Is a finite flat module which is free modulo radical free?

Question

Let $R$ be a commutative ring and $M$ is an $R$-module. The following statement is well-known.

If $M$ is finitely presented flat module, $M/J(R)M$ is free over $R/J(R)$ then $M$ is free.

Here $J(R)$ is the (Jacobson) radical of $R$. Is the statement true for finite $M$? I expect that it is not, but was not able to construct a counter-examples. Clearly, in a counter-example the ring $R$ is not Noetherian. Also, local $R$ won't work, since over a local ring any finite flat module is free.

Answer 1

Set $I := J(R)$. Suppose $M/IM$ has rank $n$ as a free $A/I$-module; let $x_{1},\dotsc,x_{n} \in M$ be elements whose images in $M/IM$ form a basis for it as a free $A/I$-module, let $N \subseteq M$ be the $A$-submodule generated by the $x_{i}$. Then $M = N$ by the usual Nakayama argument ($M/IM = (N+IM)/IM$ implies $M = N+IM$ implies $M/N = (N+IM)/N = I(M/N)$ so $M/N = 0$) so we have a surjection $\varphi : A^{\oplus n} \to M$. For all maximal ideals $\mathfrak{m}$ of $A$, the localization $M_{\mathfrak{m}}$ is a free $A_{\mathfrak{m}}$-module (by e.g. the local case) of rank $n$ (since $(M/I)_{\mathfrak{m}}$ is a free $(A/I)_{\mathfrak{m}}$-module of rank $n$), so the localization $\varphi_{\mathfrak{m}}$ is an isomorphism (the "determinant trick", see Corollary (10.4) here), hence $\varphi$ is an isomorphism.

I guess more-or-less equivalently you could also use a limit argument to say that $M$ is locally finitely presented, then use e.g. Tag 00EO to conclude that $M$ is (globally) finitely presented.

Answer 2

The result can be extended to the setting of non-commutative rings:

Prop: Let M be a finitely generated flat right module over a ring R, and let P be a projectivemodule. If M/MJ(R)is isomorphic to P/PJ(R), then M is isomorphic to P

For the proof you can check the Proposition 7.3 in the paper FLAT MODULES AND LIFTING OF FINITELY GENERATED PROJECTIVE MODULES, by Fachhini, Herbera and Skhaev (Pac. J. Math. Vol. 220, No. 1, 2005, 49-67)