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It seems to me that from $\mathbb{Z}^{n}\cong A\subseteq G \subseteq B\cong \mathbb{Z}^{n}$ it should directly follow that $G\cong \mathbb{Z}^{n}$ using the structure theorem for finitely generated abelian groups ($G$ must be finitely generated because it is a $\mathbb{Z}$-submodule of the finitely generated $\mathbb{Z}$-module $B$, and $\mathbb{Z}$ is a noetherian ring).

But my professor, in two different sources, says (after having those inclusions already) that $G$ is free of rank $n$ because the quotient group $B/A$ is finite. I assume he is arguing with additivity of the rank in short exact sequences ($G$ is free and $G/A\subseteq B/A$ is also finite, hence of rank $0$): $$ 0\to A \to G\to G/A \to 0$$

This seems to me like an unnecessary step, but since he does it in both sources I am worried that I may be missing something. Am I missing anything?

Remark 1: in case it matters, the $\mathbb{Z}$-module in question is $\mathcal{O}_{K}$, where $K$ is a number field of degree $n$ over $\mathbb{Q}$. I thought that maybe he just wants to compute the cardinality of $B/A$ because it is interesting information in itself. But why would he do it inside the proof then?

Remark 2: I also thought that he may be avoiding the structure theorem following the moto "don't use big theorems if you don't need them". So instead of using the structure theorem he would be using additivity of the rank. But again, I am afraid that this may not be the reason and that I am missing out on something.

Pedro
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    Any subgroup $H$ of a free abelian group $G$ is necessarily itself free of rank at most $\mathrm{rank}(G)$, so $G$ is necessarily free, with rank $\mathrm{rank}(A) \leqslant \mathrm{rank}(G) \leqslant \mathrm{rank}(B)$, whence $\mathrm{rank}(G) = n$. In general, a similar statement holds if $A, B, G$ are modules over a PID. Note that this does not use the full strength of any structure theorems, though the fact I cited is an important step towards classification of finitely generated modules over a PID. – Alex Wertheim Jan 23 '18 at 01:41
  • AFAIK the structure theorem for finitely generated abelian groups simply states that every finitely generated abelian group is isomorphic to $\mathbb{Z}^m\oplus\mathbb{Z_{a_1}}\oplus\cdots\oplus\mathbb{Z_{a_k}}$. How would that directly imply that $G$ is free of rank $n$? In general, over arbitrary ring $R$ a submodule of a free module need not be free. And even if it is ranks may not agree. So it all depends on what exactly you mean by the structure theorem. If you mean the stronger version (e.g. structure theorem for modules over PID) then yeah, nothing else is necassary. – freakish Jan 23 '18 at 10:18
  • @freakish aren't the structure theorem for finitely generated abelian groups and the structure theorem for finitely generated modules over $\mathbb{Z}$ the same thing? Using any of those two theorems: if $G$ has a torsion element, then so does $B$ (contradiction), so $G$ is isomorphic to $\mathbb{Z}^{m}$. Pull back a basis of $\mathbb{Z}^{m}$ to $G$ and you get that $G$ is free of rank $m$, right? And then, since they also generate $A$, $n\leqslant m$, and similarly $G\subseteq B$ implies $m\leqslant n$, hence $G$ is free of rank $n$. – Pedro Jan 23 '18 at 10:38
  • @Pedro How do you know that $n\leq m$? Generally ranks don't behave that well. For example in the world of non-abelian groups $\mathbb{F}_n\subseteq \mathbb{F}_2$ for any $n$. In the world of modules you can find a counterexamples here: https://en.wikipedia.org/wiki/Invariant_basis_number As for theorems: you would have to cite them. Different sources state different things. The common point is that a finitely generated module over PID is isomorphic to a finite product of $R/(q)$ where $(q)$ is prime. That statement is not enough. You need some additional argument for $n\leq m$. – freakish Jan 23 '18 at 10:44
  • @freakish I am not talking about non-abelian groups, I am talking about commutative algebra and modules over commutative rings, and in that link one can read "any non-zero commutative ring has the invariant basis number property", right? So what is the problem with my argument? – Pedro Jan 23 '18 at 10:55
  • @Pedro Sure, if use IBN theorem for commutative rings. :) That's a strong theorem though. The point is that the structure theorem alone is not enough. – freakish Jan 23 '18 at 10:56
  • @freakish I see. It's good to know the depth of the theorems that one uses implicitly, and I wasn't aware of the depth of that theorem. Thank you for your comment! – Pedro Jan 23 '18 at 10:59

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Let me sum up the comment discussion.

What you know for sure is that $G\simeq\mathbb{Z}^m$ for some $m$. What you don't know is the following:

Lemma. Let $R$ be a commutative ring and assume that $f:R^n\to R^m$ is a monomorphism of free $R$-modules. Then $n\leq m$.

Proof. Can be found here: $A^m\hookrightarrow A^n$ implies $m\leq n$ for a ring $A\neq 0$ $\Box$

For the simple case of $R=\mathbb{Z}$ the lemma follows from the additivity of the rank, i.e. if we have a short exact sequence

$$0\to A\to B\to C\to 0$$

and we know that $rank(B)=rank(A)+rank(C)$ then the lemma follows by taking $C:=B/\text{im}(f)$ and realizing the trivial $rank(C)\geq 0$ inequality.

So with that lemma it is easy to complete the proof: $n=m$ because $n\leq m$ and $m\leq n$ (and those inequalities follow from the lemma).

All in all: your professor seems to know what he's doing. ;)

Side note: The lemma does not hold in general non-commutative case. Rings with the property

$$R^n\simeq R^m\text{ iff }n=m$$ are known as IBN rings. You can read more about them here: https://en.wikipedia.org/wiki/Invariant_basis_number

freakish
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