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

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.

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 $$(0)$$.
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.
• Thanks for answer, but there is something I couldn't understand. Why $\mathfrak{p}_iA_{\mathfrak{m}},\mathfrak{p}_jA_{\mathfrak{m}}$ are distinct? I think both can be equal to (0), since A is not an integral domain. – winter_mute Jan 30 '20 at 15:10
• @winter_mute the correspondence $\mathfrak{q}\mapsto \mathfrak{q}A_{\mathfrak{p}}$ is a bijection between the set of all prime ideals $\mathfrak{q}$ of $A$ contained in $\mathfrak{p}$ and the set of prime ideals in $A_{\mathfrak{p}}$. This holds for arbitrary commutative ring $A$ and prime ideal $\mathfrak{p}$ of $A$. – Slup Jan 30 '20 at 15:12