Let $R$ be a ring (not necessarily commutative).

Let $A$ be a left $R$-module. When does the functor $\text{Hom}(A,-)$ preserve direct sums - in the category of left $R$-modules? For example, this is certainly true when $A$ is free and finitely generated (EDIT: or finitely generated in general, as suggested by Pierre-Yves).

Do we always need the finitely generated condition?

  • 1
    $\begingroup$ Dear Evariste: It seems clear to me that direct sums are preserved if $A$ is finitely generated. I'd be tempted to say that the converse holds. I'll think about it. Many users know this kind of things much better than I, and I hope (for you and for everybody) they they'll answer your question. $\endgroup$ Nov 2, 2011 at 12:24
  • $\begingroup$ Answered here: mathoverflow.net/questions/59282/… $\endgroup$ Dec 19, 2011 at 9:14

1 Answer 1


There have been two previous incorrect versions of the answer. I apologize for them, and thank Mariano for his friendly and efficient help. Thank you also to Evariste!

Here are the claims:

(1) If $A$ is finitely generated, then the functor $\text{Hom}_R(A,?)$ preserves direct sums.

(2) If there is an increasing sequence $(A_i)_{i\ge1}$ of proper submodules whose union is $A$, then the functor $\text{Hom}_R(A,?)$ does not preserve all the direct sums.

I don't know if such a sequence $(A_i)_{i\ge1}$ exists whenever $A$ is not finitely generated. (And I'm very anxious to know if this is true.)

Proof of (1). We have a natural $\mathbb Z$-linear injection
$$ F:\bigoplus_i\ \text{Hom}_R(A,B_i)\to\text{Hom}_R(A,\oplus_i\ B_i). $$ Moreover $g\in\text{Hom}_R(A,\oplus_i\ B_i)$ is in the image of $F$ if and only if $g(A)$ is contained into the sum of finitely many $B_i$.

In particular $F$ is bijective if $A$ is finitely generated.

Proof of (2). The natural $R$-linear map from $A$ into $\oplus\,A/A_i$ is not in the image of $F$.

EDIT. Here are three references:


Preservation of direct sums and finite generation


  • $\begingroup$ Ok, you convinced me. Thanks! $\endgroup$
    – Evariste
    Nov 2, 2011 at 13:19
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    $\begingroup$ Dear Evariste: Thanks a lot! Unfortunately, I fear that there a mistake in my post. More precisely, I think: (1) The part from "Assume $A$ is not finitely generated" is incorrect. (2) The main statement is correct. While trying to fix (1), I'm leaving the post as is for the moment. I'm really sorry... $\endgroup$ Nov 2, 2011 at 13:26
  • $\begingroup$ Hm... The most important thing for me was the question whether finitely generated is sufficient, as I really didn't know what the answer should be. So that's fine. I'll think again about what happens for non-finitely generated modules. $\endgroup$
    – Evariste
    Nov 2, 2011 at 13:33
  • $\begingroup$ Dear @Evariste: I hope it's correct now. Thanks again for having asked such an interesting question and for being so positive! $\endgroup$ Nov 2, 2011 at 13:57
  • 1
    $\begingroup$ Dear @Peter: $F$ is $\mathbb Z$-linear, and "finitely generated" means "finitely generated over $R$". $\endgroup$ Jun 6, 2012 at 8:54

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