Let $R$ be a ring, and $S$ a subring of $R$. Let $M$ be a right $R$-module, and $N$ a right $S$-submodule of $M$. If $N$ is flat (or faithfully flat) as a right $S$-module, does it then follow that the multiplication map $$ N \otimes_S R \to NR \subseteq M $$ is an isomorphism? Is flatness necessary for this to happen?
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$\begingroup$ When $N=R$ and $S=Z$, (and either if we forget the fact that $S$ is a subring of $R$ or we restrict ourselves to characteristic zero case so that $Z$ is automatically included in $R$) the condition you are looking for is that $R$ is solid (c.f. ncatlab.org/nlab/show/core+of+a+ring ) $\endgroup$– user43326Commented Aug 14, 2022 at 20:33
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No it need not be an isomorphism. Something more than flatness is required. For example, taking $M$ and $N$ to be copies of the ring $R$ you are asking for the multiplication $R\otimes_SR\to R$ to be an isomorphism. If $S$ is a field and $R$ is an associative $S$-algebra then this fails unless $R=S$.
