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It is easy to see the functor $\mathrm{Hom}(A,-)$ commutes with every arbitrary direct sum (i.e. $\mathrm{Hom} (A,\oplus_{i\in I} N_i)=\oplus_{i\in I}\mathrm{Hom}(A,N_i)$) for finitely generated module $A$ since the image of $A$ is contained in finitely many $N_i$. The converse is false, as illustrated in this answer. But I think if we impose projectivity the converse is indeed true. I would appreciate if someone could give me a hint.

Edit: Thanks for the reference provided in the comment (can be found here). However, after I checked it I can't find any assertions related to projective modules. The closest I can found is that if the base ring is Noetherian, then every module satisfies this property will be finitely generated. I am sorry if I missed something since I am not familiar with French.

Viktor Vaughn
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metaverse
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    In this question, someone provides a reference showing that, whenever $M$ projective, $\text{Hom}(M,-)$ preserves infinite direct sums iff $M$ is finitely generated. This seems to say the answer to your question is "no", so I've marked this as a duplicate. – Chris Grossack Feb 01 '23 at 06:07
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    @HallaSurvivor Hi, please see my latest edit, the paper really doesn't contain the reference for the claim. – metaverse Feb 03 '23 at 04:53

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I think I saw it now. The key fact is every projective module $A$ is a direct sum $\bigoplus_{i\in I} A_i$ of countably generated modules. So I just iteratively pick one module from each summand $A_i$ and form their direct sums. This give an increasing sequence of proper submodules $M_1\subset M_2\subset ...$ whose union is the whole module. Then the quotient map $A\to \bigoplus A/M_k$ cannot correspond to any map in $\bigoplus \mathrm{Hom}(A, A/M_k)$.

metaverse
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    This doesn't quite work as stated: maybe the $A_i$ are actually finitely generated. Of course, if the index set $I$ is infinite, then you can easily get a countable increasing sequence of proper submodules by just partitioning $I$, and if $I$ is finite, then some $A_i$ must not be finitely generated so that your argument works. – Eric Wofsey Feb 03 '23 at 14:31