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.



Normal reduced rings
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 $x^2−x=0$. 


