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.

  • $\begingroup$ By a ‘finite flat module’, I guess you mean a ‘finitely generated flat module’? $\endgroup$
    – Bernard
    Feb 19 '21 at 10:53
  • $\begingroup$ Yes, finite module is a shorter way to say finitely generated module. $\endgroup$
    – Alex
    Feb 19 '21 at 11:19

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.

  • $\begingroup$ How exactly determinant trick works in this proof? How do you know that ranks of $M_m$ are all the same? $\endgroup$
    – Alex
    Feb 20 '21 at 10:29
  • $\begingroup$ Hi, I added some more details. $\endgroup$ Feb 20 '21 at 18:21
  • $\begingroup$ So, in principle $M_m = R_m^{n(m)}$, where $n: \operatorname{Specm} \to \mathbb{N}$ is a function, but $(M/IM)_m=M_m/IM_m = R_m/IR_m \otimes_{R_m} M_m =(R_m/IR_m)^{n(m)} = (R/I)_m^n$. So $n(m)=n$, since $IR_m \neq R_m$ for any $m$. That is how I understood your proof. $\endgroup$
    – Alex
    Feb 20 '21 at 19:07
  • $\begingroup$ Yes, that was my understanding as well. $\endgroup$ Feb 20 '21 at 19:13

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