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The question is in the title. It's really hard to "show my effort" on this, because it's just coming up with an example. My effort is... thinking about it for a while? I thought about letting $R = \mathbb{Z}$ or $\mathbb{R}[x]$ but I'm having trouble getting an example of $M$ being decomposed in this way with two different $F$'s. Would love a step in the right direction. Let me know if there's other ways to improve this question by showing some kind of motivation or effort.

This comes from Steven Roman's text "Advanced Linear Algebra" as Exercise 6.5. As far as sufficient conditions goes, the fact that it's a PID helps, which is why I mentioned coming up with specific examples of PID's. Just thought I could use the Gaussian integers, too.

Moni145
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    You need a source for your question, that is all! Edit your question and let others know where your question is coming from : a textbook? A PDF? A thought after seeing a similar question whose answer you know/don't know? Remember, life is easy if you give us more information. For example, if you are reading a particular book then you can mention it. If you know any relevant facts(for example, do you know sufficient conditions for such an $F$ to be unique? That will eliminate possibilities). That's all you need to do : edit your question with any one (or more!) of these things. – Sarvesh Ravichandran Iyer Jul 09 '21 at 18:43
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    There you go! Nicely done, and +1. – Sarvesh Ravichandran Iyer Jul 09 '21 at 18:48

2 Answers2

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You know $M/M_{\operatorname{tor}}$ is free. Taka a basis $(e_1, \ldots, e_n)$ of this free module. To obtain $F$ we need to take a lift $\bar e_i$ of every $e_i$. There are $|M_{\operatorname{tor}}|$ choices for each lift, and they all give a different $F$. To see this, fix a choice $F^0$ spanned by lifts $(\bar e_1^0, \ldots, \bar e_n^0)$. An arbitrary choice of lifts is then of the form $(\bar e_i^0 + t_i)_i$ with $t_i \in M_{\operatorname{tor}}$, and we recover the $t_i$ from $F$ by looking at the primages of $e_i$ under the isomorphisms $F \to M/M_{\operatorname{tor}}$ and $F^0 \to M/M_{\operatorname{tor}}$. The difference of these preimages is $t_i$.

So there are $|M_{\operatorname{tor}}|^{\operatorname{rank}(M)}$ choices of $F$.

Bart Michels
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    This approach looks really interesting. However, I haven't been exposed to "lifts." Where might I find a resource on this, or could you explain a little bit of what that means? – Moni145 Jul 09 '21 at 19:08
  • Oh, whenever you have a quotient of modules $M/N$, a "lift" of an element $a \in M/N$ is an element of $M$ that projects to $a$. So a representative of the equivalence class. Example: $\mathbb Z/3\mathbb Z$, $-1 \in \mathbb Z$ is a lift of the element $2 \in \mathbb Z / 3 \mathbb Z$. – Bart Michels Jul 09 '21 at 19:11
  • That makes a lot of sense. Thanks a lot! – Moni145 Jul 09 '21 at 19:12
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$\Bbb Z\oplus (\Bbb Z/2)=(1,0)\Bbb Z\oplus (0,1)\Bbb Z=(1,1)\Bbb Z\oplus (0,1)\Bbb Z$