Commutative Algebra – Can a Quotient Ring R/J Be Flat Over R?

ac.commutative-algebraag.algebraic-geometryflatnessnoncommutative-algebra

If R is a ring and J⊂R is an ideal, can R/J ever be a flat R-module? For algebraic geometers, the question is "can a closed immersion ever be flat?"

The answer is yes: take J=0. For a less trivial example, take R=R1⊕R2 and J=R1, then R/J is flat over R. Geometrically, this is the inclusion of a connected component, which is kind of cheating. If I add the hypotheses that R has no idempotents (i.e. Spec(R) is connected) and J≠0, can R/J ever be flat over R?

I think the answer is no, but I don't know how to prove it. Here's a failed attempt. Consider the exact sequence 0→J→R→R/J→0. When you tensor with R/J, you get

0→ J/J2→R/J→R/J→0

where the map R/J→R/J is the identity map. If J≠J2, this sequence is not exact, contradicting flatness of R/J.

But sometimes it happens that J=J2, like the case of the maximal ideal of the ring k[tq| q∈Q>0]. I can show that the quotient is not flat in that case (see this answer), but I had to do something clever.

I usually think about commutative rings, but if you have a non-commutative example, I'd love to see it.

Best Answer

  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 $\operatorname{Spec}(A/I) \to \operatorname{Spec}(A)$ 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 $\operatorname{Spec}(A/I) \to \operatorname{Spec}(A)$ 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 $\operatorname{Spec}(A)$ 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:

  1. There exists a connected affine scheme $\operatorname{Spec}(A)$, 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. $X_{red} \to X$ 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).

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