# When commutative ring hom $A\to A/I$ flat?

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?

• If you're familiar with the algebraic geometry language you might be interested in stacks.math.columbia.edu/tag/04PV . – Captain Lama May 22 '20 at 14:25
• Thanks. (However I'm in the beginning of learning the scheme theory.) – nessy May 23 '20 at 13:57
• 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? – Geoffrey Trang May 23 '20 at 14:09

## 3 Answers

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,

• Is it true that flatness implies projectiveness (in this context)? – nessy May 23 '20 at 13:06
• 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? – nessy May 23 '20 at 13:32
• There are non-noetherian commutative rings with fp flat modules that are not projective. See here. – 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.