# Why are submodules of a free module over a semihereditary ring flat?

This question comes from a question in Passman's A course in Ring Theory. He asks to prove that a submodule of a free module, say $V$, over a von Neumann regular ring $R$ is flat. He says the reader should use the fact that $R$ is semihereditary, which I was able to prove. Now I know that finitely generated submodules of $V$ must be projective, hence flat, but am struggling to prove this for arbitrary submodules of $V$.

I have been trying to prove that submodules of free modules over von Neumann Regular rings must be finitely generated, which would give the result, but I am having no luck proving it. I am starting to think that this may be the wrong approach. I've also had no luck considering arbitrary exact sequences and showing that the short sequence of elements of the sequence tensored with $V$ is also exact. Any help would be much appreciated.

## 2 Answers

Hint:

A (left) module is the direct limit of its finitely generated (left) submodules.

• I am not fluent in the language of algebra, and have no experience with direct limits unfortunately. However, the analyst in me delights at being able to use limits. As Passman has not mentioned limits at all up until this point, I do not believe that this is his intended proof, but thank you for the hint, I will investigate further. – K.Power Jun 26 '17 at 23:24
• Well all you have to understand is that is is the union of its finitely generated submodules (which is obvious, but formalised in the context of direct limits), and the direct limit functor is exact, i. e. a direct limit of short exact sequences is short exact. As a consequence, a direct limit of flat modules is flat. One you're familiarised with the concept and language, this is not hard to prove. – Bernard Jun 26 '17 at 23:34

There is a useful flatness criterion that says that if $A$ is a right $R$-module, then $A$ is flat iff the following is true. Given an exact sequence $$0\to K\to F\to A\to 0$$ where $F$ is flat, then $K\cap FI=KI$ for every finitely generated left ideal $I$. This can be found as Proposition 3.60 in Rotman's Homological Algebra, among other places I'm sure.

Using this you can prove that if $R$ is von Neumann regular, any right (left) module is flat. To see this, let $M$ be an arbitrary right $R$-module. It is a quotient of some free module $F$, yielding an exact sequence $$0\to N\to F\to M\to 0,$$ where $N\subseteq F$. Let $I$ be a finitely generated left ideal of $R$, which is necessarily principal and generated by an idempotent, since $R$ is von Neumann regular. Say $I=Re$, with $e$ idempotent. Then $FI=Fe$ and $NI=Ne$. Using the criterion, we need $N\cap FI=NI$, that is, $N\cap Fe=Ne$.

If $n=fe\in N\cap Fe$, then $$n=fe=fe^2=ne\in Ne,$$ so $N\cap Fe\subseteq Ne$. Clearly $Ne\subseteq N\cap Fe$ since $N\subseteq F$, so $M$ is flat.

The converse is true as well. I've heard this result called Harada's Theorem, or the Auslander-Harada theorem.