Proposition

A multiplicatively closed subset $S$ of a ring $A$ is said to be saturated if

Prove that

1. $S$ is saturated $\iff A \setminus S$ is a union of prime ideals.
2. 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 $S$.
3. If $S = 1 + a$ where $a$ is an ideal of $A$, find $\overline{S}$.

Solution

1

Suppose $A \setminus S$ is the union of some prime ideals. Let $x, y \in A$.

• Suppose $xy \notin S$. Then $xy \in A \setminus S$, so $xy \in p$ for some $p \in \Spec(A)$. Thus $x \in p$ or $y \in p$, so $x \in A \setminus S$ or $y \in A \setminus S$. This implies $x \notin S$ or $y \notin S$.
• Suppose $x \notin S$ or $y \notin S$. Then $x \in A \setminus S$ or $y \in A \setminus S$. Without loss of generality, $x \in A \setminus S$. Then $x \in p$ for some $p \in \Spec(A)$. Since $p$ is an ideal, $xy \in p$. Thus $xy \notin S$. Therefore, $xy \notin S \iff (x \notin S \text{ or } y \notin S)$. In other words, $xy \in S \iff (x \in S \text{ and } y \in S)$.

On the other hand, suppose $S$ is saturated. Let $a \in A \setminus S$. Then $(a)$ is an ideal disjoint from $S$ because $xa \in S$ implies $x \in S$ and $a \in S$.

Let $\Sigma$ be the set of ideals of $A$ containing $a$ and disjoint from $S$. Then $(a) \in \Sigma$, so $\Sigma$ is nonempty. $\Sigma$ is a partially ordered set by the usual set inclusion. Let $I_1 \subset I_2 \subset \cdots$ be an ascending chain in $\Sigma$. Then $I = \cup I_i$ is an ideal of $A$ containing $a$ and disjoint from $S$. Thus we found an upper bound of the given chain. By Zorn’s Lemma, $\Sigma$ contains a maximal element $m$.

Let $x, y \in A \setminus m$. Then $m + (x), m + (y) \notin \Sigma$ because $m$ is maximal. Since $m + (x), m + (y)$ are ideals containing $a$, they cannot be disjoint from $S$. In other words, there exist $s \in (m + (x)) \cap S$ and $t \in (m + (y)) \cap S$. Then $st \in S$ because $S$ is multiplicatively closed, so $st \notin m$ because $m$ is disjoint from $S$. However, $st \in (m + (x))(m + (y)) = m + (xy)$. Therefore, $xy \notin m$, and thus $m$ is a prime ideal.

The argument above shows that, for every $a \in A \setminus S$, there exists a prime ideal $p \in \Spec(A)$ such that $a \in p \subset A \setminus S$. Therefore, $A \setminus S$ is the union of such prime ideals.

2

Let $T$ be the complement of all prime ideals $p \in \Spec(A)$ such that $p \cap S = \emptyset$. Then $S \subset T$. By Part 1, $T$ is saturated. Let $T’$ be a saturated multiplicatively closed subset such that $T’ \subsetneq T$. By Part 1, $A \setminus T’$ must be a union of some prime ideals. Since $T’ \subsetneq T$, at least one of the prime ideals that make up $A \setminus T’$ cannot be disjoint from $S$. In other words, $S \not\subset T’$. Therefore, $T$ is a smallest saturated multiplicatively closed subset. Since the intersection of two saturated multiplicatively closed subsets is a saturated multiplicatively closed subset, $T$ must be the unique one.

3

For all $p \in \Spec(A)$,

• Discussion 1
  • $p \cap (1 + a) \ne \emptyset$ if and only if there exists an $x \in p$ such that $x \in 1 + a$. In other words, $\exists x \in p, x - 1 \in a$, so $\exists x \in p, 1 - x \in a$. This is possible if and only if $1 = (1 - x) + x \in a + p$.
• Discussion 2
  • Let $\Sigma$ be a set of all proper ideals containing $p + a$. Given any chain $I_1 \subset I_2 \subset$ in $\Sigma$, $I = \cup I_i$ is an upperbound of the chain and $I \in \Sigma$ because $1 \notin I$. By Zorn’s Lemma, $\Sigma$ contains a maximal element, which must be a maximal ideal in $A$.

Let $P = \{ p \in \Spec(A) \mid p \subset A \setminus S \}$ and $M = \{ m \in \Max(A) \mid a \subset m \}$. Then we claim that $\bigcup_{p \in P} p = \bigcup_{m \in M} m$.

• $\bigcup_{p \in P} p \supset \bigcup_{m \in M} m$?
  • Let $m \in M$. Then $m$ is a prime ideal contained such that $m \subset m$ and $a \subset m$. By the argument above, $m \subset A \setminus S$. Thus $m \in P$.
• $\bigcup_{p \in P} p \subset \bigcup_{m \in M} m$?
  • Let $p \in P$. By the argument above, there exists an $m \in M$ such that $p \subset m$ and $a \subset m$. Thus $p \subset m \subset \bigcup_{m \in M} m$.

Therefore, $\bigcup_{p \in P} p = \bigcup_{m \in M} m$. Hence, $\overline{S} = A \setminus \{ m \in \Max(A) \mid a \subset m \}$.