I'm trying to show that a subset $S=\{m_1,\dots,m_k\}$ of an $R$-module $M$ generates $M$ freely if and only if $S$ is a basis for $M$.
I think I can see the 'if' - using the unique expression of a general $m\in M$ in terms of the basis $S$, for a certain $\Phi$ define $\Theta : M \to N$ by $\Theta (m) = \Theta (r_1m_1 + \dots + r_km_k) = r_1\Phi(m_1) + \dots + r_k\Phi(m_k)$, whose restriction to $S$ is $\Phi$. It can then be verified that $\Theta$ is a homomorphism. But I'm a little lost about the other direction.
Since this must hold for all maps $\Phi$, I'm thinking we need to consider a special one - the identity (into $M$), perhaps?
Relevant definitions:
All rings are commutative with unity.
$S$ generates $M$ freely if $S$ spans $M$ and any map $\Phi :S \to N$, where $N$ is an $R$-module, can be extended to an $R$-module homomorphism $\Theta :M\to N$.
Hint: Let $e_1, \ldots, e_k$ be the standard basis of $R^k$. Extend $m_i \mapsto e_i$ to a homomorphism $M \to R^k$. Show that this is a bijection and then use it to show that the $m_i$ are a basis.