Exercise 3.7

Question

A multiplicatively closed subset $S$ of a ring $A$ is said to be \emph{saturated} if
  \begin{equation*}
    xy \in S \iff x \in S~	ext{and}~y \in S.
  \end{equation*}
  Prove that
  \begin{enumerate}[label=oman*)]
    \item $S$ is saturated $\iff A - S$ is a union of prime ideals.
    \item If $S$ is any multiplicatively closed subset of $A$, there is a unique smallest saturated multiplicatively closed subset $\overline{S}$ containing $S$, and that $\overline{S}$ is the complement in $A$ of the union of the prime ideals which do not meet $S$. ($\overline{S}$ is called the \emph{saturation} of $S$.)
  \end{enumerate}
  If $S = 1 + \mathfrak{a}$, where $\mathfrak{a}$ is an ideal of $A$, find $\overline{S}$.

Amit Awasthi · Accepted Answer

\begin{proof}[Proof $i)$]
  Suppose $A \setminus S = \bigcup \mathfrak{p}_i$ for some prime $\{\mathfrak{p}_i\}$. Then, if $xy \in A \setminus S$, we have that $xy \in \mathfrak{p}_i$ for some $i$, and so $x \in \mathfrak{p}_i \subseteq A \setminus S \lor y\in \mathfrak{p}_i \subseteq A \setminus S$, i.e., $x 
otin S \lor y 
otin S$. Conversely, if $x 
otin S \lor y 
otin S$, then $x \in \mathfrak{p}_i \lor y\in \mathfrak{p}_i$ for some $i$. But then, since $\mathfrak{p}_i$ is an ideal, $xy \in \mathfrak{p}_i \subseteq A \setminus S$, i.e., $xy 
otin S$.
  \par Now suppose $S$ is saturated; we want to show that all $x 
otin S$ are in some prime ideal disjoint from $S$. We see that $(x) \cap S = \emptyset$, for $S$ is saturated, and so is a proper ideal. Moreover, $(x)^e \subseteq S^{-1}A$ is not $(1)$, for $x/1$ is not a unit in $S^{-1}A$. Then, there exists a maximal ideal $\mathfrak{m} \supseteq (x)^e$ by AM Corollary $1.4$. We see by AM Proposition $3.11iv)$ implies $\mathfrak{m}^c \subseteq A$ is a prime ideal that does not intersect $S$, and $x \in \mathfrak{m}^c$.
\end{proof}
\begin{proof}[Proof of $ii)$]
  Let $\mathcal{S}$ be the set of saturated multiplicatively closed subsets of $A$ containing $S$, which is non-empty since it contains $A$. We claim $\overline{S} = \bigcap_{S \in \mathcal{S}} S$ works. Clearly it is unique. Since $A \setminus \overline{S} = A \setminus \bigcap_{T \in \mathcal{S}} S = \bigcup_{T \in \mathcal{S}} A \setminus T = \bigcup_{T \in \mathcal{S}} \bigcup \mathfrak{p}_i$, where the last union is over the prime ideals in the complement of each $S$ shown to exist in part $i)$, and each $\mathfrak{p}_i$ does not intersect $\overline{S}$. Moreover, if there is any $\mathfrak{p}$ that does not intersect $\overline{S}$, $A \setminus \mathfrak{p} \in \mathcal{S}$, and so $\mathfrak{p} \subseteq A \setminus \overline{S}$.
\end{proof}
\begin{proof}[Solution for last statement]
  First note $\mathfrak{p} \cap S 
e \emptyset \iff \exists a \in \mathfrak{a}$ such that $1 + a \in \mathfrak{p} \iff 1 \in \mathfrak{p} + \mathfrak{a}$. By part $ii)$, then, $A \setminus \overline{S} = \bigcup \mathfrak{p}$, where the union is taken over $\mathfrak{p}$ such that $1 
otin \mathfrak{p} + \mathfrak{a}$.
  \par We claim this equals the union $\bigcup \mathfrak{m}$ taken over maximal ideals $\mathfrak{m} \supseteq \mathfrak{a}$. Clearly, $1 
otin \mathfrak{m} + \mathfrak{a}$ since $\mathfrak{m}$ is proper yet contains $\mathfrak{a}$. Likewise, if $\mathfrak{p}$ is such that $1 
otin \mathfrak{p} + \mathfrak{a}$, we can find a maximal ideal $\mathfrak{m} \supseteq \mathfrak{p} + \mathfrak{a}$ by AM Corollary $1.4$, and we see $1 
otin \mathfrak{m} + \mathfrak{a}$. Thus, $\overline{S} = A \setminus \bigcup_{\mathfrak{m} \supseteq \mathfrak{a}} \mathfrak{m}$.
\end{proof}

Exercise 3.7

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Exercise 3.7

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