# Why is the algebraic closure of $\Bbb{Q}$ not a finitely generated $\Bbb{Q}$-module?

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!

• Exhibit an infinite list of linearly independent elements of $\overline{\mathbb Q}$. – user98602 Feb 25 '15 at 16:02
• It's overkill in this case, but the Artin-Schreier theorem describes fields $k$ with $[\overline{k}:k]$ finite. In particular, they contain an element $x$ with $x^2 = -1$, which $\mathbb{Q}$ lacks. – anomaly Feb 25 '15 at 16:07

## 1 Answer

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.

• What do you mean of arbitrarily high degree? – AaAaAa Feb 25 '15 at 16:03
• The "and so" might need some justification. – quid Feb 25 '15 at 16:05
• @AaAaAa It means, that $n$ can be arbitrarily high for $x^n-2$. – Dietrich Burde Feb 25 '15 at 16:05
• Ok so I understand how $2^{1/n}$ is an element of $\overline{\Bbb{Q}}$ for every $n$, but what exactly do you mean by $2^{1/n}$ having degree $n$? – AaAaAa Feb 25 '15 at 16:12
• @AaAaAa, it means that the polynomial of least degree having $2^{1/n}$ as a root has degree $n$. For your purposes, this means that $\{2^{k/n} : k=0, \dots, n-1 \}$ is linearly independent. – lhf Feb 25 '15 at 16:16