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Let $R$ be a commutative ring, $T$ a variable, $f(T) \in R[T]$, $I=(f)$, $A = R[T]/I$ and $J$ the ideal of $R$ generated by the coefficients of $f$.

A special case of Nagata's flatness theorem (in his notations, $X=\{T\}$) says that $A$ is $R$-flat if and only if $J$ is a direct summand of $R$.

What happens if $I$ is non-principal? Is there a 'nice' result in this case?

For example, $R=\mathbb{C}[x^2,x^3]$, $A=\mathbb{C}[x]=R[T]/\tilde{I}$, $\tilde{I}$ is not principal, $\tilde{J}=R$, but $A$ is not $R$ flat; see the answer to this question ($\tilde{I}$ is generated by three polynomials).

Remark: I have read an article by one of Nagata's students concerning faithful flatness, in the same spirit of the above Nagata's flatness theorem, but unfortunately I am not able to find it now. Perhaps some ideas there are relevant for my question?

Thank you very much!

Edit: This is the article I referred to in the above remark.

user237522
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  • Non-principal ideals in this situation are less interesting. For example, if $R$ is a UFD and $R[T]/I$ is flat over $R$, $I$ a proper ideal, then $I$ must be principal. – Mohan Jun 12 '17 at 15:57
  • Thanks. Can you please explain your example? – user237522 Jun 12 '17 at 17:04
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    Let $A=R[T]$, then $A$ is also a UFD, since $R$ is. So, any ideal $I=Jf$ where $J$ has height at least two. We wish to show that if $A/I$ is $R$-flat, then $J=A$. If not, you have an exact sequence, $0\to A/J\stackrel{f}{\to} A/I\to A/fA\to 0$. But, height of $J$ is at least two implies $J\cap R\neq 0$. Then, for any $0\neq a\in J\cap R$, we get $A/I\stackrel{a}{\to} A/I$ is non-injective, contradicting flatness. Thus $J=A$ and then $I=fA$, a principal ideal. – Mohan Jun 12 '17 at 22:06
  • Truly, I had once downloaded that article of Ohm and Rush (and additional articles in that subject https://projecteuclid.org/download/pdf_1/euclid.bams/1183533044 https://www.jstor.org/stable/2038427?seq=1#page_scan_tab_contents) but I have forgotten about it...Thank you very much! – user237522 Jun 12 '17 at 23:00
  • @Mohan, please, why any ideal $I$ of $A$ is of the form $I=Jf$, where $J$ is an ideal of $A$ of height at least two and $f \in A$? I have now asked this as a new question: https://math.stackexchange.com/questions/3537316/flatness-of-fracrti-over-r-implies-that-i-is-principal-or-i-rt – user237522 Feb 06 '20 at 23:43

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