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It is known that a reduced commutative unitary Baer ring $A$ is normal iff for every prime ideal $\mathfrak p$ of $T(A)$ (here $T(A)$ is the total quotient of $A$) one has $A/(\mathfrak p\cap A)$ is integrally closed in $T(A)/\mathfrak p$.

I would be interested in an example for which one implication does not hold if the condition "$A$ is Baer" is removed. My first intuition tells me that one can find counterexamples for both implications.

Edit 1: I wish to elaborate on David's comment (which I believe should be moved as Answer). I think his comment was very beautiful and in fact the easiest example that one could probably come up with. The ring $A$ described was $$A = \{(a,b) \in \mathbb Z^2 : a+b \equiv 0 \mod 2\}$$ As mentioned, clearly $\mathbb Q^2$ is the total quotient of this ring. Now, the above observation (regarding the equivalence of normality for Baer reduced rings) I actually proved in a paper and I was having doubts in it once I saw David's answer (was ca. 10 years ago). But, nothing was actually violated. At first glance I thought that the minimal prime ideals of $A$ would consist of those restricted by the two prime ideals of $\mathbb Q^2$ (which would make $A$ Baer.. but it isn't Baer just by observing that there are no non-trivial idempotents in $A$). The trick was in how he defined $A$ so that summing the coordinates should be even. The other minimal prime ideals of $A$ are those of the form $$\{(a,b): a\in\mathfrak p\text{ and }a,b\equiv 1 \mod 2\} $$ where $\mathfrak p$ is some odd prime in $\mathbb Z$. So the minimal prime ideals of $A$ is countably infinite and not compact (otherwise there is a characterization that tells me $A$ would be Baer if it were compact). Once I adjoin the idempotents of $\mathbb Q^2$ to $A$ I get the Baer hull of $A$ and then the characterization above will hold.

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    $\begingroup$ If $A = \{ (a,b) \in Z^2\mid a+b \equiv 0 \mod 2\}$, then $A $is a unital subring of $Z \times Z$ with ring of quotients $Q \times Q$. There are only two prime ideals in the latter, and their intersections with $A$ yield $Z \subset Q$ as the factor. However, $A$ is not integrally closed in $Q \times Q$, since the obvious idempotents satisfy $x^2 - x = 0$. $\endgroup$
    – David Handelman
    Commented Aug 23, 2016 at 23:29
  • $\begingroup$ Beautiful. I am going to add an edit below my post to elaborate on what you wrote.. I need to explain why this simple but nice example shows us why Baer is important. Btw, it would be nice if this was posted as an answer, if there is no better answer I could mark it as an answer. $\endgroup$
    – Jose Capco
    Commented Aug 24, 2016 at 9:13
  • $\begingroup$ Btw, off topic.. is it me or does the mod symbol render with too much space in MathJax? (I've tried several browsers and there is too much space before mod). $\endgroup$
    – Jose Capco
    Commented Aug 24, 2016 at 9:34
  • $\begingroup$ About the space for the $\mod$ symbol; TeX (plain and AmS-TeX; I don't know about LaTeX) does that too. $\endgroup$
    – David Handelman
    Commented Aug 24, 2016 at 14:10

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If $A=\{(a,b)∈\mathbb Z^2∣a+b≡0 \mod2\}$, then $A$ is a unital subring of $\mathbb Z×\mathbb Z$ with ring of quotients $\mathbb Q×\mathbb Q$. There are only two prime ideals in the latter, and their intersections with $A$ yield $\mathbb Z⊂\mathbb Q$ as the factor. However, $A$ is not integrally closed in $\mathbb Q× \mathbb Q$, since the obvious idempotents satisfy $x^2−x=0$.

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