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Is $\mathbb Q$ finitely generated as a $\mathbb Z$-module?

If I were to take a guess, I would say no but only because I am not sure how to generate all of $\mathbb Q$ from a finite set. But is this possible?

user5826
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    A $\mathbb{Z}$ module is the same as an Abelian group. – Rene Schipperus Mar 22 '18 at 03:36
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    Hint: With finitely many generators, the denominators in the generated set stay bounded. Also, I'm quite sure this has been asked and answered here before. – Torsten Schoeneberg Mar 22 '18 at 03:39
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    Hint for a more advanced method: A finitely generated $\Bbb Z$-module is Noetherian. Now look at the $\Bbb Z$-submodules $(1/n) \subsetneq (1/n^2) \subsetneq ...$. -- To back up my other claim, here are answers to much more general questions: https://math.stackexchange.com/q/598089/96384 https://math.stackexchange.com/q/1937371/96384 – Torsten Schoeneberg Mar 22 '18 at 04:35

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Suppose $\mathbb{Q}$ is generated by $$\frac{p_1}{q_1},\ldots,\frac{p_k}{q_k}\in \mathbb{Q},$$ with $p_i,q_i\in\mathbb{Z}$ for all $i=1,\ldots,k.$ Then show that $$ \langle \frac{p_1}{q_1},\ldots,\frac{p_k}{q_k}\rangle \leq \langle \frac{1}{q}\rangle$$ where $q=q_1q_2\cdots q_k.$ Since every subgroup of a cyclic group is cyclic then $$ \langle \frac{p_1}{q_1},\ldots,\frac{p_k}{q_k}\rangle=\langle \frac{m}{n}\rangle$$ for some $m,n\in \mathbb{Z}.$ But $\frac{m}{2n}\not\in\langle \frac{m}{n}\rangle$ and $\frac{m}{n}\in \mathbb{Q},$ so we reached a contradiction.

positron0802
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