Let $R$ be a non-trivial commutative ring, hence $R$ has IBN property.
Let $M$,$N$ be free $R$-modules.
Then the tensor product $M\otimes_R N$ is free and $rnk(M\otimes_R N)=rnk(M)rnk(N)$.
Let $A,B$ be bases for $M,N$ respectively.
Then how do I prove that $a\otimes b=a'\otimes b'$ iff $a=a'$ and $b=b'$, ($a\in A, b\in B$)using only above facts?
I know that it can be shown by applying a finite sequence of natural isomorphisms to have $M\otimes_R N\cong \otimes_{(i,j)\in A\times B} R$ and this proves the question.
However, I'm curious whether it can be proven not using isomorphisms but just with the rank.
Clearly, the set $\{a\otimes b|a\in A,b\in B\}$ generates $M\otimes_RN$. If you know that the latter is free and of rank $|A|\cdot|B|$, it follows that the elements in the above set have to be distinct.