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If $R / I$ is flat then $I = I^{2}$ , and for finitely generated $I$ this is equivalent to generation by an idempotent

Statement

Let $R$ be a commutative ring and $I \subseteq R$ an ideal.

If $R / I$ is flat as an $R$ -module, then $I = I^{2}$ .
If $I = R e$ for an idempotent $e^{2} = e$ , then $R / I$ is flat.
If $I$ is finitely generated, then the first two clauses combine to the usual criterion: $R / I$ is flat if and only if $I$ is generated by an idempotent.

Facts & Assumptions

Given: A commutative ring $R$ and an ideal $I \subseteq R$ .

[L1]

Flatness is equivalent to injectivity of $J \otimes_{R} M \to M$ for every ideal $J$ (Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests).

[L2]

For every $R$ -module $M$ , there is a natural isomorphism $M \otimes_{R} (R / I) ≅ M / I M$ ( $M \otimes_{R} R / I ≅ M / I M$ naturally).

[L3]

A direct summand of a flat module is flat (Direct sums and direct summands of flat modules are flat).

[L4]

If a finite module $N$ satisfies $I N = N$ , then $(1 - e) N = 0$ for some $e \in I$ (Determinant trick for Nakayama).

Proof

technique · direct

1.1

Assume $R / I$ is flat. Apply [L1] to the ideal $I$ and the module $R / I$ . The multiplication map $I \otimes_{R} R / I \to R / I$ is injective, but it is also the zero map because every $a \in I$ acts trivially on $R / I$ . Hence $I \otimes_{R} R / I = 0.$ By [L2], this tensor product is $I / I^{2}$ , so $I = I^{2}$ .

L1L2givenalgebra

1.2

If $I = R e$ with $e^{2} = e$ , then $R ≅ R e \oplus R (1 - e),$ and the quotient $R / R e$ identifies with the direct summand $R (1 - e)$ . Since $R$ is free, hence flat, [L3] shows that $R (1 - e)$ is flat. Thus $R / I$ is flat.

L3givenalgebra

2.1

Now assume $I$ is finitely generated and $R / I$ is flat. Step 1.1 gives $I = I^{2}$ , so [L4] applied to the finite module $I$ gives $e \in I$ with $(1 - e) I = 0$ . Thus every $x \in I$ satisfies $x = e x$ , whence $I \subseteq R e$ ; the reverse inclusion follows from $e \in I$ . Moreover $(1 - e) e = 0$ , so $e^{2} = e$ . Therefore $I = R e$ is generated by an idempotent.

L4step 1.1algebra

3.1

Step 1.1 proves clause 1, step 1.2 proves clause 2 and the reverse implication in clause 3, and step 2.1 proves the forward implication in clause 3.

step 1.1step 1.2step 2.1∎

Depends on

Used by

A quotient by an idempotent ideal is flat Example
The quotient by a nonidempotent ideal is not flat Example

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, Exercise (9.9) (standard reference, not scraped)
Stacks Project, Section 10.39: Flat modules and flat ring maps (standard reference, not scraped)

If R/IR/I is flat then I=I2I=I2, and for finitely generated II this is equivalent to generation by an idempotent