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?
Best Answer
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 $R$.