# Preservation of direct sums and finite generation

I asked this question on Mathematics - Stack Exchange (MSE). Having figured out out how to handle the problem in an extremely particular case, I also posted it as an answer (in the technical sense of the term) to my own question. Getting no other answer, I thought I could post the question on MathOverflow. For the reader's convenience, here is a copy and paste of the question.

This is a follow up on this MSE question, asked by Evariste.

Let $R$ be an associative ring with one. The word "module" shall mean left $R$-module. Say that a module $A$ preserves direct sums if the functor $\hom_R(A,?)$ does.

The main question is

Does the condition that $A$ preserves direct sums imply that $A$ is finitely generated?

The converse is clear: see this MSE answer.

As observed by Mariano Suárez-Alvarez in a comment to this MSE answer, if $A$ can be written as the union of an increasing sequence $(A_n)_{n\in\mathbb N}$ of submodules, then $A$ does not preserve direct sums. [The argument is described in the answer.]

Say that $A$ is countably cofinal if it can be written as such a union. If $A$ is neither finitely generated nor countably cofinal, say that $A$ is uncountably cofinal.

[Here is the motivation for this terminology. A group which can be written as the union of an increasing sequence of subgroups is called countably cofinal, and a group which is neither finitely generated nor countably cofinal, is called uncountably cofinal. Uncountably cofinal groups have been studied by Serre, Tits, MacPherson, Bergman, and many others: see this Google Search. In particular, uncountably cofinal groups do exist.]

The second question is:

Do uncountably cofinal modules exist?

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–  Sam Gunningham Nov 19 '11 at 7:16
Dear @Sam: Thank you very much for the link. I'll study it. –  Pierre-Yves Gaillard Nov 19 '11 at 7:58
The question linked by Sam is an exact duplicate. Therefore I vote to close. I'm aware that the other question has not been fully answered yet, but we may continue there. –  Martin Brandenburg Nov 19 '11 at 9:24