is that right that free module M over R is that M can be generated by a linear independent subset A, and every element of M is a finite sum of elements of A multiplied by coefficients in R(the expression should be unique)?
is a quotient of free module free? and is a direct sum of free module free?
I think the second one is yes, but i do not know how to prove it.
thanks
Yes, an $R$-module $M$ is free if it has a basis, i.e., a linearly independent generating set.
No, the quotient of free modules need not be free. Consider the $\mathbb Z$-module $\mathbb Z$. This is clearly free and $1$ is a free generator. Similarly, $2\mathbb Z$ is a free $\mathbb Z$-module, and $2$ is a free generator, but the quotient $\mathbb Z/2\mathbb Z$ is not a free $\mathbb Z$-module. In particular, every subset of $\mathbb Z/2\mathbb Z$ is linearly dependent.
On the other hand, direct sums behave much better. If $M_\alpha$ is free for each $\alpha$ in some indexing set $I$, then the direct sum $\bigoplus_{\alpha\in I}M_\alpha$ is free. Take as basis the disjoint union $\bigsqcup_{\alpha\in I} \mathcal B_\alpha$ where $\mathcal B_\alpha$ is a basis for $M_\alpha$.