An Upper and Lower Bound for a Field Extension of $\mathbb{Q}$ by a Complex Root 
Consider the polynomial $f(x) = x^3 + \zeta x + \sqrt{3}$ in $\mathbb{C}[x],$ where $\zeta$ is a primitive third root of unity. Given a root $\alpha$ of $f(x)$ in $\mathbb{C},$ prove that $4 \leq [\mathbb{Q}(\alpha) : \mathbb{Q}] \leq 12.$

We note first that $[\mathbb{Q}(\sqrt{3}, \zeta) : \mathbb{Q}] = 4.$ Either $f(x)$ is irreducible in $\mathbb{Q}(\sqrt{3}, \zeta)[x]$ or not. Given that $f(x)$ is irreducible in $\mathbb{Q}(\sqrt{3}, \zeta)[x],$ we have that $[\mathbb{Q}(\sqrt{3}, \zeta, \alpha) : \mathbb{Q}] = 12.$ We note that $\mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt{3}, \zeta, \alpha),$ from which we conclude that $[\mathbb{Q}(\alpha) : \mathbb{Q}] \leq 12.$ Getting the lower bound is proving to be more difficult than this. I would like to say that if $f(x)$ is reducible in $\mathbb{Q}(\sqrt{3}, \zeta)[x],$ then $[\mathbb{Q}(\alpha) : \mathbb{Q}] \geq 4,$ but I am not convinced. Can anyone provide a push in the right direction?
 A: If $f$ is reducible in $K=\mathbb{Q}(\zeta,\sqrt{3})$, it either splits completely or it decomposes into a linear and a quadratic factors. In the first case $\alpha\in K$. In the second case, $\alpha$ is either the roots of the linear factor and again $\alpha\in K$, or is a root of the second factor and the extension requires a further quadratic extension of $K$. So, the lower bound is when $\alpha\in K$. 
Now, if $\alpha\in K$, we need to show that $\alpha$ is not in an (strict) intermediate extension between $K\supset H\supset \mathbb{Q}$. If that is the case then $[H:\mathbb{Q}]=2$, the only divisor of $4=[K:\mathbb{Q}]$ that is not $1$ or $4$. Then $\alpha$ is a root of a quadratic polynomial with rational coefficients. This polynomial $x^2+ax+b$ must divide $f$. The condition that the remainder is $0$ gives you $\zeta$ and $\sqrt{3}$ expressed as a rational numbers:
$$\frac{x^3+\zeta x+\sqrt{3}}{x^2+ax+b}=(x-a)+\frac{(\zeta-b+a)x+(\sqrt{3}+ab)}{x^2+ax+b}$$
Therefore
$$\begin{align}\zeta&=b-a\\\sqrt{3}&=-ab\end{align}$$
which contradicts that $\zeta$ and $\sqrt{3}$ are not rational. Therefore, $\alpha$ cannot be in a field properly contained in $K=\mathbb{Q}(\zeta,\sqrt{3})$.

Aside:
The minimal polynomial of $\alpha$ turns out to be $x^{12}-2x^{10}+ 3 x^{8}-8x^6+7x^4+3x^2+9$. Therefore, actually $[\mathbb{Q}(\alpha):\mathbb{Q}]=12$
