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Let $M$ be a proper $\Bbb Z$-submodule of $\Bbb Q.$ Can we say that $M$ is finitely generated?

I know that $\Bbb Q$ is not finitely generated as a $\Bbb Z$-module.

Please help me in this regard. Thank you very much.

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  • $\begingroup$ What about the fractions with odd denominators? $\endgroup$ – Lord Shark the Unknown Feb 14 at 7:25
  • $\begingroup$ @Lord Shark the Unknown yeah it's a infinitely generated submodule. $\endgroup$ – Dbchatto67 Feb 14 at 7:34
  • $\begingroup$ It can be finitely generated or not. In addition to other answers, consider $\mathbb{Z}$ as a $\mathbb{Z}$-submodule of $\mathbb{Q}$. Obviously, it's free with basis $\{1\}$. $\endgroup$ – stressed out Feb 14 at 7:50
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    $\begingroup$ @Dbchatto67 what's your question now? It got answered twice and you just said "Yes, I know."? $\endgroup$ – Paul K Feb 14 at 7:55
  • $\begingroup$ I have got my answer from the hint given by @Lord Shark the Unknown. $\endgroup$ – Dbchatto67 Feb 15 at 3:19
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Take a finite set of rational numbers, and express them in simplest form. The denominators of these numbers when factorized will invovle a finite number of primes.

FACT 1: If a set of rational numbers involve only a finite number of primes this way, them their sum and their products will also involve THE SAME set of primes numbers or a subset, but not a larger set. (Look at LCM)

As there are infinitely many prime numbers we can take a proper infinite subset of those prime numbers and the collection of rational numbers such that their denominators are factorizable using only those prime numbers. This submodule cannot be finitely generated and is proper.

Lord Shark's short comment mentions this fact taking the subset to be all odd primes (omitting 2, so a proper subset).

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  • $\begingroup$ I have already understood that. See my comment above. $\endgroup$ – Dbchatto67 Feb 14 at 7:39

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