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Let $R$ be a commutative ring and $M$ be a free $R$-module. Since $R$ is commutative, $R$ has the IBN property, hence the rank of $M$ is uniquely well-defined. So set $n:=\mathrm{rank}(M)$.

Let $A$ be an $R$-linearly independent subset of $M$. What is an example of $A$ such that $|A|>n$?

If $R$ is a division ring, it must be $|A|≦n$, but if $R$ is commutative, I think it is possible that $|A|>n$.

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  • $\begingroup$ To construct an example of such $A$, $R$ must not be an integral domain. $\endgroup$ – Rubertos Mar 25 '15 at 2:36
  • $\begingroup$ IBN is a property of the free modules over $R$. If $M$ is not free, what do you mean by the rank of $M$? $\endgroup$ – Qiaochu Yuan Mar 25 '15 at 3:42
  • $\begingroup$ @QiaochuYuan Oh yes, I will edit it $\endgroup$ – Rubertos Mar 25 '15 at 3:49
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No, it's not possible!

Suppose $R$ is a commutative unitary ring and $M$ a free $R$-module of rank $n$. If $A\subset M$ is a linearly independent set, then $|A|\le n$.

We can assume $M=R^n$. A linearly independent set $A\subset R^n$ with $|A|=m$ gives rise to an injective $R$-module homomorphism $\phi:R^m\to R^n$, and then $m\le n$. (For this you can find a proof here.)

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