Let $A$ be a Noetherian ring and $\mathfrak{p}_{1}, \ldots, \mathfrak{p}_{r}$ all the minimal prime ideals of $A$. Suppose that $A_{\mathfrak{p}}$ is an integral domain for all $\mathfrak{p} \in \operatorname{Spec} A$. Then
(i) $\operatorname{Ass} A=\{\mathfrak{p}_{1}, \ldots, \mathfrak{p}_{r}\}$,
(ii) $\mathfrak{p}_{1} \cap \cdots \cap \mathfrak{p}_{r}=\operatorname{nil}A=0$,
(iii) $\mathfrak{p}_{i}+\bigcap_{j \neq i} \mathfrak{p}_{j}=A$ for all $i$.
It follows that $A \simeq A / p_{1} \times \cdots \times A / \mathfrak{p}_{r}$.
Could you give me a hint to show all three conditions? For (i), what I think is as below.
For (i), if there is an embedded prime ideal $P$ in the associate prime, then $P =\operatorname{ann}(x) \supsetneq \mathfrak{p}_{i}$ for some fixed $i$. Pick $a \in P \setminus \mathfrak{p}_{i}$. Then, in $A_{\mathfrak{p}_{i}}$, $(a/1)$ is unit, thus $A_{\mathfrak{p}}=0$.
However I'm stuck with this whether I can see zero ring as an integral domain or not; if we follow this, then we can conclude (i) easily by contradiction, but I don't know Matsumura's opinion about this. Can I take the viewpoint that zero ring is not a domain in this problem?
(ii) is just followed from (i). However, I don't know how to show (iii). Could you give me a hint to show this? After showing (iii), I can conclude that $A$ is decomposed as above by the chinese remainder theorem.
I thought I've got an answer on (iii).
For (iii), notes that $\mathfrak{p}_{i}$ cannot contain the intersection; otherwise, by Proposition 1.11(ii) of Atiyah-Macdonald, $\mathfrak{p}_{i}$ contain another minimal prime ideal, contradiction. Thus there exists $x \in (\bigcap_{j \neq i}\mathfrak{p}_{j} ) \setminus \mathfrak{p}_{i}$. Then, there exists $y \in A$ such that $xy=0 \in \mathfrak{p}_{i}$, thus $y \in \mathfrak{p}_{i}$. Moreover, for any $y' \in \mathfrak{p}_{i}$, $xy' \in \bigcap_{j}\mathfrak{p}_{j} = \sqrt{(0)}$, thus $(xy')^{d}=0$ for some $d\geq1$.
Now, suppose that $\mathfrak{p}_{i} + (\bigcap_{j \neq i}\mathfrak{p}_{j} ) \subseteq \mathfrak{m} \neq A$ for some maximal ideal $\mathfrak{m}$. Then, in $A_{\mathfrak{m}}$, it suffices to show that $x/1, y/1$ are nonunits and nonzeros. Nonunits are easy; if they are, then $t(xs - s)=0$ for some $t,s \in A \setminus \mathfrak{m}$, which implies that $xst = st$. But $xst \in \mathfrak{p}_{i} \subset \mathfrak{m}$ while $st \in A \setminus \mathfrak{m}$, this is contradiction. Same argument can be applied for $y/1$. Also, they are nonzeros; if they are zero, then $\exists t \in A\setminus \mathfrak{m}$ such that $t(x-0)=0$. This means that $xt \in \mathfrak{p}_{i}$. However, since $x \not\in \mathfrak{p}_{i}$, $t \in \mathfrak{p}_{i} \subseteq \mathfrak{m}$, contradiction. Hence, $A_{\mathfrak{m}}$ is not the integral domain, which contradict the assumption that all localizations are domain.