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Compute the degree of the field extension $$\mathbb{Q}(\sqrt{5},w): \mathbb{Q} ,$$ where $ w = e^{2\pi i / 3}$.

I consider the tower of fields $\mathbb{Q} \subset \mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\sqrt{5})(w)$ now, $\mathbb{Q}: \mathbb{Q}(\sqrt{5})$ has degree $2$, so I am trying to find the degree of $\mathbb{Q}(\sqrt{5}) : \mathbb{Q}(\sqrt{5})(w)$ it is $\leq 2$ since $w $ satisfies $w^2+w+1 = 0$. I am trying to show that it is exactly $2$ - I know that $ w \notin \mathbb{Q}(\sqrt{5})$ but I don't see how I can justify it is exactly $2$ from here.

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    $\begingroup$ The degree of $w$ over $\mathbb{Q}(\sqrt{5})$ can be either $1$ or $2$. It can't be $1$ because if it were, $w$ would belong to $\mathbb{Q}(\sqrt{5}) \subseteq \mathbb{R}$, which is not the case. So it must be $2$. $\endgroup$ – Gro-Tsen May 11 '16 at 15:05
  • $\begingroup$ @Gro-Tsen could you explain why if the degree is 1 it implies $w \in \mathbb{Q}(\sqrt{5})$? $\endgroup$ – james1395 May 11 '16 at 15:19
  • $\begingroup$ If $K\subseteq L$ is a field extension and $x\in L$ is of degree $1$ over $K$ then $[K(x):K]=1$ so $K(x)=K$ so $x\in K$; or, if you prefer, $x$ is the root of a degree $1$ polynomial over $K$, so it is in $K$. $\endgroup$ – Gro-Tsen May 11 '16 at 16:57
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Since $[\Bbb Q(\sqrt 5)(w) : \Bbb Q(\sqrt 5)]$ is $\leq 2$ (as $w^2 + w + 1 = 0$), it is either $1$ or $2$, and it is $1$ iff $w \in \sqrt{5}$. Since $\Bbb Q(\sqrt 5) \leq \Bbb R$ but $w \not \in \Bbb R$, we must have $[\Bbb Q(\sqrt 5)(w) : \Bbb Q(\sqrt 5)] = 2$.

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  • $\begingroup$ could you explain why if the degree is 1 it implies $w \in \mathbb{Q}(\sqrt{5})$? $\endgroup$ – james1395 May 11 '16 at 15:19
  • $\begingroup$ This says that as a vector space over $\Bbb Q(\sqrt{5})$, $\Bbb Q(\sqrt{5})(w)$ is $1$-dimensional, that is, that $\Bbb Q(\sqrt{5})(w) = \Bbb Q(\sqrt{5})$. $\endgroup$ – Travis May 11 '16 at 15:24

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