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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.