Is noetherian normal ring a finite direct product of normal domains?

Question

Let A be a Noetherian normal ring, that is, the localization of A at every prime is an integral domain, and is integrally closed in its field of fractions. I want to see A is a finite product of normal domains.

If $\mathfrak{p}_1,\dots,\mathfrak{p}_r$ are the minimal prime ideals of A. I can understand that; $$\bigcap_{i=1}^r \mathfrak{p}_i=\sqrt{(0)}=(0). $$ But I can't confirm that $\mathfrak{p}_i$ are coprime in pairs (I'm trying to use Chinese Remainder Theorem).

Why should there ideals be coprime in pairs? Thank you.

Answer 1

Suppose that for some $i\neq j$ ideal $\mathfrak{p}_i+\mathfrak{p}_j$ is proper in $A$. Then there exists a maximal ideal $\mathfrak{m}$ such that $\mathfrak{p}_i+\mathfrak{p}_j\subseteq \mathfrak{m}$. Now the localization $A_{\mathfrak{m}}$ has at least two distinct minimal prime ideals $\mathfrak{p}_iA_{\mathfrak{m}}$ and $\mathfrak{p}_jA_{\mathfrak{m}}$. On the other hand it is an integral domain. This is contradiction, since integral domain has a unique minimal prime ideal - namely $(0)$.

Remark.

This shows a little bit more. Suppose that $A$ is a noetherian ring such that $A_{\mathfrak{p}}$ is an integral domain for every $\mathfrak{p}\in \mathrm{Spec}\,A$. Then $A = A_1\times...\times A_n$, where $A_1,...,A_n$ are integral domains.