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(This is inspired by this question and asked out of pure curiosity.)

Let $R$ be a commutative ring. Let $M$ and $N$ be $R$-modules. Then, $N$ is called $M$-flat if whenever $M'\rightarrow M$ is a monomorphism of $R$-modules, then the induced morphism $M'\otimes_RN\rightarrow M\otimes_RN$ is a monomorphism, too. One can show that an $R$-module $N$ is flat (i.e., $M$-flat for every $M$) if and only if it is $R$-flat.

Let's call an $R$-module $M$ self-flat if it is $M$-flat, i.e. if whenever $M'\rightarrow M$ is a monomorphism of $R$-modules, then the induced morphism $M'\otimes_RM\rightarrow M\otimes_RM$ is a monomorphism, too.

In the aforementioned question examples were given that show that not every module is self-flat. On the other hand, it is readily checked that flat modules, injective modules, monogeneous modules, and semisimple modules are self-flat.

What are other classes of self-flat modules?

(A quick search among the usual suspects (Lazard's Autour de la platitude, Bourbaki's AC, Lam's Lectures on rings and modules, etc.) did not turn up anything about this notion.)

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    $\begingroup$ This sounds like a fun one Fred. First toy case might be: what about $M=(x,y)$, with $R=k[[x,y]]$? $\endgroup$ Nov 23, 2018 at 17:56
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    $\begingroup$ The first paragraph of the question is wrong as now written. To correct it, the last two sentences in the first paragraph should read: "Then, $N$ is called $M$-flat if whenever $M'\to M$ is a monomorphism of $R$-modules, then the induced morphism $M'\otimes_RN\to M\otimes_RN$ is a monomorphism, too. One can show that an $R$-module $N$ is flat (i.e., $M$-flat for every $M$) if and only if it is $R$-flat." $\endgroup$ Nov 23, 2018 at 20:03
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    $\begingroup$ (With the above correction in mind.) If $M'$ is a submodule in $M$ and $N$ is $M$-flat, then $N$ is also $M'$-flat. It follows that if $R$ is a submodule in $M$ and $N$ is $M$-flat, then $N$ is flat. In particular, as $k[[x,y]]\cong xk[[x,y]]\subset (x,y)\subset k[[x,y]]$, the $k[[x,y]]$-module $M=(x,y)$ is not self-flat (since it is not flat). $\endgroup$ Nov 23, 2018 at 20:08
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    $\begingroup$ Leonid: nice observation! $\endgroup$ Nov 23, 2018 at 20:50
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    $\begingroup$ Obviously, one can replace $R$ by $R/ann(M)$, so we can assume $M$ is faithful. Also from that observation, flat modules over quotients of $R$ are self flat. What about the converse? Leonid's comment shows that if $R$ embeds into $M$, $M$ must be flat, so it is almost a converse. $\endgroup$ Nov 23, 2018 at 21:22

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