I have a weird vision that comes from reading a [paper](http://www.informaworld.com/smpp/content~db=all~content=a780098125) 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 $a R[a] \cap R \ne 0$.



Integrally closed factor rings and projective modules
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](https://stacks.math.columbia.edu/tag/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 $\phi \in A/P$.


