I'm trying to show that $\Bbb{Q}$ is not a finitely generated $\Bbb{Z}$-module.

Assume to the contrary that $$\Bbb{Q}=\Bbb{Z}\dfrac{a_1}{b_1}+...+\Bbb{Z}\dfrac{a_n}{b_n}$$ where $a_i,b_i\in\Bbb{Z}$. Let lcm$(b_1,...,b_n)=c$ and $d_i=\dfrac{c}{b_i}$. Then $$\Bbb{Q}=\dfrac{d_1a_1}{c}\Bbb{Z}+...+\dfrac{d_na_n}{c}\Bbb{Z}$$ Now let $p$ be a prime that does not divide $c$. Then, there are $x_1,...,x_n\in\Bbb{Z}$ such that $$\dfrac{1}{p}=\dfrac{d_1a_1x_1+...+d_na_nx_n}{c}$$ and hence $p(d_1a_1x_1+...+d_na_nx_n)=c$, which means that $p$ divides $c$, contradiction.

That was suspiciously easy. What am I missing?

  • $\begingroup$ @user26857 Thank you. $\endgroup$ – Xena Dec 21 '14 at 9:04
  • $\begingroup$ Please, how do you know that c has a prime divisor, what if $b_1$ is prime and all other $b$'s are 1 ? $\endgroup$ – James Well Nov 22 '17 at 22:15
  • $\begingroup$ @JamesWell It is clear that at least one $b_i$ must be bigger than $1$, say $b_1 > 1$. Then $c=\operatorname{lcm}(b_1, ..., b_n) \geq b_1>1$, so $c$ is also bigger than $1$. Any number which is bigger than $1$ must have a prime divisor, so $c$ has a prime divisor. $\endgroup$ – Prism Feb 20 at 10:00

It is in fact that easy. You can make it even simpler by noting you can conclude $\mathbb Q \subseteq \langle \frac{1}{c} \rangle$ from your assumption. Since $\frac{1}{2c}\in\mathbb Q$ we have $\frac{1}{2c}=\frac{k}{c}$ for some $k\in\mathbb Z$, but there is no $k\in\mathbb Z$ such that $2k=1$.


Everything is correct.

Only suggestion I have is that once you argue that your generators can have a common denominator you might as well call them $\frac{e_i}{c}$ instead of $\frac{d_ia_i}{c}$. It just makes the equations look a bit simpler. In fact you could even argue that $\frac{1}{c}$ generates all your generators so it suffices to show that $\mathbb Z[\frac{1}{c}] \neq \mathbb Q$. The final part of your argument will be the same but you'll only have one generator in your equations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.