3 Rings and Modules of Fractions

Definition

Let $A$ be a ring. We say a subset $S\subseteq A$ is multiplicatively closed if $1\in S$ and $S$ is closed under multiplication.

Definition

For multiplication closed $S\subseteq A$, define a relation $\equiv$ on $A\times S$, where $(a,s)\equiv (b,t)$ iff $(at-bs)u=0$ for some non-zero $u\in S$.

We claim that $\equiv$ is an equivalent relation. It suffices to check transitivity. Suppose $(a,s)\equiv (b,t)$ and $(b,t)\equiv (c,u)$, then $(at-bs)v=(bu-ct)w=0$ for some $v,w\in S$. Notice that $(au-cs)tvw=atvuw-ctwsv=bsvuw-buwsv=0$ and we have done.

Definition

Let $a/s$ be the equivalent class of $(a,s)$. Let $S^{-1}A$ be the set of equivalent classes, that is, $S^{-1}A:=A\times S/\sim$. One can define a ring structure by $a/s+b/t=(at+bs)/st$ and $a/s\cdot b/t=ab/st$. We call $S^{-1}A$ the ring of fractions of $A$ w.r.t. $S$.

Remark. We need to check the computation is well-defined. See here.

Fact. We have a map $A\to S^{-1}A:a\mapsto a/1$. It is NOT necessarily injective. For example, if $0\in S$, then $S^{-1}A=0$. e.g. $A=\mathbb{Z}$ and $S=\{0,1\}$.

important

Let $I\subseteq A$ be an ideal. Easy fact: $I$ is prime iff $A-I$ is multiplication closed. (if $a,b\notin I$, then $ab\notin I$)

In this case for $S=A-p$, denote $S^{-1}A$ be $A_p$, and call it the localization of $A$ at $p$.

Fact: $A_p$ is a local ring. Indeed, consider $m=\{a/s:a\in p\}\subseteq A_p$, which is an ideal. For any $b/t\notin m$, $b\in A-p$ and $t/b\in A_p$ and so $b/t\in (A_p{)}^{\times }$. By ^ysw57c, $m$ is the unique maximal ideal.

Fact: $A_p/m\simeq \mathrm{Frac}(A/p)$. See here.

also important

Let $f\in A$. Let $S=\{1,f,\cdots ,f^n,\cdots \}$. Then denote $S^{-1}A$ by $A_f$. Note that if $0\in S$, $S^{-1}A=0$.

• $A=\mathbb{Z}$, $f=p$, $S=\{p^n{\}}_{n⩾0}$. $S^{-1}A=\{a/p^n{\}}_{n⩾0,a\in \mathbb{Z}}$. Here we do not use ${\mathbb{Z}}_p$ here.
• $A=\mathbb{Z}$, $S=\mathbb{Z}-(p)$, then get ${\mathbb{Z}}_{(p)}=\{a/b:(b,p)=1\}$.

Proposition

Let $S$ be a multiplicative closed set, and let $g:A\to B$ be a ring homomorphism such that $g(s)\in B^{\times }$ for all $s\in S$. Then there exists unique $h$ such that

is commute.

\begin{proof} Define $h(a/s)=g(a)/g(s)\in B$. We only check it is well-defined. If $a/s=a^{\prime }/s^{\prime }$, we aim to show $g(a)/g(s)=g(a^{\prime })/g(s^{\prime })$. Since $(a{s}^{\prime }-a^{\prime }s)t=0$ for some $t\in S$, one have $(g(a{s}^{\prime })-g(a^{\prime }s))g(t)=0$. Thus $g(a{s}^{\prime })=g(a^{\prime }s)$ by $g(t)\in B^{\times }$. \end{proof}

Definition

Let $M$ be an $A$-module. For a multiplicative closed set $S$, define relation on $M\times S$,

$$(m,s)=(m^{\prime },s^{\prime })⟺\exists t\in S,t(s{m}^{\prime }-s^{\prime }m)=0$$

which is an equivalent relation. Define $S^{-1}M=\{m/s:m\in M,s\in M\}$ as the set of equivalent classes.

Fact. $S^{-1}M$ is an $S^{-1}A$-module via $a/s\cdot m/t=am/st$. Also there is a $A$-module homomorphism $M\to S^{-1}M,m\mapsto m/1$.

Notation. If $S=A-p$, denote $M_p$; if $S=\{f^n:n⩾0\}$, denote $M_f$.

Easy Fact. $u:M\to N$ is an $A$-module homomorphism induces an $S^{-1}A$-module homomorphism. $S^{-1}u:S^{-1}M\to S^{-1}N,m/s\to u(m)/s$.

Proposition

The operation $S^{-1}$ is exact on $A$-modules. That is, if $M^{\prime }\stackrel{f}{\to }M\stackrel{g}{\to }{M}^{\prime \prime }$ is a exact sequence of $A$-module, then $S^{-1}{M}^{\prime }\stackrel{S^{-1}f}{\to }{S}^{-1}M\stackrel{S^{-1}g}{\to }{S}^{-1}{M}^{\prime \prime }$ is also exact.

\begin{proof} We aim to show $\mathrm{im}(S^{-1}f)=\ker (S^{-1}g)$. Since $(S^{-1}g)\circ (S^{-1}f)=S^{-1}(g\circ f)=0$, one have $\mathrm{im}(S^{-1}f)\subseteq \ker (S^{-1}g)$.

On the other hand, suppose $m/s\in \ker (S^{-1}g)$, then $g(m)/s=0/1$ in $S^{-1}{M}^{\prime \prime }$. It deduces that $g(m)t=0$ for some $t\in S$ and so $g(tm)=0$, $tm\in \ker g=\mathrm{im}f$. Assume that $tm=f(m^{\prime })$ for some $m^{\prime }\in M^{\prime }$. So $m/s=tm/ts=f(m^{\prime })/ts\in \mathrm{im}(S^{-1}f)$. \end{proof}

Proposition

There exists unique $S^{-1}A$-module isomorphism

$$f:S^{-1}A{\otimes }_{A}M\simeq S^{-1}M,a/s\otimes m\mapsto am/s.$$

\begin{proof} Consider $S^{-1}A\times M\to S^{-1}M$ by $(a/s,m)\mapsto am/s$. It is clearly $A$-bilinear, so there exists a unique map $f:S^{-1}A{\otimes }_{A}M\to S^{-1}M$. It is easy to check $f$ is $S^{-1}A$-linear and surjective.

To show $f$ injective, let $f(x)=0$. Note that $x={\sum }_{i=1}^n{a}_i/s_i\otimes m_i=1/(s_1\cdots s_n)\otimes m$ for some $m$. It deduces that $f(x)=f((1/s)\otimes m)=m/s=0$. Then there exists $t\in S$ such that $tm=0$ and so $x=(1/s)\otimes m=(1/st)\otimes (tm)=0$. \end{proof}

Corollary

$S^{-1}A$ is a flat $A$-module.

\begin{proof} By ^092787 and ^d61f0b. \end{proof}

Proposition

Let $M,N$ be $A$-modules. Then $S^{-1}M{\otimes }_{S^{-1}A}{S}^{-1}N\simeq S^{-1}(M{\otimes }_{A}N)$.

\begin{proof} By ^d61f0b, we have

$$\begin{array}{rl}{S}^{-1}M{\otimes }_{S^{-1}A}{S}^{-1}N& \simeq (M{\otimes }_A{S}^{-1}A){\otimes }_{S^{-1}A}(N{\otimes }_A{S}^{-1}A)\\ & =M{\otimes }_{A}N{\otimes }_A{S}^{-1}A\\ & =(M{\otimes }_{A}N){\otimes }_A{S}^{-1}A\\ & =S^{-1}(M{\otimes }_{A}N)\end{array}$$

and we finish the proof. \end{proof}

Example. $M_p{\otimes }_{A_p}{N}_p=(M{\otimes }_{A}N{)}_p$.

Local Properties

Definition

A property $P$ of a ring $A$ (or of an $A$-module $M$) is said to be a “local property”, if $A$ (or $M$) has $P$ $⟺$ $A_p$ (or $M_p$) has $P$ for any $p\in \mathrm{Spec}A$.

Proposition

Let $M$ be a $A$-module. TFAE:

• $M=0$;
• $M_p=0$ for all prime ideal $p$;
• $M_m=0$ for all maximal ideal $m$.

\begin{proof} i)→ii)→iii) is trivial.

iii)→i) If $M\ne 0$, take nonzero $x\in M$ and then $\mathrm{Ann}(x)\subseteq A$ is a proper ideal and $\mathrm{Ann}(x)\subseteq m$ for some maximal ideal $m$.

Consider $M\to M_m=M{\otimes }_A(A-m{)}^{-1}A,x\mapsto x\otimes 1$. Since $M_m=0$, $x/1=0$. Then there exists $t\in A-m$ such that $tx=0$ and so $t\in \mathrm{Ann}(x)\subseteq m$, contradicting with $t\notin m$. \end{proof}

Proposition

Let $\varphi :M\to N$ be a homomorphism between $A$-modules. TFAE:

• $\varphi$ is injective;
• ${\varphi }_p:M_p\to N_p$ is injective for all prime ideal $p$;
• ${\varphi }_m:M_m\to N_m$ is injective for all maximal ideal $m$.

Similar statements hold after replacing “injective” by “surjective” or “bijective”.

\begin{proof} i)→ii) Recall $M_p\simeq M{\otimes }_A{A}_p$ and by ^8p663h.

ii)→iii) Trivial.

iii)→i) Let $P=\ker \varphi$. Then $0\to P\to M\to N$ is left exact. For any maximal ideal $m$, $A_m$ is flat $A$-module yields that $0\to P_m\to M_n\to N_n$ is left exact. It deduces that $P_m=0$ and so $P=0$ by ^663ccb. \end{proof}

Proposition

Let $M$ be an $A$-module. TFAE:

• $M$ is a flat $A$-module;
• $M_p$ is a flat $A_p$-module for all prime ideal $p$;
• $M_m$ is a flat $A_m$-module for all maximal ideal $m$.

\begin{proof} i)→ii) Given injective homomorphism $X\hookrightarrow Y$ of $A_p$-module, we aim to show $X{\otimes }_{A_p}{M}_p\to Y{\otimes }_{A_p}{M}_p$ is injective. Note that $X{\otimes }_{A_p}{M}_p=X{\otimes }_{A_p}(A_p{\otimes }_{A}M)=X{\otimes }_{A}M$ and $X{\otimes }_{A}M\to Y{\otimes }_{A}M$ is injective. Now we finish the proof.

ii)→iii) is trivial.

iii)→i) Suppose $N\hookrightarrow P$ is an injective homomorphism of $A$-modules. By ^766550, it suffices to check injective after localization at any maximal ideal $m$. By ^udklm4, it is same as $N_m{\otimes }_{A_m}{M}_m\to P_m{\otimes }_{A_m}{M}_m$, which is injective by iii). \end{proof}

Extended and Contracted Ideals in $S^{-1}A$

Let $f:A\to S^{-1}A,a\mapsto a/1$. Let $C$ be the set of contracted ideals in $A$, that is,

$$C=\{f^{-1}(\beta )={\beta }^c:\beta \subseteq S^{-1}A\text{ is an ideal}\}.$$

Let $E$ be the set of extended ideals in $S^{-1}A$, that is,

$$E=\{{\alpha }^e={⟨\alpha ⟩}_{S^{-1}A}:\alpha \subseteq A\text{ is an ideal}\}.$$

Fact. ${\alpha }^e=S^{-1}\alpha =\alpha {\otimes }_A{S}^{-1}A$.

Proposition

Part 1:

• If $\beta \subseteq S^{-1}A$ is an ideal, then $\beta =({\beta }^c{)}^e$, that is, $\beta$ is an extended ideal.
• Let $\alpha \subseteq A$ be an ideal, then ${\alpha }^{ec}={\cup }_{s\in S}(\alpha :s)$. Also ${\alpha }^e=S^{-1}A$ iff ${\alpha }^{ec}=A$ iff $\alpha \cap S\ne \varnothing$. Recall that $(\alpha :s)=\{x\in A:xs\in \alpha \}$.
• If $C$ is the set of contracted ideals in $A$, then $\alpha \in C$ iff $\forall s\in S$, $s$ is not zero divisor in $A/\alpha$ iff $(\alpha :s)=\alpha$ for all $s\in S$.

Part 2:

• $P\to S^{-1}P$ gives a $1$-$1$ correspondence between

  $$\{\text{prime ideals }p\subseteq A,p\cap S=\varnothing \}\leftrightarrow \{\text{prime ideals of }{S}^{-1}A\}.$$

• The operation $S^{-1}$ commutes with $+$, $\times$, $\cap$, radical.

\begin{proof} i) By definition, ${\beta }^{ce}\subseteq \beta$. On the other hand, for any $x/s\in \beta$, one have $x/1=s\cdot x/s\in \beta$ and $x\in {\beta }^c$. It deduces that $x/1\in {\beta }^{ce}$ and so $x/s=1/s\cdot x/1\in {\beta }^{ce}$. Now we finish the proof.

ii) Note that $x\in {\alpha }^{ec}=(S^{-1}\alpha {)}^c$ iff $x/1=a/s$ for some $a\in \alpha$ and $s\in S$ iff $(xs-a)t=0$ for some $t\in S$. It yields that $xst=at\in \alpha$ as $a\in \alpha$ and $x\in (\alpha :st)$. Thus, ${\alpha }^{ec}\subseteq {\cup }_{s\in S}(\alpha :s)$. Conversely, if $x\in (\alpha :s)$, then $sx=a\in \alpha$ and $x=a/s\in S^{-1}\alpha$ and so $x\in ({\alpha }^e{)}^c$.

For the “iff” statement,

• ${\alpha }^{ec}=A$ iff $1\in {\alpha }^{ec}$ iff $1\in (\alpha :s)$ for some $s\in S$ iff $s\in \alpha$ for some $s\in S$ iff $\alpha \cap S\ne \varnothing$.
• Also
  • ${\alpha }^e=S^{-1}A$ yields $({\alpha }^e{)}^c=A$.
  • ${\alpha }^{ec}=A$ yields $({\alpha }^{ec}{)}^e=A^e=S^{-1}A$. Also note that $({\alpha }^e{)}^{ce}={\alpha }^e$ by i) and then ${\alpha }^e=S^{-1}A$.

iii) Recall that $\alpha \in C$ iff $\alpha =({\alpha }^e{)}^c$, it is a HW. See Proposition 1.17.

By ii), ${\alpha }^{ec}={\cup }_{s\in S}(\alpha :s)$, where $(\alpha :s)\supseteq \alpha$. Hence $\alpha \in C$ iff $\alpha ={\alpha }^{ec}$ iff $(\alpha :s)=\alpha$ for all $s\in S$ iff $\forall s\in S$, $s$ is not zero divisor in $A/\alpha$.

iv) Step 1. For a given prime ideal $p$ with $p\cap S=\varnothing$, we show $S^{-1}p$ is prime. Since $0\to p\to A\to A/p\to 0$ is exact, $0\to S^{-1}p\to S^{-1}A\to S^{-1}A/p\to 0$ is also exact by ^092787 and ^d61f0b. It deduces that $S^{-1}A/S^{-1}p\simeq S^{-1}A/(S^{-1}A{\otimes }_{A}p)\simeq S^{-1}A{\otimes }_{A}A/p$, and it suffices to show $S^{-1}(A/p)$ is an integral domain.

It is easy to check $S^{-1}(A/p)\simeq {\overline{S}}^{-1}(A/p)$ where $\overline{S}=\mathrm{im}(S\to A/p)$. Note that $P\cap S=\varnothing$ yields that $0\notin \overline{S}$. Since $A/p$ is an integral domain and $\overline{S}$ is a subset without $0$, one have ${\overline{S}}^{-1}(A/p)\subseteq \mathrm{Frac}(A/p)$ and so $S^{-1}(A/p)$ is an integral domain.

Step 2. We show the map $p\to S^{-1}p$ is injective map. For any $s\in S$, $s$ is not a zero divisor of $A/p$ because $A/p$ is an integral domain and $S\cap p=\varnothing$. By iii), $p\in C$ and then $p^{ec}=p$ by Proposition 1.17.

Step 3. We show the above map is surjective. That is, given a prime ideal $\subseteq S^{-1}A$, it is of form $S^{-1}p$. By i), for $q\subseteq S^{-1}A$ prime, then $q=q^{ce}$. So it suffices to show $q^c$ is prime in $A$. Recall that for a ring homomorphism $\theta :G\to H$ and prime ideal $\delta \subseteq H$, one have ${\theta }^{-1}(\delta )\subseteq G$ prime. Indeed, $\theta$ induces a map $\mathrm{Spec}H\to \mathrm{Spec}G$. Now $q^c=f^{-1}(q)$ for $f:A\to S^{-1}A$. Finally, note $q^c\cap S=\varnothing$, otherwise $S^{-1}(q^c)=q^{ce}=S^{-1}A$.

v) This says

• $S^{-1}M+S^{-1}N=S^{-1}(M+N)$ as $S^{-1}M=M\otimes S^{-1}A$ and so for $N,M+N$.
• $S^{-1}(M\oplus N)=S^{-1}A{\otimes }_A(M\oplus N)$
• $S^{-1}(M\cap N)=S^{-1}M\cap S^{-1}N$, whose proof is as follows.

  • Note that $S^{-1}(M\cap N)=S^{-1}A{\otimes }_A(M\cap N)$ and $S^{-1}M\cap S^{-1}N=(S^{-1}A{\otimes }_{A}M)\cap (S^{-1}A{\otimes }_{A}N)$.

  • Consider the left exact sequence $0\to M\cap N\stackrel{\iota }{\to }M\oplus N\stackrel{\phi }{\to }P$ where $\iota (x)=(x,x)$ and $\phi (m,n)=m-n$.

  • Tensor with the flat $A$-module $S^{-1}A$, and there is a exact sequence

    $$0\to (M\cap N){\otimes }_A{S}^{-1}A\to (M{\otimes }_A{S}^{-1}A)\oplus (N{\otimes }_A{S}^{-1}A)\to P{\otimes }_A{S}^{-1}A$$

    Then conclude.

• To show $S^{-1}(r(I))=r(S^{-1}I)$:
  • For any given $x/s\in r(S^{-1}I)$, there is $(x/s{)}^n\in S^{-1}I$ for some $n$ and $x^n/s^n=i/t$ for some $i\in I$ and $t\in S$. Then $(uxt{)}^n\in I$, $uxt\in r(I)$ and $uxt/uts=x/s\in S^{-1}(r(I))$.
  • Conversely, for any given $x/s\in S^{-1}(r(I))$, $x^n\in I$ and $s\in S$, one have $x^n/s^n\in S^{-1}I$ and so $x/s\in r(S^{-1}I)$.

Now we finish the proof. \end{proof}

Remarks. “Many fields.”

• Let $p\subseteq A$ be a prime ideal. Then $A_p$ is a local ring, where $p^e=(A-p{)}^{-1}p$ is the unique maximal ideal. It deduces that $A_p/p^e$ is the residue field of $A_p$. In fact, $\mathrm{Frac}(A/p)\simeq A_p/p^e$. ^4vq1ey
  • RHS equals $(A-p{)}^{-1}A/(A-p{)}^{-1}p=A_p/(p{A}_p)=A_p{\otimes }_A(A/p)=(A-p{)}^{-1}(A/p)$, because $p{\otimes }_A{A}_p\to A{\otimes }_A{A}_p\to A/p{\otimes }_A{A}_p\to 0$ is exact.
  • as argument in ^961e0a iv), it is isomorphic to $(\overline{A-p}{)}^{-1}(A/p)=\mathrm{Frac}(A/p)$, where $\overline{A-p}=\mathrm{im}(A-p\to A/p)=A/p\setminus \{0\}$.
  • Examples.
    • $(p)\subseteq \mathbb{Z}$, ${\mathbb{Z}}_{(p)}=\mathbb{Z}/(\mathbb{Z}-p\mathbb{Z})$ and ${\mathbb{Z}}_{(p)}/p{\mathbb{Z}}_{(p)}\simeq {\mathbb{F}}_p=\mathrm{Frac}(\mathbb{Z}/p)$
    • $(0)\subseteq \mathbb{Z}$, ${\mathbb{Z}}_{(0)}=\mathbb{Q}$ and $\mathbb{Q}/(0)=\mathbb{Q}=\mathrm{Frac}(\mathbb{Z}/(0))$.
    • $A=\mathbb{Z}[x]$, $p=(p)$ and $A_{(p)}=\mathbb{Z}[x]/(\mathbb{Z}[x]-p\mathbb{Z}[x])$. We have $A_{(p)}/pA(p)\simeq {\mathbb{F}}_p(x)$ and $\mathrm{Frac}(\mathbb{Z}[x]/(p))=\mathrm{Frac}({\mathbb{F}}_p[x])={\mathbb{F}}_p(x)$.
• Even if $A$ is integral, $\mathrm{Frac}(A/p)$ in general is not $\mathrm{Frac}(A)$.
• Let $A$ be an integral domain, and let $p\subseteq A$ be a prime ideal. Then $A_p=(A-p{)}^{-1}A\hookrightarrow \mathrm{Frac}(A)$ and so $A_p$ is an integral domain.
  • In fact, $\mathrm{Frac}(A_p)=\mathrm{Frac}(A)$.
    • Proof. $A\hookrightarrow A_p\hookrightarrow \mathrm{Frac}(A)$ is integral domain, which yields that $\mathrm{Frac}(A_p)=\mathrm{Frac}(A)$.
    • Example. $\mathrm{Frac}({\mathbb{Z}}_{(p)})=\mathrm{Frac}(\mathbb{Z})=\mathbb{Q}$.

Proposition

Let $M$ be a finitely generated $A$-module. Then $S^{-1}(\mathrm{Ann}(M))=\mathrm{Ann}(S^{-1}M)(\ast )$.

\begin{proof} Let $\{m_1,\cdots ,m_n\}$ be generators of $M$. For $(\ast )$, $\subseteq$ is obvious, because $a$ kills all $m\in M$ yields that $a/s$ kills $m/t$ for all $m/t\in S^{-1}M$.

Now for $\supseteq$. For $0\ne a/s\in \mathrm{Ann}(S^{-1}M)$, we have $a/s\cdot m_i/1=0$ for all $i$. It deduces that $t_{i}a{m}_i=0$ for some $t_i\in S$. Then $(({\prod }_{i=1}^n{t}_i)a)m_i=0$ for all $i$ and $a({\prod }_{i=1}^n{t}_i)\in \mathrm{Ann}(M)$ is not zero. Therefore, $a/s=(a\prod t_i)/(s\prod t_i)\in S^{-1}(\mathrm{Ann}(M))$. \end{proof}

Proposition

Let $A\to B$ be a ring homomorphism, and let $p\subseteq A$ be a prime ideal. Then $p$ is a contraction of a prime $q\subseteq B$ iff $p^{ec}=p$. Remark that even in this case, $p^e$ might not be prime.

\begin{proof} "→" As $p=q^c$, we have $p^{ec}=(q^c{)}^{ec}=q^c=p$ by Proposition 1.17.

"←" Now suppose $p^{ec}=p$. Let $S=\mathrm{im}((A-p)\to B)$, which is multiplicatively closed in $B$. Consider

$$A\stackrel{f}{\to }B\to S^{-1}B,p\mapsto p^e\mapsto S^{-1}{p}^e.$$

We claim that $p^e\cap S=\varnothing$. If $x\in p^e\cap S$, then $f^{-1}(x)\cap A-p\ne \varnothing$. But $f^{-1}(p^e)=p^{ec}=p$ by assumption, which is a contradiction.

This implies $S^{-1}{p}^e\subseteq S^{-1}B$ is a proper ideal: Otherwise, $1/1=a/s$ where $a\in p^e$, and $1/1=a/s$ means $s^{\prime }(s-a)=0$ and $s^{\prime }s=s^{\prime }a\in S\cap p^e=\varnothing$.

Now there exists a maximal ideal of $S^{-1}B$ which containing $S^{-1}(p^e)$. By ^961e0a, this maximal ideal is of form $S^{-1}m$ with $m\cap S=\varnothing$ and $m\subseteq B$ is prime.

Claim that $m^c=p$. Indeed, $S^{-1}m\supseteq S^{-1}(p^e)$ yields $m\supseteq p^e$ and so $m^c\supseteq p^{ec}=p$. Since $m\cap S=\varnothing$, one have $f^{-1}(m\cap S)=\varnothing$, $m^c\cap (A-p)=\varnothing$ and $m^c\subseteq p$. Therefore, $m^c=p$. \end{proof}

Remark

In ^961e0a (iv), the preimage of a maximal ideal in $S^{-1}A$ under the localization map is not necessarily maximal in $A$. For instance, consider the canonical map $A_p\to S^{-1}{A}_p$. The ideal $p{A}_p$ is the unique maximal ideal of $S^{-1}{A}_p$, but its preimage in $A$ is the prime ideal $p$, which may not be maximal.

Conversely, if $m\subseteq A$ is a maximal element in the set  $\{\mathfrak{p}\subseteq A:\mathfrak{p}\cap S=\varnothing \}$, then its extension $m^e\subseteq S^{-1}A$ is a maximal ideal. Indeed, suppose otherwise: there exists a prime ideal $q=n^e\in S^{-1}A$ such that $q⊋m^e$. Then its contraction $n⊋m$ in $A$ would contradict the maximality of $m$ in the set of primes disjoint from $S$.

In summary, in $1$-$1$ correspondence in ^961e0a (iv),

$$\{\text{prime ideals of }p\subseteq A,p\cap S=\varnothing \}\leftrightarrow \{\text{prime ideals of }{S}^{-1}A\}.$$

this induces a $1$-$1$ map

$$\{\text{maximal elements above}\}\leftrightarrow \{\text{maximal ideals in }{S}^{-1}A\}.$$

But a maximal element in LHS is not necessary a maximal ideal in $A$.