# When every module is a scalar extension?

Let $A \subseteq B$ be commutative noetherian domains. Of course, if $M$ is an $A$-module, then $M \otimes_A B$ is a $B$-module. I am curious to know if there exist additional conditions on $A$ and $B$, such that every $B$-module $N$ is necessarily of the form $M \otimes_A B$ for some $A$-module $M$.

I do not mind to assume one or more of the following additional conditions: $A$ is a UFD (but I do not want to assume that $B$ is a UFD). $A$ is regular. $B$ is a complete intersection ring (but I do not want to assume that $B$ is regular). $B$ is a faithfully flat $A$-module. $B$ is a free $A$-module.

I once ran into a paper (unfortunately I cannot find it now) which calls such $N$ extendable (maybe that paper answers my curiosity?).

• A module of the form $M\otimes B$ is sometimes called "relative free" (and a direct summand of such a module is called "relative projective"). You're asking for coniditions when every module is relative free. If $A$ is a field, then this means that every $B$-module is free which is more or less equivalent to being a principle ideal domain. – Johannes Hahn May 10 '15 at 16:58
• @JohannesHahn: what do you mean by "more or less equivalent"? If $B$ is a nonzero ring that is not a field and $I$ is a maximal ideal of $B$ then $B/I$ is a $B$-module that is not free. – grghxy May 10 '15 at 17:04
• Thank you both. Please can you recommend books/papers dealing with relative free modules? – user237522 May 10 '15 at 18:08
• Hochschild, Relative homological algebra, Trans. Amer. Math. Soc. 1956. – Vinteuil May 11 '15 at 8:55

An example: if $B$ is the henselization of a local ring $A$, then for any finite type $B$-module $N$ there exists a finite type $A$-module $M$ such that $N$ is a direct summand of $M\otimes_AB$. You can find this result and some similar others in http://arxiv.org/pdf/0707.4197v3.pdf