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Let $G$ be a finitely generated abelian group whose free part has rank $r$. I know that every subgroup $H$ is finitely generated and has free part of rank $s\leq r$, and that also $G/H$ is finitely generated with free part of rank $r-s$.

I was wondering if this extends also to arbitrary modules, or at least to modules over a PID. I can't find it anywhere.

Is it true? If so, where can I find this theorem?

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This does not extend to arbitrary modules. For example, if $M = \Bbb Z\oplus \Bbb Z$, then $G=M$ is a rank-1 free $M$-module, but letting $H=\Bbb Z$, we have $G/H=\Bbb Z$, and neither one has any free rank. Other non-trivial examples exist.

However, for modules over a PID this does hold, as someone commented while I was typing up a proof of this fact from my head.

My proof:

If $G$ is a finitely generated module, then it can be decomposed as a direct sum of cyclic submodules. How can I derive from this fact that a submodule $H$ of $G$ is finitely generated?

Suppose free-rank($G$) is $r$, and $H$ is a submodule of $G$, with free-rank($H$)=$s$, say $e_1, \cdots, e_s$ generate the free part of $H$. Since they are, by construction, linearly independent, then they can be completed to a set {$e_i$} of $r$ linearly-independent elements of $G$. Since for these $e_i$ generate a torsion-free module, then for $i>s$, $e_i\notin H$. Then $H/\langle e_i\rangle\leq G/\langle e_i\rangle$ is a torsion module, and so finite, since $G$ is finitely generated. So $H/\langle e_i\rangle$ is finite, thus finitely generated, and $H\leq\{e_1, \cdots, e_s\}\oplus H/\langle e_i\rangle$ is finitely generated.

Supposing $H$ is finitely generated, then also $H$ can be decomposed in a direct sum of cyclic modules. From this, how can I deduce the quotient (which is obviously finitely generated) has free part of rank $r−s$?

To deduce that the free part has rank $r−s$, simply look at the quotient $G_i/H_i$ and add. For $\{e_1, \cdots, e_r\}$, this rank will be 0, and for {$e_{r+1}, \cdots, e_s$} it will be 1; for the torsion parts of $G$ it will be 0.

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  • $\begingroup$ Thank's for the answer. But I don't understand what the structure theorem has to do with my statement. If $G$ is a finitely generated module, then it can be decomposed as a direct sum of ciclic submodules. How can I derive from this fact that a submodule $H$ of $G$ is finitely generated? Supposing $H$ is finitely generated, then also $H$ can be decomposed in a direct sum of ciclic modules. From this, how can I deduce the quotient (which is obviously finitely generated) has free part of rank $r-s$? $\endgroup$ – User Dec 31 '15 at 11:52
  • $\begingroup$ If $G=\oplus G_i$, and $H$ is a submodule of $G$, then we can look at $H_i$:=$\Pi_{G_i}H$ ($\Pi$=projection); then $H = \oplus H_i$. Since $R$ is a PID, then each $H_i$ will be generated by 1 element, and so $H_i$ will be finitely generated. To deduce that the free part has rank $r-s$, simply look at the quotient $G_i/H_i$, and add. Note in what cases this rank will be 1, and in what cases it will be 0. $\endgroup$ – Alex Jan 1 '16 at 14:50
  • $\begingroup$ At first sight your reasoning seemed true to me, but indeed it isn't! It's not true that $H=\oplus H_i$; you can only say that $H\leq\oplus H_i$. For example, let $G$ be $\Bbb Z\oplus \Bbb Z = G_1\oplus G_2$. Let $H$ be the submodule generated by $(2,1)$. Then $H_1=2\Bbb Z\leq G_1$ and $H_2=\Bbb Z\leq G_2$. Here $H\neq H_1\oplus H_2$! How can the proof be modified then? $\endgroup$ – User Jan 2 '16 at 0:19
  • $\begingroup$ Sorry I missed your comment: I would have answered it even without the bounty. What I said is correct: you just have to start with the "right" basis of $G$ (the one that includes the free vectors of $H$). So if $H=(2, 1)$ then your $G$ would be decomposed as, say, $<(2, 1), (1, 2)>$, and my argument would work. In either case, since there is a bounty, then I typed up the full story above; let me know if this does not answer your question. $\endgroup$ – Alex Jan 5 '16 at 4:32
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For modules over a PID $R$ you can try to prove the following.

Lemma: The rank of a finitely generated module $M$ is the dimension of $M \otimes_R \text{Frac}(R)$, where $\text{Frac}(R)$ is the fraction field of $R$.

This is a straightforward corollary of the structure theorem. Since tensoring with $\text{Frac}(R)$ is exact, a short exact sequence of finitely generated modules becomes a short exact sequence of $K$-vector spaces, and then you can reduce to the corresponding statement for vector spaces.

Over an arbitrary commutative ring $R$ it's unclear what one ought to mean by the rank of a finitely generated module. There's a good answer to this question under the additional hypothesis that $M$ is projective and a pretty good answer, although it does not spit out a single number, under the additional hypothesis that $M$ is flat.

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