A multiplicatively closed subset $S$ of a ring $A$ is said to be saturated if: $$ xy\in S \Leftrightarrow x\in S \text{ and } y\in S $$

Exercise 3.7(i): $S$ is saturated if and only if $A-S$ is a union of prime ideals.

$(\Leftarrow)$: Let $A-S$ be a union of prime ideals $\{ \mathfrak{p}_i \}_{i\in I}$. $$ S = A-(A-S) = A-\bigcup_{i\in I} \mathfrak{p}_i = \bigcap_{i\in I} A\setminus \mathfrak{p}_i $$ Let $xy\in S$ then $xy\in A\setminus \mathfrak{p}_i$ for all $i\in I$. Let $\mathfrak{m}_i$ be the maximal ideal in the local ring $A_{\mathfrak{p}_i}$ and consider the maps: $$ A \overset{\mu}{\to} A_{\mathfrak{p}_i} \overset{\pi}{\to} A_{\mathfrak{p}_i} / \mathfrak{m}_i $$ Thus, $\pi(\mu(xy)) \neq 0$ since $\mu(xy)$ is a unit. Therefore, $\pi(\mu(x)), \pi(\mu(y)) \neq 0$. Suppose that $x\in \mathfrak{p}_i$ (any $i$) for a contradiction, then $(x)\subset \mathfrak{p}_i$ and hence $(x)_{\mathfrak{p}_i} \subset \mathfrak{p}_i^e = \mathfrak{m}_i$; therefore, $\pi(\mu(x)) = 0$. Contradiction. Therefore, $x\in S$; similarly for $y$.

($\Rightarrow$): Let $x \in A$ be a unit, then $x\cdot 1/x = 1\in S$. Therefore, $x\in S$. Therefore, $A\setminus S$ contains only non-units. Consider: $$ U := \bigcup_{\mathfrak{p} \text{ prime} \\ \mathfrak{p} \subset A\setminus S} \mathfrak{p} $$ I claim that $A\setminus S \subset U$. Let $x\in A\setminus S$, then $x\in \mathfrak{m}$ for some maximal ideal $\mathfrak{m}$ (since $x$ is a non-unit). Therefore, $(x)\subset \mathfrak{m}$ and $S^{-1}(x) \subset S^{-1}\mathfrak{m}$ where $S^{-1}\mathfrak{m}$ is a prime ideal in $S^{-1}A$ by the correspondence theorem. The contraction of $S^{-1}\mathfrak{m}$ back into $A$ is a prime ideal which contains $x$ and is contained in $A\setminus S$. Therefore, $x\in U$.

I hope I used the correspondence theorem correctly. Please let me know if my reasoning is sound. Thank you.

EDIT: I think I have not guaranteed that $\mathfrak{m}$ does not meet $S$, so this proof might not work as stated. I will try to fix it tomorrow morning. Please let me know your thoughts anyway.


1 Answer 1


Let $A-S$ be a union of prime ideals. Suppose $xy\in S$, but $x\notin S$. Then $x\in A-S$, which is a union of prime ideals, and hence $x\in\mathfrak{p}\subseteq A-S$. So $xy\in\mathfrak{p}$ too. So $xy\in A-S$. Contradiction.

Let $S$ be saturated. Let $x\in A-S$. You showed $x$ is not a unit. So $x$ remains a non-unit in $S^{-1}A$. Then $x$ is in a maximal (hence prime) ideal in $S^{-1}A$. So $x$ is in a prime ideal in $A-S$ by the correspondence.


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