Suppose $M$ is a finitely generated left module over a ring $R.$ 


We define the **rank** of $M$ as the minimal number of generators of $M.$ 


If in addition $M$ is free, then we define the **free-rank** of $M$ as minimal cardinality of a basis of $M.$ 


It is clear that $\text{rank}(M)\leq\text{free-rank}(M)$. 



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> I want to know an example when $\text{rank}(M)<\text{free-rank}(M)$. 
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If $R$ is commutative, then they are equal; so $R$ must be non-commutative.



Rank versus free-rank of a module
There are abelian groups $A$ such that $A\cong A\oplus A \oplus A$ but $A\not\cong A\oplus A$.


Let $E=\operatorname{End}(A)$. The functor $F=\operatorname{Hom}(A,-)$ is an equivalence from the category of finite direct sums of copies of $A$ to the category of finitely generated free $E$-modules.


So $E\oplus E$ is a module with free rank $2$ but is a direct summand of $E\cong E\oplus E\oplus E$, and so has rank $1$.


For an example where the ring has the IBN property:


For any $n>2$ there is an abelian group $A\_n$ such that $A\_n^n\cong A\_n$ but $A\_n^k\not\cong A\_n$ for $1<k<n$. Let $E\_n=\operatorname{End}(A\_n)$ and $E=\prod\_{n>2}E\_n$.


Then $E^2$ has free rank $2$ but rank $1$.


