Let $\overline{\Bbb{Q}}$ be the algebraic closure of $\Bbb{Q}$. I am trying to show that $\overline{\Bbb{Q}}$ is not finitely generated as a $\Bbb{Q}$-module, however I do not know where to go with this.
I have tried assuming that $\overline{\Bbb{Q}}$ is finitely generated by $x_1, \ldots , x_r$ and then saying that if $\alpha \in \overline{\Bbb{Q}}$ then there is an $f \in \Bbb{Q}[X]$ such that $f(\alpha) = 0$, but I cannot see where to go from here to obtain a contradiction.
Any help is much appreciated, thanks!
A finitely generated $\mathbb Q$-module is a finite-dimensional $\mathbb Q$-vector space.
There are elements in $\overline{\mathbb Q}$ of arbitrarily high degree and so $\overline{\mathbb Q}$ cannot be a finite-dimensional $\mathbb Q$-vector space.
For instance, $x^n-2$ is irreducible over $\mathbb Q$ for every $n$ and so $2^{1/n}$ is an element of $\overline{\mathbb Q}$ of degree $n$. The $\mathbb Q$-subspace generated by $2^{1/n}$ has dimension $n$.
Note that this proves that $\overline{\mathbb Q} \cap \mathbb R$ is not a finitely generated $\mathbb Q$-module. You don't even need to consider complex numbers.