Is there any conditions for an ideal $I$ that assures the canonical map $A\to A/I$ is flat?

Here's my try:

(1)It's obvious when $I=(0)$ or $I=A$.

(2)Since $I\otimes_A A/I=0$, it can't be faithfully flat unless $I=0$.

(3)If $I$ contains non zero divisor $a\in I$, then multiplying $a$ is injective as a map $\lambda_a \colon A\to A$. Tensoring flat $A$-Module $A/I$, $\lambda_a\otimes 1 \colon A/I \to A/I$ (which is a zero map) must be injective and $I=A$.

(4)If $I=rad(A)$, the nilradical of $A$, then $A\to A/I=A_{rad}$ is flat iff $I=0$. Let $A\to A/I$ be flat. You can take $\mathfrak{p}\in SpecA$ if $A$ is nonzero. $A_\mathfrak{p} \to A/I\otimes_A A_{\mathfrak{p}} = {A_{\mathfrak{p}}}_{rad}$ is flat, and is local ring hom, so it is faithfully flat, then injective. It is also surjective and bijective, so $rad(A_\mathfrak{p})=0.$ Then $rad(A)=I=0$.

Can somebody help me?

  • $\begingroup$ If you're familiar with the algebraic geometry language you might be interested in stacks.math.columbia.edu/tag/04PV . $\endgroup$ – Captain Lama May 22 '20 at 14:25
  • $\begingroup$ Thanks. (However I'm in the beginning of learning the scheme theory.) $\endgroup$ – nessy May 23 '20 at 13:57
  • $\begingroup$ For (2), $I \otimes_{A} A/I=0$ if and only if $I$ is idempotent. Why must an ideal with a flat quotient be idempotent? $\endgroup$ – Geoffrey Trang May 23 '20 at 14:09

Suppose that $R/I$ is a flat $R$-module. Then, I claim that $R \to R/I$ must be a localization.

Indeed, define $S$ to be $\{s \in R\mid \exists t \in R \, (st-1) \in I\}$. Then, clearly, there is an induced surjective ring homomorphism $\varphi:S^{-1}R \to R/I$.

To show that $\varphi:S^{-1}R \to R/I$ is also injective, suppose that $\varphi(\frac{a}{s})=0$. This means that for some $t \in R$, $(st-1) \in I$ and $at \in I$. Now, showing that $a$ must annihilate at least one element of $S$ is where we need to use flatness.

Since $R/I$ is a flat $R$-module, for any ideal $J$ of $R$, $I \cap J=IJ$. In particular, $at \in I \cap aR$, hence $at \in I(aR)=aI$. This means that $at=ai$ for some $i \in I$. Since $(t-i)s-1=(st-1)-is \in I$, $(t-i) \in S$. Also, $a(t-i)=at-ai=0$, so $a$ annihilates at least one element of $S$. This means that $\frac{a}{s}=0 \in S^{-1}R$. Hence, $\varphi$ is also injective, and so it is an isomorphism.

Conversely, any localization of $R$ is a flat $R$-module. Hence, a quotient of $R$ is a flat $R$-module if and only if it is a localization of $R$.


Since $R/I$ is finitely generated over $R$, it's flat if and only if it's projective. We can then use this answer from Mathoverflow to deduce that $I$ is principal generated by an idempotent.

Edit: This requires an assumption that $R$ is noetherian, or that $R/I$ is finitely presented,

  • $\begingroup$ Is it true that flatness implies projectiveness (in this context)? $\endgroup$ – nessy May 23 '20 at 13:06
  • $\begingroup$ If $R$ is Noetherian (or more generally, coherent) ring, it can be proved that finitely generated flat module is projective (since finitely presented flat implies projective). However does it hold in general? $\endgroup$ – nessy May 23 '20 at 13:32
  • $\begingroup$ There are non-noetherian commutative rings with fp flat modules that are not projective. See here. $\endgroup$ – Zeek May 23 '20 at 14:01

I found a nice page: https://stacks.math.columbia.edu/tag/04PQ

It seems the perfect classification of "pure ideals"(= ideals which makes $R\to R/I$ flat) is unknown.


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