# Proof verification: Atiyah-Macdonald Exercise 3.7(i)

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.

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.