TITLE: Rank of free submodules of a free module over a commutative ring.
QUESTION [7 upvotes]: Free modules over a commutative ring $R$ with $1$ have well-defined rank. 

I have been wondering if there is a ring $R$ such that there are free modules $M'\subset M$ with $\operatorname{rank}(M')>\operatorname{rank}(M)$ for a while, but can't come up with examples or a proof of the contrary in general. 

All I know is that this never happens if $R$ is a domain, since you can consider $M'$ and $M$ as vector spaces over the fraction field of $R$.

REPLY [8 votes]: This can't happen. See Corollary 5.11 of Keith Conrad's notes on exterior powers:
http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf