\mathfrak q)$ and $\dfrac qs\in i^{-1}(\mathfrak q)S^{-1}R$, i.e. $\mathfrak q\subset i^{-1}(\mathfrak q)S^{-1}R$. The reverse inclusion, again, is obvious.