# Prove the set of all $m \times n$ matrices over $R$ is a free $R$-module

How to prove the set of all $m \times n$ matrices over $R$ is a free $R$*-module* with a basis of $mn$ elements?

Hint: Try to find a basis with $m\cdot n$ elements.

• Hi alexander, I am trying like this (what you think and what else to prove it is a free $R$-modules: $M_m ._n$, the set of $m \times n$ matrices. Let the standard basis for $M_m,n$ is the set $[B_ij |i=1,2,...,m ; j=1,2,...,n ]$ where $Bij$ is the $m \times n$ matrix whose entries are all zeros, except for the $ij$ entry which is 1. This set has $mn$ elements, therefore $dim(Mm,n) = mn$
– user91036
Sep 7 '13 at 2:40
• yep. The matrices $B_{ij}$ give you the right basis. You just need to show that this is indeed a basis (the set is linearly independent and its span is the entire space) and then you're done. The existence of a basis characterizes a free module so this is all you need to show. Sep 7 '13 at 3:25

Hint: How would you do it if $R$ was a field, i.e. how would you show that the set of $m \times n$ matrices over a field $F$ is an $F$-vector space of dimension $mn$? The proof in your case is identical.