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Let us call a ring homomorphism $f\colon R\rightarrow S$ von Neuman regular if it has the property that for every left $S$-module $M$, the left $R$-module $f^*M$ is flat.

In particular, $\mathrm{id}_R$ is von Neumann regular if and only if every left $R$-module is flat, i. e. $R$ is von Neumann regular in the usual sense.

Question: Has this notion been studied? Does anyone know a reference?

The rings here are associative, unital and not necessarily commutative.

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  • $\begingroup$ Out of curiosity, does the flatness of $f^\ast M$ imply anything about the flatness of $M$? I have no mastery of flatness, especially under this transfer. $\endgroup$
    – rschwieb
    Oct 11, 2018 at 15:11
  • $\begingroup$ @rschwieb: It means that tensoring with $M$ preserves exactness for exact sequences which are induced up from $R$. But not any exact sequence is of this form. I don't know if anything stronger is true. $\endgroup$
    – nikola karabatic
    Oct 12, 2018 at 8:18
  • $\begingroup$ Flatness of $f^*M$ does not tell you anything about $M$ in general, as can be seen by taking $R$ to be absolutely flat (e.g., a field). $\endgroup$
    – Fred Rohrer
    Oct 12, 2018 at 10:56

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