4
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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 $R$.

$\endgroup$
2
$\begingroup$

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,

$\endgroup$
3
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.