Is $R/I$ flat over $R$ when $I$ is the nilradical of $R$ and $I=I^2$?

Question

This is a flowup of this question. To be precise, let $R$ is a commutative unitary ring with nilradical $I$. In the above URL, it is proved that if $R/I$ is a flat $R$-module, then $I=I^2$.

My question is whether the converse holds in general. Thanks !

If $I$ is finitely generated, yes, because by Nakayama $I=0$. — Martin Brandenburg, Sep 09 '13 at 14:34
One can also work in the case $I$ is finitely generated but not necessarily $\sqrt R$. Assuming $I^2=I$, use the determinant trick to prove that $I=(e)$ for an idempotent $e$, and then it easily follows that $R/I$ is flat. — Karl Kroningfeld, Sep 20 '13 at 02:27

score 4 · Accepted Answer · edited Apr 13 '17 at 12:58

Not in general, no.

Let $R$ be a zero dimensional local ring, and suppose that its maximal ideal $I$ satisfies $I^2 = I$. Then $R/I$ is flat over $R$ only if $I = 0$. (See this MO answer.)

E.g. if $S$ is a one-dimensional local domain with maximal ideal $\mathfrak m $ such that $\mathfrak m^2 = \mathfrak m$ and $x \in \mathfrak m$ is non-zero, then $R := S/(x)$ is a zero-dimensional local ring whose maximal ideal $I$ satifies $I^2 = I$.

E.g. Let $S = \overline{\mathbb Z}_p$ denote the integral closure of $\mathbb Z_p$ in $\overline{\mathbb Q}_p$, and take $s = p$. Then $R = \overline{\mathbb Z}_p/(p)$ gives a concrete example.

Is $R/I$ flat over $R$ when $I$ is the nilradical of $R$ and $I=I^2$?

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