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In 'Commutative Ring Theory' by Matsumura the definition of normal ring is as follows:

A ring $R$ is called normal if for every prime ideal $\mathfrak p\subset R$, $R_{\mathfrak p}$ is an integrally closed domain.

I know that a domain is integrally closed if and only if localisation at every prime ideal gives an integrally closed domain, i.e., it is normal.

I want to have an example of a 'non-domain' which is normal. Also is there any equivalent criteria for a ring (not necessarily domain) to be a normal ring like in the domain case?

Thank you in advance.

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    $\begingroup$ I wonder to which "equivalent criteria... to be normal ring like in the domain case" are you referring to? $\endgroup$ – user26857 Sep 15 '17 at 7:53
  • $\begingroup$ @user26857 I wrote that in the question: A domain $R$ is normal iff $R_\mathfrak p$ is integrally closed domain (definition) iff $R$ is integrally closed domain (Atiyah Macdonald prop. $5.13$ ) $\endgroup$ – user276115 Sep 15 '17 at 14:33
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The product of normal rings is normal and never a domain.

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  • $\begingroup$ To prove product of two normal rings is normal: Let $A$ and $B$ are two normal rings. Prime ideals of $A\times B$ is of the form $P\times B$ or $A\times Q$ where $P$ and $Q$ are prime ideals of $A$ and $B$ respectively. Then $(A\times B)_{P\times B}=A_P$, hence it is integrally closed. Similarly for the other case. Is this correct? $\endgroup$ – user276115 Sep 15 '17 at 15:33
  • $\begingroup$ To nitpick, the product of normal rings can be a domain if one of them is the zero ring. $\endgroup$ – Eric Wofsey Sep 15 '17 at 18:37
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    $\begingroup$ @EricWofsey, on prime numbered years my rings are required to have non-zero identities. $\endgroup$ – Mariano Suárez-Álvarez Sep 15 '17 at 19:01
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For example $F\times F$ for a field $F$.

Or any of these at DaRT.

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