Confusion about the splitting field of $x^3-7$

Question

I'm asked to find the degree of the splitting field of $x^3-7$ over the rationals.

The roots are $\sqrt[3] 7e^{\frac{2\pi ik}{3}},\ k=0,1,2$. Explicitly, $$x_1=\sqrt[3] 7,\\ x_2=\sqrt[3] 7 \bigg(-\frac{1}{2}+i\frac{\sqrt 3}{2}\bigg),\\ x_3=\sqrt[3] 7 \bigg(-\frac{1}{2}-i\frac{\sqrt 3}{2}\bigg).$$

The splitting field is an extension that contains all roots. I see at least two essentially different possibilities of building it. First, I could adjoin $\sqrt[3] 7, i,\sqrt 3$. Such an extension will contain all roots. On the other hand, I could adjoin $\sqrt[3] 7, i\sqrt 3$. This extension also contains all the roots. I believe the two extensions have different degrees. So how should I understand which degree I should find to begin with? Which extension is the splitting field?

Answer 1

The splitting field $K$ of $x^3-7$ is $\mathbb Q(\sqrt[3]7, \omega)$, where $\omega=-\frac{1}{2}+i\frac{\sqrt 3}{2}$ is a primitive cubic root of the unity, a root of $x^2+x+1$. There is no need to decompose $\omega$ into $i$ and $\sqrt 3$.

It is true that $L =\mathbb Q(\sqrt[3]7, i, \sqrt 3)$ contains all the roots of $x^3-7$, but $L$ is not the smallest such field. That is $K$.

Answer 2

You have $x^3-7$ and $\alpha=\sqrt [ 3 ]{ 7 }$ is a clear root of $x^3-7$. Then after factoring and applying the quadratic formula $($if needed$)$ one factors $x^3-7=(x-\alpha)(x-\alpha\zeta )(x-\alpha\zeta ^2)$ where $\zeta $ is a complex cube root of unity. $\zeta ^2+\zeta +1=0$ and $\zeta \notin\mathbb{R}$ hence $\notin\mathbb{Q}(\alpha)$, so splitting the field has the degree $3\cdot2=6$. In fact the splitting field is $\mathbb{Q}(\alpha,\zeta)$.