# $\Bbb{Q}$ is not a finitely generated $\Bbb{Z}$-module

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?

• @user26857 Thank you. – Xena Dec 21 '14 at 9:04
• Please, how do you know that c has a prime divisor, what if $b_1$ is prime and all other $b$'s are 1 ? – James Well Nov 22 '17 at 22:15
• @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. – Prism Feb 20 at 10:00

## 2 Answers

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.