Relations between field extensions of $\mathbb{Q}$

Question

We were the following related questions, with $\omega$ as the third root of unity:

1. Determine the relation between the following field extension: $\mathbb{Q}(\sqrt{3}\omega)$, $\mathbb{Q}(\sqrt{3}+\omega)$ and $\mathbb{Q}(\sqrt{3},\omega)$.
2. Calculate $[\mathbb{Q}(\sqrt{3},\omega):\mathbb{Q}]$.
3. Determine the minimal polynomial for $\sqrt{3}+\omega$ over $\mathbb{Q}$.

For the first the following relations seem obvious: $\mathbb{Q}(\sqrt{3}+\omega)\subset\mathbb{Q}(\sqrt{3},\omega)$ and $\mathbb{Q}(\sqrt{3}\omega)\subset\mathbb{Q}(\sqrt{3},\omega)$. Then we can also calculate that $(\sqrt{3}\omega)^3/3=\sqrt{3}$ so $\sqrt{3}\in \mathbb{Q}(\sqrt{3}\omega)$. As well as $(\sqrt{3}\omega)^4/3=\omega$ so $\omega\in \mathbb{Q}(\sqrt{3}\omega)$. From this follows then $\mathbb{Q}(\sqrt{3},\omega)\subset \mathbb{Q}(\sqrt{3}\omega)$. So $\mathbb{Q}(\sqrt{3},\omega)= \mathbb{Q}(\sqrt{3}\omega)$ is the relation. For the other relation, I couldn't quite find such an equality. I found that since $\omega := -1/2 + \sqrt{3}i/2$ this implies $\mathbb{Q}( \sqrt{3},\omega)= \mathbb{Q}( \sqrt{3},i)$. I then calculated that $(\sqrt{3}+\omega)^2+(\sqrt{3}+\omega) -2=3i$ so $ i \in \mathbb{Q}(\sqrt{3}+\omega)$. I couldn't calculate $\sqrt{3}$, so that's where that question ended for me.
Now onto the second question where I used the fact that $\mathbb{Q}( \sqrt{3},\omega)= \mathbb{Q}( \sqrt{3}, i)$ following with the product formula so that $[\mathbb{Q}(\sqrt{3},\omega):\mathbb{Q}]=4$. For the third part, I just tried calculating the powers for $(\sqrt{3}+\omega)$ and subsequently for $(\frac{\sqrt{3}-1}{2} + \frac{\sqrt{3}i}{2})$ but got lost in the calculation so that I couldn't get any polynomial from it such that $(\sqrt{3}+\omega)$ is a root.

Answer 1

For the third part let $x=\sqrt{3}+\omega$, rewrite it to $x-\sqrt{3}=\omega$ and cube it. This gives $$ x^3+9x-1 =3\sqrt{3}(x^2+1). $$ Then square it. This yields \begin{align} 0 & =x^6 - 9x^4 - 2x^3 + 27x^2 - 18x - 26\\ & =(x^2 - 2x - 2)(x^4 + 2x^3 - 3x^2 - 4x + 13). \end{align} Both polynomials are irreducible. Obviously $x$ is not a root of the quadratic polynomial, since you already know that $[\Bbb Q(x):\Bbb Q]=4$. So the minimal polynomial of $\sqrt{3}+\omega$ is $$ x^4 + 2x^3 - 3x^2 - 4x + 13. $$

Answer 2

This is too long for a comment.

Here is an alternative to Dietrich Burde's answer.

This does not require factorization of a sixth-degree polynomial.

Let $x=\sqrt3+\left(\frac{-1}2+\frac{\sqrt3}2i\right).$

Then $x+\frac12=\sqrt3\left(1+\frac12i\right).$

Squaring both sides, $x^2+x+\frac14=3\left(1+\frac{-1}4+i\right)=3\left(\frac34+i\right)=\frac94+3i,$

so $x^2+x-2=3i.$

Again squaring both sides, $x^4+x^2+4+2x^3-4x^2-4x=-9$,

so $x^4+2x^3-3x^2-4x+13=0$.