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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?

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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 $[\mathbb{Q}(\alpha):\mathbb{Q}]=12$

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  • $\begingroup$ Excellent. During my original attempt of the problem, I had worked out exactly what you stated in the first paragraph, and I had thought about whether or not $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt{3}, \zeta),$ but as you correctly pointed out, we simply need to show that $\alpha$ is not contained in a degree two extension of $\mathbb{Q}.$ Thank you very much for your time. $\endgroup$ – Dylan_Carlo_Beck Aug 2 '18 at 6:06
  • $\begingroup$ Can you elaborate on how you obtain the minimal polynomial? I am fascinated. $\endgroup$ – Dylan_Carlo_Beck Aug 2 '18 at 6:08
  • $\begingroup$ @Dylan_Carlo_Beck That was done by a computer. $\zeta$ and $\sqrt{3}$ are expressed in radicals, and so can be the roots of a cubic. So, you can compute the three values of $\alpha$. Then the minimal polynomial can also be found algorithmically by searching for rational combinations between its powers. No something worth doing by hand. $\endgroup$ – user580373 Aug 2 '18 at 10:30
  • $\begingroup$ I just typed up my solution to the problem, and I noticed that we should have $\zeta = b - a^2.$ Of course, the contradiction is reached all the same, but I just figured I'd point that out. $\endgroup$ – Dylan_Carlo_Beck Aug 2 '18 at 18:35

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