Let $A$ be a ring, and let $M$ be a right $A$-module. Then $M$ is flat if and only if for each left $A$-module $N$ we have that $Tor^1_A(M,N) = 0$. Becasuse $Tor$ commutes with filtered direct limits, and since every left module is such a limit of finitely presented $A$-modules, $M$ is flat if and only if $Tor^1_A(M,N) = 0$ for any finitely presented right $A$-module $N$.

Recall that $N$ is called an $FP_3$-module if there is an exact sequence

$F_2 \to F_1 \to F_0 \to N \to 0$

with $F_0,F_1,F_2$ finitely generated free $A$-modules. My question is: if we know that $Tor^1_A(M,N) = 0$ for all $FP_3$-modules $N$, can we also conclude that $M$ is flat?

  • $\begingroup$ Not that it matters, but I think $FP_2$ is more standard terminology ($FP_0$ is finitely generated, $FP_1$ is finitely presented, ...). $\endgroup$ Jun 8, 2015 at 18:27
  • $\begingroup$ By the way, this is probably a silly question (it only occurred to me five minutes ago, so I haven't thought very deeply about it), but must a ring that's not von Neumann regular have any non-projective $FP_2$ modules? $\endgroup$ Jun 8, 2015 at 18:34

1 Answer 1


Answering my own question in comments:

Let $k$ be a field, and $A=k\oplus V$, where $V$ is an infinite-dimensional square-zero ideal. Then I think $A$ has no non-projective $FP_3$-modules (using your terminology), as the kernel of any map between finitely generated free modules, that's not a surjection onto a direct summand, has infinitely generated kernel.

So every $A$-module $M$ satisfies your condition, but there are non-flat modules.

  • $\begingroup$ Thank you Prof. Rickard. Your understanding of homological algebra is so deep! $\endgroup$
    – FlatModule
    Jun 9, 2015 at 8:10

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