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Fact A : If $E_i$ is a directed system of $R$-modules, and $F$ is another $R$-module. Then

$$\varinjlim(F\otimes E_{i})=F\otimes(\varinjlim E_{i}).$$

Fact B : "Every module is a direct limit of its finitely generated submodules."

How does Fact A implies Fact B?

Serge Lang claims this in his Algebra book.

user26857
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Babai
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  • possible duplicate of Tensor products commute with direct limits – rschwieb Aug 05 '14 at 13:25
  • @rschwieb: I am not asking for the proof of the expression I just wrote. I am asking how does it follow from that "Any module is a direct limit of its finitely generated submodule." – Babai Aug 05 '14 at 13:27
  • You have a natural map $lim E_i \rightarrow E$ coming from the inclusions. This natural map is an isomorphism if and only if it is an isomorphism at all prime ideals (under localization). So you tensor both sides with $R_p$ and see that $lim {E_i}_p = (lim E_i)_p$ and thus reduce to proving the result for $R$ a local ring. Not an answer by any means, but I'm sure it's a good reduction to make. – Cass Aug 05 '14 at 13:30
  • Hi Susobhan : I understand, but that's why it's marked as "possible." Probably not enough people will agree, and your question will stay :) Honestly from what I've seen, I don't see any strong link between this fact about f.g. modules and the proposition in any of the proofs I've seen. Why do you think the connection is important? – rschwieb Aug 05 '14 at 13:31
  • @rschwieb- Lang claims this in his algebra book. – Babai Aug 05 '14 at 13:33
  • @Cass: Is $E_{i}\bigotimes_{R} R_{p}= {(E_{i})}_p$? – Babai Aug 05 '14 at 13:45
  • ^^^^^ Yes it is. – Cass Aug 05 '14 at 13:49
  • I understand what you have written. But still I dont understand how to reduce it to local ring case. By the way is it true that the finitely generated submodules of $E_p$ are $(E_i)_{p}$'s. – Babai Aug 05 '14 at 13:57
  • EDIT: Well I'm not so sure I can answer the original question anymore, but to the comment question above: Yes they are. To see this, observe that a finitely generated submodule of $E_p$ is the same thing as a map $(R_p)^r \rightarrow E_p$. Just by writing down elements, it's easy to construct a lift of this map to $R^r \rightarrow E$. So every f.g. submodule of $E_p$ comes from an f.g. submodule of $E$. – Cass Aug 05 '14 at 14:22
  • I think reducing to local ring case is clear to me now. Suppose we assume for R a local ring it is proved. Let us now consider R to be any ring. We need to show $(lim E_i){p} \cong E{p}$. But $ (lim E_i)p=lim {E_i}_p=E{p}$. (The last equality holds because $(E_{i}){p}'s $ are the finitely generated $R_p$ submodules of $E{p}$ . And we assumed we have proved for the local ring case already.) Is that correct? – Babai Aug 05 '14 at 14:36
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    Yes, that's the reduction. My thought right now is to further reduce to the case of $R=k$, a field, but I'm not having so much luck managing it. – Cass Aug 05 '14 at 14:39
  • If you can please write it as an answer. – Babai Aug 05 '14 at 14:39
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    Just as a note, by the way, I don't really see why any of this is necessary. The map $f:lim E_i \rightarrow E$ is injective because direct limit preserves injectivity. $f$ is also surjective because every element $e$ of $E$ can be put into a finitely generated submodule, namely $Re$. So $f$ is an isomorphism. Maybe there's some other way with this Fact, but the proof I just gave is about as straight ahead as anyone could hope. – Cass Aug 05 '14 at 14:56
  • The injectivity is not clear to me. $E_i\hookrightarrow E$'s are injective will only say $E_i\rightarrow lim(E_i)$ is injective. Why should $lim(E_i)\rightarrow E$ be injective? – Babai Aug 05 '14 at 15:07
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    Take the trivial directed system $F_i = E$ for all $i$ over the same index set as $E_i$. Then, we have injective maps $E_i \rightarrow F_i$ for all $i$ and these are compatible with directed system maps. Take direct limit on both sides. lim is left (and right) exact, so you get an injective map $lim E_i \rightarrow lim F_i$. But $lim F_i$ is clearly $E$. – Cass Aug 05 '14 at 15:14
  • @user26857 It's on the fifth line of Page 604- Algebra (Revised Third Edition) by Serge Lang – Babai Aug 07 '14 at 10:00

1 Answers1

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I am 100% sure that Fact A and Fact B cannot be proven from another.

[So in some sense, I don't give an answer here, but rather the meta-answer that there is no answer.]

Martin Brandenburg
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    You've misread his question. He is not trying to prove the tensor product fact (call it Fact A) using the Fact B = "Modules Are Limits of F.G. Submodules". He is trying to prove Fact B using Fact A. He already believes Fact A is true. But in any case, as I said in a comment, I agree that it is not necessary or even useful to appeal to Fact A in proving Fact B. – Cass Aug 06 '14 at 01:13
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    You are right, I've misread the question. But as for the original version, is was quite easy to misunderstand it ... the new version is more precise. I've edited accordingly. – Martin Brandenburg Aug 06 '14 at 13:10