The proof that I know of the theorem goes like this:
- Any module $M$ is a quotient of a free module $F$ (over any ring).
- Any submodule $K$ of a free module $F$ over a PID $R$ is a free module, so in particular the kernel of the above quotient map is free.
- For any free submodule $K$ of a free module $F$ over a PID $R$ we can find a $y\in K, x\in F$ and $a\in R$ such that $F=\left
\oplus F'$ and $K=\left \oplus K'$ with $K'=K\cap F'$. Furthermore, $\left$ is an ideal maximal among images of $K$ under homomorphisms $F\to R$. - If $K$ is finitely generated (which would follow from $F$ being finitely generated), we iterate this construction, which gives us sequences $x_i\in F$, $y_i\in K$ and $a_i\in R$ such that $F=F'\oplus_{i} \left
$ and $K=\oplus_{i}\left $ where $a_ix_i=y_i$ and $a_i$ divides $a_j$ for $i $ by $\left $ being maximal among images of $\oplus_{i\leq j}a_j$ under homomorphisms from $\oplus_{i\leq j}\left \oplus F'$ to $R$). - Taking quotients, we obtain that $M=F'\oplus_i R/(a_i)$ where $a_i$ divides $a_j$ if $i
My question is why does this break down if we drop the assumption in 4 that $K$ is finitely generated? Does iterating the process transfinitely no longer give us a basis for $K$? If so, why not?