Question

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 aR[a] ∩ R ≠ 0

Answer 1

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 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 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

where the 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 so that it is faithfully flat over R. It now follows that S=R by the following standard argument.

Suppose is in S, where x,y are in R. Then x is in yS and yS 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 R = yR: since S is faithfully flat over R we get by changing base a faithfully flat map for any ideal I

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

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.