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?



Can you test flatness on $FP_3$-modules?
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.


