Submodules and quotients of finitely generated modules over a field are free

Question

Given $M$ finitely generated module over a field $K$, we want to show every submodule of $M$ and every quotient of $M$ is free.

I only have the vaguest idea that this must use the fundamental theorem of finitely generated abelian groups, but not much beyond that.

Answer 1

Every module over a field is free.

It has nothing to do with finite generation.

It is equivalent to showing that every vector space has a basis, which is proven everywhere.

Answer 2

As Ben points out, modules over a vector space are free: if $P$ is a submodule and $Q$ is a quotient module then $P, Q$ are modules over $K$ and hence free.

But note that we are using properties of $K$ here. Being a submodule or a quotient module are sort of irrelevant. For instance $\mathbf{Z}$ is a free $\mathbf{Z}$-module but most of its quotients: $\mathbf{Z}/n\mathbf{Z}$ are not free. Likewise, $\mathbf{Z}[x]$ is a free $\mathbf{Z}[x]$-module but it has a submodule $(2, x)$ which is not free.

Also, the ring $\mathbf{Z}/4\mathbf{Z}$ as a module over itself has $\mathbf{Z}/2\mathbf{Z}$ as both a submodule and a quotient module.