Surjection from a free R-module onto an R-module

Question

I've been stuck on this question for a while. It is part of an assessed assignment, so I would not like a complete answer, but just some sort of hint to get me in the right direction.

Let M be an R-module. Show that there is a surjection from a free R-module onto M.

Any ideas?

Answer 1

Hint: Put $S=\{m_{i}\}_{i\in I}$ as generator set of $M$ and take the free $R-$module $R^{(I)}$.

Answer 2

Assuming $1\in R$, let $M'$ be the free $R$ module generated by the elements of $M$. In other words, the elements of $M'$ are (formal) finite linear combinations of the elements of $M$, with coefficients from $R$. For instance, if $R=\Bbb Z$ and $M$ is the cyclic group on three elements (for clarity called $A,A^2$ and $A^0$), then the elements of $M'$ look like $$ n_1A^0+n_2A^1+n_3A^2 $$ and addition and equality between two such elements are both evaluated term by term (like a polynomial).

Now look at the map $\phi:M'\to M$ given by taking any polynomial-like linear combination as above and evaluating it as an element in $M$ (i.e. what you get if you change the formal "$+$" above into the group operation from $M$). To continue with our example, the element $1\cdot A^0+(-3)\cdot A^1+2\cdot A^2$ is sent to $A^1$.

This $\phi$ is clearly surjective, since any $m\in M$ is the image of $1\cdot m\in M'$.