Show that $\mathbb{Q}(\sqrt{3},\sqrt{3}, \sqrt{3},…)$ is algebraic over $\mathbb{Q}$ but not a finite extension.

Show that $$\mathbb{Q}(\sqrt{3},\sqrt{3}, \sqrt{3},...)$$ is algebraic over $$\mathbb{Q}$$ but not a finite extension.

I think for the algebraic part, since for every simple extension, each of those elements can be adjoined and each of these simple extensions has a minimal polynomial that cannot be reduced in $$\mathbb{Q}$$. For example, the simple extension $$\mathbb{Q}(\sqrt{3}, \sqrt{3})(\sqrt{3})$$ has minimal polynomial $$x^{8}-3$$. And since each simple extension has an increasingly large degree, the degree of the simple extensions over the previous extension gets larger for each attachment. But I am not sure how to express this formally...

For the infinite degree part, I was thinking because the set $$\left \{\sqrt{3},\sqrt{3}, \sqrt{3},...\right \}$$ is linearly independent?

Any element $$\alpha$$ of $$F$$ is a rational expression in the numbers adjoined. As such, it can involve only finitely many of the $$\sqrt[2^k]3$$. If in such an expression, $$\sqrt[2^n]3$$ is one with maximal $$k$$, then all other $$\sqrt[2^k]3$$ are powers of $$\sqrt[2 k]3$$. It follows that $$\alpha\in\Bbb Q(\sqrt[2^n]3)$$ and $$\alpha$$ is algebraic.

A different approach for the infinity part: By Eisenstein, the polynomial $$X^{2^n}-3$$ is irreducible. Hence $$[F:\Bbb Q]\ge[\Bbb Q(\sqrt[2^n]3):\Bbb Q]\ge 2^n$$, where $$n$$ is arbitrariy. It follows that $$[F:\Bbb Q]$$ is inifnite.

• Thanks for the insight! I didn't realize I could use Eisenstein's criterion. – numericalorange Dec 7 '18 at 18:07

Let $$L = \mathbb Q(3^{\frac 1{2^n}})$$.

Why is $$L$$ algebraic over $$\mathbb Q$$? It is because the generating set of $$L$$ is $$\mathbb Q \cup \{3^{\frac 1{2^n}}\}$$, which are all contained in $$\bar{\mathbb Q}$$, the algebraic closure of $$\mathbb Q$$. Therefore, $$L \subset \bar {\mathbb Q}$$, which by definition of the algebraic closure implies it is algebraic over $$\mathbb Q$$.

However, $$L$$ contains subfields of the form $$K_n$$, where $$K_n = \mathbb Q(3^{\frac 1{2^n}})$$. One can conclude that $$[K_n : \mathbb Q] = 2^n$$, since $$x^{2^n} - 3$$ is irreducible via the Eisenstein criterion(shifted).

Then, suppose $$L$$ were finite, then $$[L : \mathbb Q] < \infty$$, but by the tower property it is equal to $$[L : K_n][K_n : \mathbb Q]$$ for each $$n$$. In particular, $$[L : \mathbb Q]$$ is a multiple of $$2^n$$ for all $$n$$, clearly impossible if it is finite.