Due to the examples given in the answer to this question, I know that the conclusion is of course incorrect. But by reading Kaplansky's proof of theorem 1 in this paper and replacing every occurrence of the word "countably" by "finitely" there (both in the statement and the proof), I'm not able to observe where the proof fails. Maybe it's some silly mistake that I overlooked, but it also maybe some mistake one easily commits without even realizing it and leads to a wrong conclusion. Anyway, thanks in advance for this seemingly naive question.
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In the last paragraph of Kaplansky's proof, the construction could yield an infinite number of non-zero $x_{ij}$, even if each $M_i$ is finitely generated. The infinite matrix he produces will have finitely many non-zero entries in each row, but could have infinitely many non-zero rows.
answered Apr 22, 2014 at 15:48
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$\begingroup$ But in this case, I think there's only finitely many rows to be considered, corresponding to a finite generating set of the initial $M_i$. $\endgroup$– Hua WangCommented Apr 22, 2014 at 15:55
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2$\begingroup$ No. The point is that you each time you add an $x_{ij}$, that may not be in either $P$ or $Q$, your current list of elements may not suffice to generate the components of $x_{ij}$ in $P$ and $Q$, so you then have to go on to include more elements to generate these, and then you need to deal with these new elements, and then ... $\endgroup$ Commented Apr 22, 2014 at 16:04
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$\begingroup$ Ah, I see it now. The stupid mistake I made is that I took condition (4) for granted, while in fact it's not true in general. Thanks a lot. $\endgroup$– Hua WangCommented Apr 22, 2014 at 16:49