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I'm studying chapter 7.2 Algebraic extensions in Abstract Algebra by S. Lovett, and I'm stuck with an exercise problem.

Let $S = \{ \sqrt[n]{2} : n \in \mathbb{Z}$ with $n \geq 2 \}$. Prove that $\sqrt{3} \notin \mathbb{Q}[S]$.

Here is my argument.

Suppose that $\sqrt{3} \in \mathbb{Q}[S]$.

Since $\mathbb{Q}[S] = \cup \mathbb{Q}[\sqrt[n]{2}]$, there exists an $n$ such that $\sqrt{3} \in \mathbb{Q}[\sqrt[n]{2}]$.

If $n$ is odd, $[\mathbb{Q}[\sqrt[n]{2}]:\mathbb{Q}]=[\mathbb{Q}[\sqrt[n]{2}]:\mathbb{Q}[\sqrt{3}]][\mathbb{Q}[\sqrt{3}]:\mathbb{Q}]$ and $[\mathbb{Q}[\sqrt{3}]:\mathbb{Q}]=2$, so we have a contradiction.

Now I'm stuck at proving $\sqrt{3} \notin \mathbb{Q}[\sqrt[n]{2}]$ for $n$ even.

I appreciate any help on this part or suggestion of another approach on whole problem.

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If you can show the statement for $n$ a power of $2$, then you can argue with degrees to take care of all other $n$. Proof by induction works here. Let $\beta_k = \sqrt[2^k]{2}$. Suppose $\sqrt{3} \not\in \mathbb{Q}(\beta_k)$ where $k \leq m$. Now suppose $\sqrt{3} \in \mathbb{Q}(\beta_{m+1})$. Then there exist $a_0,a_1 \in \mathbb{Q}(\beta_m)$ such that $$\sqrt{3} = a_0 + a_1\beta_{m+1}.$$ Square both sides and try to get a contradiction.

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  • $\begingroup$ Thanks a lot. I have thought about why proving for a power of 2 takes care of all other even n. Let $n=2^{k} m$ for some $m$ odd. If $\mathbb{Q}[\sqrt{3}]$ is contained in $\mathbb{Q}[\sqrt[n]{2}]$, then let $K_{1} = \mathbb{Q}[\sqrt{3}]$ and $K_{2} = \mathbb{Q}[\sqrt[2^{k}]{2}]$. Then $[K_{1}K_{2}:\mathbb{Q}] \leq 2^{k+1}$, but since $[\mathbb{Q}[\sqrt[n]{2}]:\mathbb{Q}]=2^{k} m$ we must have $[K_{1}K_{2}:\mathbb{Q}]=2^{k}$, and this implies that $\mathbb{Q}[\sqrt{3}]$ must be contained in $\mathbb{Q}[\sqrt[2^{k}]{2}]$. I'm not sure I'm thinking in a correct way. $\endgroup$ – Ehtelerosa Sep 28 '16 at 8:13
  • $\begingroup$ We know that $[K_1K_2 : \mathbb{Q}] = 2^{k+1}$ because by the above argument $\sqrt{3} \not\in K_2$. So $[K_1K_2 : K_2] = 2$ and $K_1K_2$ is strictly between $K_2$ and $\mathbb{Q}(\sqrt[n]{2})$. But this is impossible, because $[\mathbb{Q}(\sqrt[n]{2}) : K_2] = m$ is odd. $\endgroup$ – Ethan Alwaise Sep 28 '16 at 22:25

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