# Proving that $\mathbb{A}=\{\alpha \in \mathbb{C}: \alpha \text{ is algebraic over } \mathbb{Q} \}$ is not a finite extension

It is true that all finite field extensions are algebraic. It is not true however that all algebraic extensions are finite.

In lectures we were given the example of the field extension

$$\mathbb{A}=\{\alpha \in \mathbb{C}: \alpha \text{ is algebraic over } \mathbb{Q} \}$$

I want to prove to myself that this is indeed a non-finite algebraic extension.

I have shown that $\mathbb{A}$ is a subfield of $\mathbb{C}$. Now I consider roots of the equation $x^n-2$ for $n \geq 1$. Clearly such roots are in $\mathbb{A}$ by definition. I think that if I show that for a given $k$, the root of $x^k-2$ (say $\psi_k$) does not lie in $\mathbb{Q}(\psi_1,...,\psi_{k-1})$, then I have proven the extension is not finitie (since this will hold for all $n$).

Can someone please confirm this reasoning and suggest how the proof could be finished.

Your idea is correct, and can be formalized quickly as follows. The polynomial $X^n-2$ is irreducible over $\mathbb Q$. Let $\alpha$ be a root. Then $[\mathbb Q(\alpha):\mathbb Q]=n$. But $\mathbb Q(\alpha)\subseteq \mathbb A$, and so $\mathbb A$ contains subfields of arbitrarily large degree over $\mathbb Q$, and so itself must be infinite dimensional over $\mathbb Q$, as needed.
• @Mathmo This is a very neat proof. Make sure you prove why $x^n-2$ is irreducible. – LASV Nov 5 '13 at 0:06