Does this proposition hold even if it's not finitely generated?
I think it does, since $M$ isomorphic to the direct sum of $M_i$, $M_i$ isomorphic to the direct sum of $A$.$(m_1,m_2,\ldots,m_n)\to(a_1,a_2,\ldots a_n)$. hence $M$ isomorphic to the direct sum of $A$.
But my friend tell me it doesn't hold without explaining.
Could you please tell me if the proposition hold without finitely generated?
Thanks in advance
Sure, it is always true that a free module (under this definition) will be isomorphic to $A^{(I)}$ for some index set $I$, no matter if the set $I$ is finite or infinite.
Suppose you are given $M=\bigoplus_{i\in I} M_i$ with isomorphisms $\phi_i:M_i\to A$. This can be composed with the map that injects $A$ into the $i$th position in $\bigoplus_{i\in I}A$. Abusing notation, use $\phi_i$ to denote this composition, so that $\phi:M_i\to \bigoplus _{i\in I} A$.
Then you can verify that $\phi: M\to \bigoplus_{i\in I}A$ given by $\phi(\sum m_i)=\sum\phi_i (m_i)$ is an isomorphism.
One thing worth worth pointing out here is the difference between notations $A^n$ and $A^{(n)}$. Usually $A^n:=\prod_{i\in I}A$, and as you probably know $\prod_{i\in I}A\ncong \bigoplus_{i\in I}A$ in general. But $A^I\cong A^{(I)}$ when $I$ is a finite set, so it's safe to use $A^n$ to mean either the product or the sum since $n$ indicates the number of copies is finite.