A free submodule of a free module having greater rank the submodule

examples-counterexamples commutative-algebra modules

Solution 1:

The answer is yes: I advise you to look up to the MO question https://mathoverflow.net/questions/30860/ranks-of-free-submodules-of-free-modules, where you can find several proofs.

However, if you are interested in the non-commutative case, there exists the following notion: a ring $R$ is said to have the IBN (invariant basis number) property if $R_R^n\cong R_R^m$ implies $n=m$, for every positive integers $n,m$. Well, there do exist not IBN rings. Also, it is possible to find a ring $R$ and an $R$-module $M_R$ such that $M_R\cong M_R^n$ for every $n\geq 1$. This gives you examples of isomorphic modules of different "rank" (in fact, the rank is well defined only for commutative rings). I hope I have not digressed too much.

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