# Quotient ring of the zero-divisor ideal is a flat module

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.

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)