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made notation consistent with the question
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Jeremy Rickard
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Answering my own question in comments:

Let $k$ be a field, and $R=k\oplus V$$A=k\oplus V$, where $V$ is an infinite-dimensional square-zero ideal. Then I think $R$$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 module$A$-module $M$ satisfies your condition, but there are non-flat modules.

Answering my own question in comments:

Let $k$ be a field, and $R=k\oplus V$, where $V$ is an infinite-dimensional square-zero ideal. Then I think $R$ 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 module $M$ satisfies your condition, but there are non-flat 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.

Source Link
Jeremy Rickard
  • 35.2k
  • 2
  • 110
  • 151

Answering my own question in comments:

Let $k$ be a field, and $R=k\oplus V$, where $V$ is an infinite-dimensional square-zero ideal. Then I think $R$ 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 module $M$ satisfies your condition, but there are non-flat modules.