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Let $\phi: A \rightarrow B$ be a ring homomorphism. Here $A$ and $B$ are commutative rings. Suppose we have an $A$-module $M$, and we hope to construct a $B$-module. A traditional way is via base change, namely construct the module $M_B:= B \otimes_A M$. Then this is a $B$-module.

Yet in the posts

MO 300531: https://mathoverflow.net/questions/300531/adjoints-of-scalar-extension-and-scalar-coextension?rq=1

and

MSE 2780723: Induced representation: which adjoint is it?

People also constructed a $B$-module $M^B := \text{Hom}_A(B, M)$ and call this a scalar coextension (as in the math.SE post quoted above answered by @Qiaochu Yuan). My question is: What is the $B$-module structure on $M^B$? It is indeed an $A$-module, but I looked up many posts and books but have not found any definition of it. So could someone provide me with some references or give me the definition of it?

My guess: Let $f \in M^B := \text{Hom}_A(B, M)$ and $b \in B$, then the $A$-module homomorphism $b \cdot f$ is defined by sending any element $r \in B$ to $f(br)$. Is this definition correct? It seems that there should be a correct definition, since people usually regard (or define) this as the right adjoint to the functor of the restriction of scalars, i.e. view a $B$-module $N$ as an $A$-module via $\phi$, by defining $a \cdot x := \phi(a) \cdot x$ for any $a \in A$ and $x \in N$.

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    $\begingroup$ Your guess is correct. $\endgroup$
    – Hanno
    Sep 16, 2020 at 13:00
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    $\begingroup$ Coextension of scalars makes sense with no commutativity assumptions and then your guess is not correct; you should multiply on the right if you want a left module. $\endgroup$ Sep 16, 2020 at 16:30
  • $\begingroup$ @QiaochuYuan Thank you so much! $\endgroup$
    – Hetong Xu
    Sep 18, 2020 at 2:40
  • $\begingroup$ @Hanno Thank you for your comments! $\endgroup$
    – Hetong Xu
    Sep 18, 2020 at 2:40

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