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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.

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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.

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    $\begingroup$ @Mathmo This is a very neat proof. Make sure you prove why $x^n-2$ is irreducible. $\endgroup$ – LASV Nov 5 '13 at 0:06
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    $\begingroup$ @Luis, that is trivial, Eisenstein Criterion at p=2. $\endgroup$ – Nicky Hekster Nov 5 '13 at 23:42
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    $\begingroup$ I know it is trivial. But there is certainly something to be said. Especially because Eisenstein's is not trivial. $\endgroup$ – LASV Nov 6 '13 at 0:20

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