Determine the degree of another field extension.

I am looking for more details to part of a solution to this question How do I find a splitting field $x^8-3$ over $\mathbb{Q}$?. I would like to determine the degree of the splitting field for $x^8-3$ over $\mathbb{Q}$. It is easy to see that the splitting field is $\mathbb{Q}(\zeta,\sqrt{3})$, where $\zeta$ is a primitive 8th root of unity and $\sqrt{3}$ is the positive real $8$th root of $3$. Also, it is easy to see that $[\mathbb{Q}(\zeta):\mathbb{Q}]=4$ and $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=8$ by noticing that $x^4+1$ and $x^8-3$ are irreducible, respectively, over $\mathbb{Q}$. I see then, by the tower law, it remains only to find either of $[\mathbb{Q}(\zeta,\sqrt{3}):\mathbb{Q}(\zeta)]$,$[\mathbb{Q}(\zeta,\sqrt{3}):\mathbb{Q}(\sqrt{3}]$. In an answer to the aforementioned question, it is stated that "It is not difficult to see that $\zeta\not\in\mathbb{Q}(\sqrt{3})$, hence K is of dimension 4⋅8=32 over $\mathbb{Q}$. How did they conclude that $[\mathbb{Q}(\zeta,\sqrt{3}):\mathbb{Q}(\sqrt{3}]=4$ since $\zeta\not\in\mathbb{Q}(\sqrt{3})$? It seems this only means $[\mathbb{Q}(\zeta,\sqrt{3}):\mathbb{Q}(\sqrt{3}]\not= 1$.

I believe, then, that it is sufficient to show that $x^4+1$ is irreducuble over $\mathbb{Q}(\sqrt{3})$; to this end, one can see that $x^4+1$ factors over $\mathbb{R}\supset \mathbb{Q}(\sqrt{3})$ as $(x^2-\sqrt{2}+1)(x^2+\sqrt{2}+1)$. Thus, $x^4+1$ is irreducuble over $\mathbb{Q}(\sqrt{3})$ if and only if $\sqrt{2}\not\in\mathbb{Q}(\sqrt{3})$. However, the usual trick of showing that the square of any element in $\mathbb{Q}(\sqrt{3})$ is not equal to $2$ is very difficult, since $\mathbb{Q}(\sqrt{3})$ is an 8-dimensional vectorspace over the rationals! I pray that I am making things more difficult than necessary...

Thanks in advance to any help or insight you may afford.

• Why is this downvoted? – Spanky Mar 18 '14 at 15:59

No one has answered my question, and now I have a solution! The solution below is a little different than what I suggested in the OP, as I believe it is more concise, but it essentially boils down to showing that $\sqrt{2}\not\in \mathbb{Q}(\sqrt{3})$:
Since $\zeta=(1+i)\frac{\sqrt{2}}{2}$, it is immediate that $\mathbb{Q}(\sqrt{2},i)\subset \mathbb{Q}(\zeta)\subset(\sqrt{2},i)$, hence $\mathbb{Q}(\zeta,\sqrt{3})=\mathbb{Q}(\sqrt{2},i,\sqrt{3})$. Then, from the tower lemma, $$[\mathbb{Q}(\sqrt{2},i,\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},i,\sqrt{3}):\mathbb{Q}(\sqrt{2},\sqrt{3})][\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}):\mathbb{Q}].$$
It is easy to see that $[\mathbb{Q}(\sqrt{2},i,\sqrt{3}):\mathbb{Q}(\sqrt{2},\sqrt{3})]=2$ since $\mathbb{Q}(\sqrt{2},\sqrt{3})\subset \mathbb{R}$ implies $1<[\mathbb{Q}(\sqrt{2},i,\sqrt{3}):\mathbb{Q}(\sqrt{2},\sqrt{3})]$ and $i$ is a root of $x^2+1 \in \mathbb{Q}(\sqrt{2},\sqrt{3})[x]$ implies $[\mathbb{Q}(\sqrt{2},i,\sqrt{3}):\mathbb{Q}(\sqrt{2},\sqrt{3})]\leq 2$. Also, it is easy to see $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=8$ because $x^8-3$ is irreducible (Eisenstein with $p=3$); that is, $x^8-3$ is the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}$. Finally, we can see $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{3})] \leq 2$ since $\sqrt{2}$ is a root of $x^2-2$, which is a polynomial over $\mathbb{Q}(\sqrt{3})$. To see $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{3})] \not= 1$, it suffices to show $\sqrt{2} \not\in \mathbb{Q}(\sqrt{3})$. If $\sqrt{2}\in \mathbb{Q}(\sqrt{3})$, we have again by the tower lemma $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$. We know $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=8$ and $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$, so $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}(\sqrt{2})]=4$. Thus, the minimal polynomial $m_{\sqrt{3}}(x)$ of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$ is a monic polynomial of degree $4$, and since this minimal polynomial must divide $x^8-3$, we know that in $\mathbb{Q}(\sqrt{3})$, $$m_{\sqrt{3}}(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4),$$ where each $\alpha_i=\sqrt{3}\zeta^{n_i}$, each $n_i=0,\dots,7$. Then the constant term of $m_{\sqrt{3}}(x)$ is of the form $\sqrt{3}\zeta^n$ for some integer $n$, and since $m_a(x)$ is a polynomial over $\mathbb{Q}(\sqrt{2})\subset \mathbb{R}$, the constant term of $m_a(x)$ must be $\pm \sqrt{3}$. It is easy to see that $\sqrt{3}\not\in \mathbb{Q}(\sqrt{2})$ (suppose $3=(a+b\sqrt{2})^2$ and derive a contradiction). This contradiction implies $\sqrt{2} \not\in \mathbb{Q}(\sqrt{3})$. Since this is the case, $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{3})]=2$, so we can conclude that $[\mathbb{Q}(\sqrt{2},i,\sqrt{3}):\mathbb{Q}]= (2)(2)(8)=32$.