$\mathbb Q$ is obviously a $\mathbb Z$-module, however, it is not finitely generated. I can't figure out why.
If $\mathbb Q$ is finitely generated, then there are $x_i \in \mathbb Q$ such that $\mathbb Q = \sum_{i=1}^n \mathbb Z x_i$.
I know that I have to construct an element in $\mathbb Q - \sum_{i=1}^n \mathbb Z x_i$, but I can't figure it out.
Thanks for the help.
On
Suppose you had a finite generating set $\frac{a_1}{b_1},...,\frac{a_m}{b_m}$; we can assume all denominators are positive.
Now think about forming a sum of those fractions. You can always group the fractions and rewrite your sum as an integer linear combination $$n_1 \frac{a_1}{b_1} + ... + n_m \frac{a_m}{b_m} = \frac{n_1 a_1}{b_1} + ... + \frac{n_m a_m}{b_m} $$ Now ask, what can the denominator of this sum be? It's never going to be larger than the product $b_1 \cdot ... \cdot b_m$. So no matter what sum you form, you're never going to get the rational number $$\frac{1}{b_1 \cdot ... \cdot b_m+1} $$
Consider the chain of subgroups $$ \langle 1\rangle\subsetneq \left<\frac{1}{2}\right>\subsetneq \left<\frac{1}{4}\right>\subsetneq \dots\subsetneq \left<\frac{1}{2^n}\right>\subsetneq \left<\frac{1}{2^{n+1}}\right>\subsetneq \dotsb $$
Alternatively: suppose $x_1,x_2,\dots x_n\in\mathbb{Q}$. Then you can write $$ x_i=\frac{a_i}{b},\quad i=1,2,\dots,n $$ for some integers $a_1,\dots,a_n$ and $b>0$.
Then $1/(2b)\notin\langle x_1,\dots,x_n\rangle$, otherwise $$ \frac{1}{2b}=\frac{a_1c_1+a_2c_1+\dots a_nc_n}{b} $$ with $c_1,c_2,\dots,c_n\in\mathbb{Z}$.