Let $R$ be an associative ring with 1. Let $M$ be a product if infinitely many copies of $R$, viewed as a left $R$-module. Is $M$ locally free? More precisely: is it true that every finitely generated submodule of $M$ is contained in a free submodule?
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1 Answer
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No.
Let $R$ be the ring of eventually constant sequences of integers, and let $e_i\in R$ be the sequence whose $i^{th}$ term is 1, with all other terms zero.
Then if $I$ is the annihilator of an element of a free $R$-module, either
- there is some $N$ so that $e_i\in I$ for all $i>N$, or
- there is some $N$ so that $e_i\not\in I$ for all $i>N$.
But the element $x=(e_1,e_3, e_5, \dots)$ of the countable product of copies of $R$ is annihilated by $e_{2i}$ but not by $e_{2i+1}$ for every $i$, and so the submodule generated by $x$ cannot be a submodule of a free module.
answered Jun 12 at 16:11
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1$\begingroup$ I don't agree that every element of $R$ is annihilated by $e_1-e_{i+!}$ for sufficiently large $i$. $\endgroup$ Commented Jun 13 at 9:39
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$\begingroup$ @PeterKropholler You make a good point! I think the example does work, but of course my justification is nonsense. I’ll delete the answer in a while if I don’t manage to fix it quickly. $\endgroup$ Commented Jun 13 at 11:40
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$\begingroup$ @PeterKropholler I think (or at least hope!) that I’ve edited my answer so that it is correct. $\endgroup$ Commented Jun 13 at 12:24
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$\begingroup$ Yes I believe your answer. Clearly some condition would be required on R. I wonder if products of free modules are locally free for Noetherian rings? $\endgroup$ Commented Jun 16 at 6:37