This question was asked in my assignment on modules and I couldn't solve this particular question.
Let $V$ be a free $A$-module of infinite rank. Then $|V | =|A|\cdot\operatorname{rank}_A V= \sup \{ |A|, \operatorname{rank}_A V\}$.
Thoughts: I have been following module theory from Algebra by Thomas Hungerford and this is a masters level course. V is a free A-module implies that there exists an infinite basis set {$ v_1,..., v_n,... $} . But I don't have any ideas on how will I prove the given result.
Can you please outline a proof?
Let $\{e_i: i \in I\}$ be a set of basis of $A$. Then every element can be uniquely written as $$\sum_{j \in J} a_j e_j$$ where $a_j \in A - \{0\}$, and $J$ is a finite subset of $I$.
Fix a finite non-negative integer $n$. We count the number of such elements with $\lvert J \rvert = n$. The number of ways to choose $J$ is $$\binom{\lvert{I\rvert }}{n} = \lvert{I \rvert}$$ while given $J$, the number of ways to choose $a_j$ is $$(\lvert{A\vert} - 1)^n.$$ Therefore, the total number of such elements with $\lvert J \rvert = n$ is $$(\lvert{A\vert} - 1)^n \cdot \lvert{I \rvert} = \lvert{A\rvert} \cdot \lvert{ I\rvert}$$ since $I$ is infinite.
Finally, taking the union over all non-negative $n$, we obtain that the cardinality of $V$ is $$\lvert{\mathbb{N}\rvert} \cdot \lvert{A\rvert} \cdot \lvert{ I\rvert} = \lvert{A\rvert} \cdot \lvert{ I\rvert}.$$