1
$\begingroup$

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.

$\endgroup$
1
2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.