modules - Dimension of the rationals over the integers

What is the dimension of $\mathbb Q$ when it is seen as a module over the integers $\mathbb Z$ (with the usual definitions of addition and multiplication)?

Initially I thought that the dimension ought to be 2, because each rational is uniquely defined by a pair of integers. But then I started looking for a base of size 2, and saw that it doesn't work:

Suppose we have the following base: $\{{p_1 \over q_1},{p_2 \over q_2}\}$. Then for every integers $n_1$, $n_2$:

$$n_1 {p_1 \over q_1} + n_2 {p_2 \over q_2} = \frac{n_1 p_1 q_2 + n_2 p_2 q_1}{q_1 q_2}$$

It is obvious that a rational number with a denominator of $q_1 q_2 + 1$ cannot be represented as such a linear combination. Therefore the given set is not a base.

By a similar argument, no finite set can be a base.

On the other hand, the following countable set is a base:

$$\{ {1 \over 1}, {1 \over 2}, {1 \over 3}, ... \}$$

so the dimension of $Q$ over $Z$ is $א_0$.

Is my conclusion correct?

Answer

No set of more than one rational number is independent, so there is no basis, the rationals are not a free module over the integers.

However, your argument that there is no finite generating set is correct.