$x^3 -2$ is irreducible over $\mathbb{Q}$ of degree $3$. And $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$
However, $x^3-1$ is reducible, it equals $(x-1)(x^2 + x + 1)$. So the degree of the polynomial $x^2+x+1$ satisfied by $\omega = \frac{-1 + \sqrt{3} i}{2}$ is $2$. But the vector space basis is $\{1,\omega, \omega^2\}$ so $[\mathbb{Q}(\omega):\mathbb{Q}] = 3$.
Have I made a mistake or can the degrees not match?