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Let $A$ be a commutative unitary ring. In the D. G. Northcot's Multilinear Algebra it is claimed that if a proper ideal $I$ of $A$ contains a zero non-divisor element, say $a$, then the $A$-module $\frac{A}{I}$ is non-flat. Indeed, consider the short exact sequence $$ 0 \xrightarrow{} A \xrightarrow{\cdot a} A \xrightarrow{\pi_{(a)}} \frac{A}{(a)} \xrightarrow{} 0, $$ where leftmost nontrivial morphism is defined to be a multiplication by $a$, which must be injective as $a$ is not a zero-divisor and has exactly the ideal $(a)$ as its image. Tensoring with $\frac{A}{I}$ will nullify the left morphism, but the module $A \otimes \frac{A}{I} \cong \frac{A}{I} \neq 0$, as $I$ is proper. So, the new sequence can't be exact, and $\frac{A}{I}$ can't be flat.

I am wondering if the contrary is true. So, what I'm looking for is

an example of the ideal $I$ such that any $a \in I$ is a zero divisor but the quotient $\frac{A}{I}$ is not flat.

Personally, I don't think that the contrary is true, because determining flatness of the quotient seems to be a far more delicate question. For example, see this discussion on mathoverflow: https://mathoverflow.net/questions/208/can-a-quotient-ring-r-j-ever-be-flat-over-r

On the other hand, I don't know there to start looking for examples. Thanks in advance.

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Let $A$ be a ring and $\epsilon \in A$ such that $\epsilon^2=0$ and moreover, $\epsilon x= 0 \implies x\in (\epsilon)$. I don't know if this works in general, so I'll take specifically $A=\mathbb{Z/4}, \epsilon = 2$ or $A=k[\epsilon] = k[X]/(X^2)$ where $k$ is an integral domain.

Then, with $I=(\epsilon)$ we have a short exact sequence $0\to I\to A\to I\to 0$ where the first map is inclusion and the second one is multiplication by $\epsilon$.

Tensoring with $A/I$ yields $0\to I\otimes A/I \to A/I\to I\otimes A/I\to 0$.

In the first case, with $\mathbb{Z/4}$, this is $0\to\mathbb{Z/2\to Z/2 \to Z/2\to 0}$, which can't be exact.

In the second case, with $k [\epsilon]$, this is $0\to k\to k\to k\to 0$ which can't be exact (if it were, you could tensor it with the field of fractions of $k$ for instance, and it couldn't be exact then)

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  • $\begingroup$ I was afraid that the answer would be to convoluted but this answer is elementary and understandable. Thanks! $\endgroup$ – Nik Pronko May 27 at 18:17

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