If P is a R-module, what is the difference between the Free modules F(P) and P? I have $P$ is finitely generated $R$-module. I am trying to construct a surjection from  $F(P)$ to $P$. 
It seems like like elements of $F(P)$ are in $P$ but the converse is not nessicarally true. $x\in F(P)\implies \exists ! r_i\in R-\{0\} \exists ! u_i\in P \ s.t. x=\sum r_iu_i$. So we always have $F(P)\subseteq P$ but it is not always true that elements of $P$ can be uniquely written as sums of elements of $R$ and $P$. exp. $0\in \mathbb{Z}_2\text{ and } 0 =2*1=4*1$. Am I thinking about this correctly?
Also, if this is the case, how can I always get a surjection from $F(P)$ to $P$? 
I think it goes something like this, $P$ is generatet by, say, $\{u_i\}_1^n$. But these $u_i$ may not be linearly indepednet. So if $p\in P$ then $p$ may have more than one such representation as linear combinations of the $u_i$ one of these representations must be in $F(P)$. Am I on to something here? 
 A: As far as I know $F(P)$ denotes the free $R$-module generated by the set $P$, which will ultimately look like $\bigoplus_{p\in P}R$, each copy of $R$ corresponding to a single element of $P$.
This will be the case even if $P$ is already a free $R$ module. For example, let's let $R=\mathbb Q$ and $P=R$. Then $P$ is already a projective $R$-module, and $F(P)=\bigoplus_{q\in \mathbb Q} \mathbb Q$.
I do not recall having ever seen a module used as generators for a free module. Generally the generating set is much smaller, often finite. For this reason, I suspect you may be misreading or confusing two things you have seen.
But anyhow, back to the original question.
It is very easy to see a surjection from $F(P)\to P$. You can choose literally any images in $P$ for the elements of the generating set, and then you are guaranteed a unique homomorphism that carries the generators to those images. Obviously here you'd probably just map $p\mapsto p$, and then the universal property says there's a homomorphism from $F(P)\to P$ that matches your choice. QED.
