ac.commutative algebra - Can a quotient ring R/J ever be flat over R?

1) If A is arbitrary and I is an ideal of finite type such that A/*I* is a flat A-module, then V(I) is open and closed. In fact, A/*I* is a finitely presented *A*-algebra and thus Spec(A/I)->Spec(A) is a flat monomorphism of finite presentation, hence an étale monomorphism, i.e., an open immersion (cf. EGA IV 17.9.1).

2) If A is a noetherian ring then A/*I* is flat if and only if V(I) is open and closed (every ideal is of finite type).

3) If A is not noetherian but has a finite number of minimal prime ideals (i.e., the spectrum has a finite number of irreducible components), then it still holds that A/*I* is flat iff Spec(A/I)->Spec(A) is open and closed. Indeed, there is a result due to Lazard [Laz, Cor. 5.9] which states that the flatness of A/*I* implies that I is of finite type in this case.

4) If A has an infinite number of minimal prime ideals, then it can happen that a flat closed immersion is not open. For example, let A be an absolutely flat ring with an infinite number of points (e.g. let A be the product of an infinite number of fields). Then A is zero-dimensional and every local ring is a field. However, there are non-open points (otherwise Spec(A) would be discrete and hence not quasi-compact). The inclusion of any such non-open point is a closed non-open immersion which is flat.

The example in 4) is totally disconnected, but there is also a connected example:

5) There exists a connected affine scheme Spec(A), with an infinite number of irreducible components, and an ideal I such that A/*I* is flat but V(I) is not open. This follows from [Laz, 7.2 and 5.4].

[Laz] Disconnexités des spectres d'anneaux et des préschémas (Bull SMF 95, 1967)

Edit: Corrected proof of 1). An open closed immersion is not necessarily an open immersion! (e.g. X_red->X is a closed immersion which is open but not an open immersion.)

Edit: Raynaud-Gruson only shows that flat+finite type => finite presentation when the spectrum has a finite number of associated points. Lazard proves that it is enough that the spectrum has a finite number of irreducible components. Added example 5).