I didn't know what to put on my search query so it's possible that this has already been asked before. If so, please give me a link to the answer and I'll delete this question.

In linear algebra over fields, it is easy to see that if a matrix has more columns than rows, then there is a nontrivial relation between the columns.

This result is easily expandable to commutative integral domains, since they can be embedded into fields, and I would think that proving the same result for division rings and hence for any integral domains in general wouldn't be such a problem.

However I have read somewhere that this holds for matrices over arbitrary rings, but couldn't find a proof. Intuitively, this seems normal because removing integrity causes more things to be zero, more easily (for instance if the ring has trivial multiplication, then this is obviously true), and so should make creating non trivial relations easier.

But when the ring is not "regular" enough, a theory of dimension over this ring is harder to put into place and so it wouldn't have surprised me that this result were actually false.

So my question is : is this theorem true for arbitrary rings, and if so, how do you prove it ?


Your question is the same us:

Can a free module of rank $n$ contain a linearly independent set with more than $n$ elements?

In the commutative case, this cannot happen. See https://mathoverflow.net/questions/30860/ranks-of-free-submodules-of-free-modules

In the non-commutative case, on the other hand, it can happen, and there are rings R for which the free module of rank 1 is isomorphic to the free module of rank 7, for example. The key word you want to google for it "IBN".


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