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.

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    $\begingroup$ If you're working over a field, isn't $M$ just a $K$-vector space? $\endgroup$ – Ben West Aug 25 '17 at 2:10
  • $\begingroup$ Oh, I think I see, so the fact that $K$ is a field means $M$ itself is free, so of course every submodule is free. Is there anything extra to prove every quotient is free? $\endgroup$ – BMac Aug 25 '17 at 2:19
  • $\begingroup$ I guess it depends on how much linear algebra you want to take for granted. Showing a quotient of $M$ is again a $K$-vector space would imply it has a basis. $\endgroup$ – Ben West Aug 25 '17 at 2:22
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    $\begingroup$ @BMac Being a submodule of a free module does not imply freeness, in general. $\endgroup$ – rschwieb Aug 25 '17 at 2:27

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.

  • $\begingroup$ Don't we need finitely generated to assume the vector space is a free $K$-module? $\endgroup$ – BMac Aug 25 '17 at 11:58
  • $\begingroup$ @BMac With the axiom of choice (Zorn's Lemma) we can show that every vector space has a basis. $\endgroup$ – Trevor Gunn Aug 25 '17 at 12:14
  • $\begingroup$ @BMac No, as I said: it has nothing to do with finite generation. [...] every vector space has a basis. That implies every $K$ module is free. $\endgroup$ – rschwieb Aug 25 '17 at 12:33
  • $\begingroup$ So the fact that the module is finitely generated, and the fact that we're talking about submodules and quotients--all this information is just extra padding that distracts from the real question? $\endgroup$ – BMac Aug 28 '17 at 17:47
  • $\begingroup$ @BMac apparently. $\endgroup$ – rschwieb Aug 28 '17 at 17:53

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.

  • $\begingroup$ Okay, so there's a lot less to this problem than I thought. I did ask rschwieb up above whether the finite generation does matter though, because I thought we needed that to declare the modules were free over our field. $\endgroup$ – BMac Aug 25 '17 at 12:00

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