2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Exhibit an infinite list of linearly independent elements of $\overline{\mathbb Q}$. $\endgroup$ – user98602 Feb 25 '15 at 16:02
  • $\begingroup$ 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. $\endgroup$ – anomaly Feb 25 '15 at 16:07
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What do you mean of arbitrarily high degree? $\endgroup$ – AaAaAa Feb 25 '15 at 16:03
  • $\begingroup$ The "and so" might need some justification. $\endgroup$ – quid Feb 25 '15 at 16:05
  • $\begingroup$ @AaAaAa It means, that $n$ can be arbitrarily high for $x^n-2$. $\endgroup$ – Dietrich Burde Feb 25 '15 at 16:05
  • $\begingroup$ 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 $n$? $\endgroup$ – AaAaAa Feb 25 '15 at 16:12
  • $\begingroup$ @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. $\endgroup$ – lhf Feb 25 '15 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.