I am working through Rotman's Galois Theory, and I came across an example that confused me a bit. Here is a screenshot of the problem:
I am not sure, why the degree of $\mathbb{Q}(\omega)/\mathbb{Q}$ is only 2. $\omega$ if I am not mistaken is a cube root of unity, and I assumed that adjoining it would give an extension of degree 3.
When viewed as a vector space of $\mathbb{Q}$ does $\mathbb{Q}(\omega)$ not have {$1,\omega,\omega^{2}$} as a basis?
It is clear to me however, that $\omega$ is indeed a root of $x^{2}+x+1$, for $\omega^{2}+\omega+1 = -\frac{1}{2}+\frac{i\sqrt{3}}{2}-\frac{1}{2}-\frac{i\sqrt{3}}{2}+1=0 $
In light of this, is the reason that {$1,\omega,\omega^{2}$} is not a basis because $\omega^{2} = -\omega-1$? The basis I proposed was not linearly independent?
Finally, is there a general way of determining the degree of $\mathbb{Q}(\omega)/\mathbb{Q}$, where $\omega^{n}=1$?
No. The minimal polynomial of $\omega$ over $\mathbf Q$, $x^2+x+1$, has degree $2$, hence $\,[\mathbf Q(\omega):\mathbf Q]=2$. A basis is $(1,\omega)$. Indeed: \begin{align*} \omega^2 &= -\omega-1\\ \omega^3&= -\omega^2-\omega=-2\omega-1\\ \text{&c.} \end{align*}