Linear independent set of the right cardinality is a basis.

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In the case of vector space over a field this result is trivially true. If we assume the IBN property for commutative rings, is there a simple proof to this result for a free module $A^n$ over a commutative ring $A$?

(Note that for any free module $A^n$ over a commutative ring $A$, every generator set of $A$ with cardinality $n$ is a basis. Let’s assume we can also use this result.)

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This is not true. Take $A=\mathbb{Z}$ and $n=1$. Then the set $\{2\}$ is linearly independent in $\mathbb{Z}$, but it is not a basis.