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.

  • $\begingroup$ 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. $\endgroup$ – winter_mute Jan 30 '20 at 15:10
  • $\begingroup$ @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 $A$. $\endgroup$ – Slup Jan 30 '20 at 15:12
  • $\begingroup$ Oh, I missed such a simple thing! Thanks for the answer. $\endgroup$ – winter_mute Jan 30 '20 at 15:18

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