I am currently working on the following problem from Lang's Algebra:
- Let $M$ be a finitely generated torsion-free module over $\mathfrak{o}$. Prove that $M$ is projective. [Hint: Given a prime ideal $\mathfrak{p}$, the localized module $M_\mathfrak{p}$ is finitely generated torsion-free over $\mathfrak{o}_\mathfrak{p}$, which is principal. Then $M_\mathfrak{p}$ is projective, so if $F$ is finite free over $\mathfrak{o}$, and $f: F \rightarrow M$ is a surjective homomorphism, then $f_\mathfrak{p}: F_\mathfrak{p} \rightarrow M_\mathfrak{p}$ has a splitting $g_\mathfrak{p}: M_\mathfrak{p} \rightarrow F_\mathfrak{p}$, such that $f_\mathfrak{p} \circ g_\mathfrak{p} = \text{id}_{M_\mathfrak{p}}$. There exists $c_\mathfrak{p} \in \mathfrak{o}$ such that $c_\mathfrak{p} \not \in \mathfrak{p}$ and $c_\mathfrak{p}g_\mathfrak{p}(M) \subset F$. The family $\{c_\mathfrak{p}\}$ generates the unit ideal $\mathfrak{o}$ (why?), so there is a finite number of elements $c_{\mathfrak{p}_i}$ and elements $x_i \in \mathfrak{o}$ such that $\sum x_ic_{\mathfrak{p}_i} = 1$. Let $$g = \sum x_i c_{\mathfrak{p}_i} g_{\mathfrak{p}_i}.$$ Then show that $g: M \rightarrow F$ gives a homomorphism such that $f \circ g = \text{id}_M$.]
I have managed to prove the parts of the hint up to the point where $f_\mathfrak{p}$ is introduced. I don't know what this map is and can't find it anywhere, but I would imagine that it is somehow induced by $f$. Does anyone know what this notation means? Could it be that Lang simply wants us to construct such a map $f_\mathfrak{p}$ from $f$?