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 $a R[a] \cap R \ne 0$.

  • $\begingroup$ I am a little confused by what you mean by an integral element of R. Do you mean an element of the total quotient ring satisfying a monic polynomial with coefficients in R or something along those lines? Also semiprime in this situation really just means reduced right? $\endgroup$ Nov 6, 2009 at 0:10
  • $\begingroup$ semiprime in this situation just means reduced, yes. An integral element need not be an element of the total quotient ring, if this is too confusing you may as well think of an integral element to be an element "a" belonging to an over-ring of R such that R[a] is a finitely generated R-module... I don't even know the answer even if "a" were in the total quotient ring, so if that helps lets try to first suppose 'a' is in the total quotient ring. All elements of the total quotient are essential elements, so we may start with that. $\endgroup$
    – Jose Capco
    Nov 6, 2009 at 0:53
  • $\begingroup$ Im thinking in the lines of a counter-example.. one that is as simple as possible.. does Z[T√2,T] satisfy our property? this integral domain is not integrally closed and its total quotient is Q(√2,T) .. but does this ring satsify this property for projectivity? $\endgroup$
    – Jose Capco
    Nov 6, 2009 at 1:03

2 Answers 2


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 $\phi \in A/P$.

  • $\begingroup$ I needed to refresh my memory (this question was posted 10 years ago). On first impression, I agree with what you write and after so many years I will mark this as answered! I would probably prove the Lemma in a different way (by assuming that $b_i$ is not zero and arriving to the contradiction, because the last sentence of the proof had me thinking), but that is a matter of taste. I was wondering if there is any community doing research in this direction nowadays. I switched to algebraic geometry but if there are more research interest in this area, then I would consider looking at it again. $\endgroup$
    – Jose Capco
    Mar 15, 2021 at 8:26
  • $\begingroup$ @JoseCapco I have no idea if there are people doing research in this strand of commutative algebra! But I can tell you how I arrived at your question: a long time ago I was interested in understanding how to generalize aspects of multiplicative ideal theory (think Gilmer's innovations in commutative ring theory over a domain) to ugly rings with gross zero divisors and also in understanding how to think about birational geometry over nasty schemes. The lemma about projective and rational maps that I mentioned is, I think, a basic building block in these investigations. $\endgroup$ May 12, 2021 at 5:27

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.

  • $\begingroup$ I am about to digest your proof. But I thought to give a comment first. You wrote in the beginning that b is essential over R.. b is supposed to be belonging to an over-ring of R. And as far as I know essentiality requires injectivity... I thought it was clear that b is supposed to belong to an over-ring of R. You could mean one element that belongs to the product of factor over minimal primes whose projection to R/P is b, but which of this element preserves essentiality and integrality for both R and R/P? more comments later $\endgroup$
    – Jose Capco
    Nov 7, 2009 at 10:07
  • $\begingroup$ It is probably not quite what you are after then. I guess it is reasonable to restrict the notion of essential to injective morphisms in which case a different argument would be required or a counterexample might exist (I interpreted over-ring as algebra, requiring injectivity did occur to me but I figured essential should make sense in the same generality as integral). $\endgroup$ Nov 7, 2009 at 10:30
  • $\begingroup$ Nevertheless, your proof does use some arguments that is interesting to me (like using flatness of S to argue that yS∩R = yR). I think it is providing me something to work on. Regarding essentiality, its a general term in Category theory: Let C be a categry and then a monomorphim a-->b (here we see injectivity!) is said to be essential iff for any morphism b-->c such that a-->b-->c is a monomorphism one has b-->c is a monomorphism. For Crings this is equivalent to saying that any non-zero ideal of the overring, from a commutative ring extension of R, restricts to a nonzero ideal of R. $\endgroup$
    – Jose Capco
    Nov 7, 2009 at 12:37
  • $\begingroup$ ok.. I'm a bit lost here. How did you used flatness to show that yS∩R = yR .. I know yS=yR@S but who does this show yR=yS∩R?? If your proof is correct, then the conjecture would at least hold for an integral domain. I hope the question isn't so silly. $\endgroup$
    – Jose Capco
    Nov 7, 2009 at 17:31
  • $\begingroup$ I'll edit in that part of the argument - I was a bit slack saying how this works one really uses the fact that S over R is actually faithfully flat not just flat $\endgroup$ Nov 7, 2009 at 22:23

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