Integrally closed factor rings and projective modules I have a weird vision that comes from reading a paper by Raphael and Desrochers..
Let $R$ be commutative unitary semiprime ring such that for any integral and essential element $a$ of $R$, $R[a]$ is a projective $R$-module. I conjecture that for any minimal prime ideal $P$ of $R$, one has $R/P$ is an integrally closed domain.
Does anyone have a counter-example to this? 
PS: In case someone is unfamiliar.. $a$ is an essential element of $R$ iff $a R[a] \cap R \ne 0$.
 A: Would it be particularly surprising if this were true? If I have understood what you mean correctly (so that since you want to consider integrality in full generality a version of essentialness relative to the image of $R$ is required) I believe I have a proof:
First observe that we can reduce to $R$ local. Indeed one can check normality locally and this preserves minimality of primes (when they survive the localization) and so we can assume $R$ is a local reduced commutative ring with unit.
Now let $P$ be a minimal prime ideal in $R$ and consider the composite $R \rightarrow R/P \rightarrow k(P)$ and suppose that $b$ is an integral element in $k(P)$ over $R/P$ and hence over $R$. As in Jose's comment we know that $b$ is essential over $R/P$ and hence over $R$ (I don't see how to make sense of essential for non-injective morphisms otherwise, maybe I am being dense here). So by hypothesis $R/P[b]$ is a finitely generated projective $R$-module and so is free since $R$ is local. But $\operatorname{Ann}(R/P[b])$ is clearly at least $P$ so $P=0$ (since $R$ is reduced) and $R$ is in fact a domain.
I next claim that in fact $R$ is integrally closed in its field of fractions $K(R)$. To see this lets denote by $S$ the integral closure of $R$ in $K(R)$. Then
$S = \operatorname{colim} R[\alpha_1,\ldots,\alpha_n]$
where the $\alpha_i$ vary over all the integral elements. By hypothesis each of the modules occurring in the colimit is free so $S$ is flat over $R$. In particular, it is flat, finite, and $R \subseteq S$ so that it is faithfully flat over $R$. It now follows that $S=R$ by the following standard argument.
Suppose $a = \frac{x}{y}$ is in $S$, where $x,y$ are in $R$. Then $x$ is in $yS$ and $yS \cap R = yR$ by faithful flatness (we prove this below) so that $y$ divides $x$ in $R$ also. In particular $a$ is in $R$.
Proof that $yS \cap R = yR$: since $S$ is faithfully flat over $R$ we get by changing base a faithfully flat map for any ideal $I$, $R/I \rightarrow S\otimes_R R/I \cong S/IS$
which is injective (since faitfully flat maps are always injective  - this follows by using the fact that the kernel of the functor on module categories given by base changing is trivial). In particular we have that $IS \cap R = I$.
In fact I think this gives something stronger. We have shown that the localization at each maximal ideal is a normal domain so in particular $R$ is normal. If $R$ is noetherian it follows that it is a product of finitely many normal domains.
A: The answer is yes.
The key observation is that rational projective extensions are trivial.
Recall an extension of commutative rings $A \subseteq B$ is called rational when for each $b \in B$, the ideal $(A :_A b)$ is dense(=has trivial annihilator) in $A$ and $(A :_A b)b \not= 0$.
Lemma Let $A \subseteq B$ a rational extension of rings and $A \subseteq C$ an extension of rings exhibiting $C$ as a projective $A$-module.  For any amalgamation $D$ of the span $C \supseteq A \subseteq B$ the multiplication map $B \otimes_A C \rightarrow D$ is injective.
Proof Let me just sketch the case that $C$ is a free $A$-module; the generalization to projectives is straightforward.  Let $c_i$ be a basis for $C$.  If an element $\sum b_i \otimes_A c_i$ maps to $0$, i.e. $\sum b_i c_i = 0$ in $D$, then choose a dense ideal $I$ of $A$ such that $I b_i \subseteq A$.  For each element $a \in I$, we get $\sum (a b_i) c_i = 0$.  Since $c_i$ is a basis, deduce $a b_i = 0$.  Thus $b_i I = 0$, which implies $b_i = 0$.
Lemma If $A \subseteq B$ is a rational extension of rings exhibiting $B$ as a projective $A$-module, then $A = B$.
Proof: By the previous lemma, $B \otimes_A B \rightarrow B$ is injective, i.e. $B$ is an epimorphism.  Note also that $B$ is a faithfully flat $A$-module (it is locally free of nonzero rank).  So $A \subseteq B$ is a faithfully flat epimorphism, which is surjective [Stacks Lemma 04VU]
From here there are a number of ways to proceed.
One route is to argue as follows:  Suppose that $A$ is reduced  and for each ring extension $A \subseteq B$, we have that $A[b]$ is projective whenever $bA[b] \cap A \not= 0$ and $b$ is integral over $A$.  In particular this hypothesis will apply to every integral element of the extension $A \subseteq Q_{max}(A)$, where $Q_{max}(A)$ denotes the maximal rational extension of $A$, which is a self-injective reduced ring.  By the previous lemma, this implies that $A$ is integrally closed in $Q_{max}(A)$.  For one thing this implies that $A$ and $Q_{max}(A)$ have the same set of idempotents, so one readily checks that $A$ is Baer because $Q_{max}(A)$ is. Now we know that $A$ is a Baer ring that is integrally closed in $T(A)$, it's total ring of fractions (which is von Neumann Regular).
So we could finish with the following.
Lemma: Let $A$ be a reduced ring such that $T(R)$ is von Neumann Regular and $A$ is integrally closed in $T(R)$.  Then for each minimal prime $P$ of $A$, $A/P$ is integrally closed.
Proof Let $\bar{\phi} \in \kappa(P)$ be integral over $A/P$.  Lift to a relation $\phi^n + \sum a_i \phi^i = \psi \in PT(A)$. Since $T(A)$ is a VNR, write $\psi = e u$ where $u$ is a unit and $e$ is an idempotent of $T(A)$.  Hence $(1 - e) \phi$ is integral over $A$,  and by the integrally closed assumption $(1 - e) \phi \in A$.  Note also that $e \in A$ by the integrally closed assumption, so $1 - e \equiv 1$ modulo $P$.  Hence $\phi \in A/P$.
