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I did some editing over at free module, under the section on submodules of free modules. I don’t have Rotman’s book before me, so I can’t check whether he assumes the commutativity hypothesis for proposition 2, but I put it in to be safe. (Actually, I’ll bet it’s needed, since we have to be careful around invariant basis number which holds for commutative rings.) The proof that I added does use this hypothesis.
Also, I deleted the remark that this is the Nielsen-Schreier theorem in the case $R = \mathbb{Z}$, since NS refers to groups as opposed to abelian groups.
Thanks. Yeah, I guess I omitted the “commutative” there.
Concerning NS: the Wikipedia entry makes it sound as if Dedekind’s statement that subgroups of free abelian groups are free is regarded as part of the Nielsen-Schreier theorem. That’s how we got into all this in the first place. But it might not be the right way to look at it, after all.
Yeah, I’m not sure. But thanks for alerting me to the WP article; I’ll have a look and think about it some; happy to reinstate the remark if it all seems to fit together.
Edit: Okay, I see the historical remark in the Wikipedia article, that NS is a non-abelian analogue of Dedekind’s theorem, so I’ll go ahead and reinstate something (edit: written before I saw your last comment, so scratch that). It would indeed be nice to derive this result as a corollary of NS. The example that worried me when I commented earlier on $p^{-1}(H)$ is where we take $S$ to be a two-element set, with $p: F(S) \to F(S)^{ab} = F(S)/[F(S), F(S)] = \mathbb{Z}x \oplus \mathbb{Z}y$ the projection, and with $H$ the subgroup generated by $x+y$. It looked as though the inverse image $p^{-1}(H)$ might be infinitely generated; at least I didn’t see any relations between elements of the form
$y^m x y^{-m} x^n y x^{-n}$all of which map to $x+y$.
happy to reinstate the remark if it all seems to fit together.
Ah, no need to do that. I tried to take care to crosslink the NS entry with the pid-entry, that should be sufficient.
Thanks for all your work on this. Very much appreciated.
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