# Free $A$-module isomorphic to a direct sum of copies of $A$?

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?

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.
If you interpret $\underbrace{A\oplus\cdots\oplus A}_n$ as $\bigoplus_{i\in I}A$ for infinite $n$ then you're right back at the definition.