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There is a nice result concerning flat modules over a ring:

If every finitely generated submodule of a module $M$ is flat, then $M$ is flat.

However, the proof I've read in Rotman's Homological Algebra is quite messy IMHO. In fact, it violates the golden rule of tensor products: never use its construction, use its universal property!!

I attach the proof to the end of the post. The "mess" is in the lemma.

So the question is:

  • Is there a cleaner proof, also as elementary? (Rotman has just barely defined flat modules at that point, plus proven a few elementary properties)

(I'm thinking perhaps this proof can be translated to use the universal property...)

If you know a proof that is not as elementary, please do post it anyway, it should be interesting.

  • If you think not, then could you give an heuristic* as to why you think there is not a cleaner proof? i.e., what is the fundamental reason (if there is any) of necessarily having to deal with the construction of the tensor product in the proof of this theorem?

*I'm not proficent in the usage of the word "heuristic", I apologize in advance if it's improperly used ;)

the proof

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I don't have the book on me, what exactly has he covered up until this point? Are you aware that the direct limit functor is exact and that all modules are direct limit of their f.g. submodules? – Alex Youcis Jan 12 '12 at 22:04
@Alex: No, (co)limits are a couple of chapters after this. He has covered projective and injective modules, and their elementary properties. Here's the link to Google Books, maybe you'll find it useful… . But as I added in a recent edition, I am interested in other proofs, even if they are not as "elementary" (I know I'm abusing the word, please bear with me). – Bruno Stonek Jan 12 '12 at 22:06
I'm in a rush now, but I'll try to come back and answer this later (assuming no one else has, which I am sure they will have). That said, if I recall there are results before this that Rotman opts for the messier approach using the construction (e.g. his proof that coproduct commute with tensor) instead of the slick proof. I think the reason he goes for this messier approach, is that to apply the universal property you'd have to construct maps which require the exact machinery he's trying to avoid. – Alex Youcis Jan 12 '12 at 22:10
I personally think the golden rule is to not define maps out of the tensor product using the construction. After that, you take what you can get! – Dylan Moreland Jan 12 '12 at 22:16
up vote 4 down vote accepted

Let $\mathcal F$ be the poset of all finitely generated submodules of $M$, ordered by inclusion, so that $M=\varinjlim\limits_{F\in\mathcal F}F$.

Let $$0\to N_1\to N_2\to N_3\to 0 \qquad\qquad(\star)$$ be an exact sequence of left modules, and let us suppose that all elements of $\mathcal F$ are flat, so that for each $F\in\mathcal F$ the sequence $$0\to F\otimes N_1\to F\otimes N_2\to F\otimes N_3\to 0$$ is exact.

Since the direct limit of short exact sequences is an exact sequence, taking the limit of these sequences as $F$ varies in $\mathcal F$ provides us with an exact sequence $$0\to\varinjlim\limits_{F\in\mathcal F}(F\otimes N_1)\to\varinjlim\limits_{F\in\mathcal F}(F\otimes N_2)\to\varinjlim\limits_{F\in\mathcal F}(F\otimes N_3)\to0$$

Now, the functor $(-)\otimes N_i$ commutes with direct limits —because it is a left adjoint— so for each $i$ we have $$M\otimes N_i=\varinjlim\limits_{F\in\mathcal F}(F\otimes N_i)$$ so the above exact sequence is in fact isomorphic to one of the form $$0\to M\otimes N_1\to M\otimes N_2\to M\otimes N_3\to0$$

After checking that all maps are what they should be, we see that tensoring $(\star)$ with $M$ preserved exactness, so $M$ is flat.

Note 1. The only information on the tensor product that one needs to do all this is that it is a left adjoint—the rest is simply generic general nonsense.

Note 2. Throughout direct limit in what I wrote means directed direct limit.

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¡Gracias! Both you and Alex used limits, so I went to that section in the book, and ta-da, he does exactly this proof in corollary 5.35, p.248. But eh, I'm just a few days into my month-and-a-half study of homological algebra through Rotman, I'm not thinking of limits yet ;) Anyway, the question is still open for another slick proof that does not use limits! – Bruno Stonek Jan 12 '12 at 22:41
@BrunoStonek There is probably a slightly more compact proof that doesn't use limits, but I think without limits you're not going to be happy. This problem absolutely begs for the use of limits since limits, almost by definition, are useful to pass from finitary objects (e.g. f.g. submodules) to the general case. – Alex Youcis Jan 13 '12 at 11:04

As I stated in the comment I think the reason he goes for the explicit construction, is that to use the universal property he'd have to use a lot of the machinery which he is seeming to avoid.

The slickest proof uses Lazard's theorem which says that flat modules are precisely the direct limit of f.g. free modules. Indeed, if $\{M_\alpha\}$ denotes the set of f.g. submodules of $M$ then it's a common fact that $\displaystyle \varinjlim_\alpha M_\alpha=M$, and by Lazards theorem for each $\alpha$ we have that $\displaystyle M_\alpha=\varinjlim_\beta F_{\alpha,\beta}$ for some f.g. free module $F_{\alpha,\beta}$. Then,

$$M=\varinjlim_\alpha\;\;\varinjlim_\beta F_{\alpha,\beta}=\varinjlim_{(\alpha,\beta)}F_{\alpha,\beta}$$

and so $M$ itself is a direct limit of f.g. free modules, and so flat.

EDIT: Although, in retrospect, I'm trying to recall if you need this result to prove Lazard's theorem!

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Thanks, this is slick for sure, even if Lazard's theorem is not exactly elementary IMHO. In any case, this theorem is also in Rotman, but we didn't do it on my course, so I didn't have it in mind. See my comment on Mariano's answer for another reference. – Bruno Stonek Jan 12 '12 at 22:43
I'm not sure I understand this proof. What are the maps between the different $F_{\alpha,\beta}$ in that last directed limit? – Omar Antolín-Camarena Nov 6 '12 at 19:30

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