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In Hilton's A Course in Homological Algebra Page 21. $A$ is a ring with unit which is not necessarily commutative. Here is Propsition 3.4:

Prop3.4: Let $B$ be an $A$-module and $\{A_{j}\}_{j\in J}$ be a family of $A$-modules. Then there is an isomorphism $$ \eta: \operatorname{Hom}_{A}(\oplus_{j\in J}A_{j},B)\rightarrow\prod_{j\in J}\operatorname{Hom}_{A}(A_{j},B). $$

I am just confused what the isomorphism actually means, since we can only prove it's a group isomorphism and both may not have a module structure. Is the author's writings are not quite rigorous or the abbreviation is reasonable?

user1729
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richer_Ge
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  • Why "may not have a module structure"? – Hagen von Eitzen May 25 '21 at 15:03
  • Suppose $A_{i},A_{j}$ denotes A-module. Because the ring A may not be commutative, we can't give $Hom_{A}(A_{i},A_{j})$ a standard A-module structure $\lambda\psi(a):=\psi(\lambda(a))$ just directly checking the module conditions. – richer_Ge May 25 '21 at 15:10
  • To be precise, it should be an isomorphism of $\mathbb Z$-modules. You're right! – azif00 May 25 '21 at 15:12
  • If either $B$ or all the $A_i$s are bimodules, then the $\operatorname{Hom}_{\mathfrak{M}_A}$ (or in ${}_A\mathfrak{M}$) have a natural $A$-module structure and you have an isomorphism of $A$-modules. Otherwise, you only have a $\mathbb{Z}$-module isomorphism. – user10354138 May 25 '21 at 15:12
  • I think I can understand it now! Thank you. – richer_Ge May 25 '21 at 15:14
  • isomorphism of abelian groups may be? – Jackozee Hakkiuz May 25 '21 at 15:14

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