I'm looking through a proof and I'm trying to tell if it uses a certain fact. If it doesn't, then I think I've figured out a homework problem. Here is the lemma and the proof of it:
Lemma: Every short exact sequence [of $R$-modules] $$0\rightarrow N \rightarrow M\xrightarrow{g} P \rightarrow 0 $$ with $P$ free, splits.
Proof: Let $\{e_i\}$ be a basis for $P$. Since $g$ is surjective by exactness, we can find $m_i\in M$ such that $g(m_i)=e_i$. Define $h: P\to M$ as $h(e_i)=m_i$. By $R$-linearity, we can expand the definition to all elements of $P$: $h(\sum_i r_ie_i)=\sum_i r_im_i$. Since $g(h(e_i))=e_i$, it follows, again by linearity, that $g(h(\sum_i r_ie_i))=\sum_i r_ie_i$. Hence, the sequence splits.
I'm trying to see if this would work if $P$ were only a finitely generated $R$-module that is not necessarily free. It seems to me like this argument is not using linear independence of $\{e_i\}$, so I would modify the argument by saying "Let $\{p_1,\dots,p_n\}$ span $P$", and write $h(m_i)=p_i$. Then I think all the other arguments would still make sense. Am I thinking about this correctly? Is there something I'm missing?
Linear independence is needed to guarantee that the map is well-defined, although this is glossed over in the text. If there are multiple ways to express some $p\in P$ as a linear combination of basis elements, then the way $h$ is defined here may give "multiple values" for what $p$ should be.
As an example, let $R = P = \Bbb{C},$ and take $e_1 = 1, e_2 = 2.$ Let $M = \Bbb{C}^2$ and let $g : \Bbb{C}^2\to\Bbb{C}$ be given by $g(\alpha,\beta) = \alpha + \beta.$ We can see that $e_1$ and $e_2$ span $\Bbb{C},$ and we have $g(1,0) = e_1$ and $g(0,2) = e_2,$ but if we define $h(e_1) = (1,0)$ and $h(e_2) = (0,2),$ then there are two distinct possibilities for $h(0,2) = h(2e_1):$ it could be $(2,0)$ or $(0,2).$