TITLE: A result of flat modules that *needs* to deal with the *construction* of the tensor product?
QUESTION [5 upvotes]: 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 ;)

REPLY [6 votes]: 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.