What Is the Basis of the Splitting Field of $x^3 - 2$ over $\mathbb Q$?

Question

Let $\zeta$ denote the primitive $3$rd root of unity, and the roots of the polynomial are $ \sqrt[3]2, \zeta \sqrt[3]2, \zeta^2 \sqrt[3]2$. From this, I cannot see how the splitting field has degree $6$ over $\mathbb Q$. However, I can prove that $K = \mathbb Q(\zeta, \sqrt[3]2)$ is the splitting field of the polynomial. Furthermore, it has degree $6$ over $\mathbb Q$ because:

1. The minimal polynomial of $\sqrt[3]2$ over $\mathbb Q$ is $x^3 - 2$. So the basis of $\mathbb Q(\sqrt[3]2) / \mathbb Q$ is $\{ \, 1, \sqrt[3]2, \sqrt[3]4 \, \}$.
2. The minimal polynomial of $\zeta$ over $\mathbb Q(\sqrt[3]2)$ is $x^2 + x + 1$. So the basis of $K / \mathbb Q(\sqrt[3]2)$ is $\{ \, 1, \zeta \, \}$.

Therefore, the basis of $K/ \mathbb Q$ is $\{ \, 1, \sqrt[3]2, \sqrt[3]4, \zeta, \zeta \sqrt[3]2, \zeta \sqrt[3]4 \, \}$. This algorithm follows the proof in the multiplicity $[ \, L : K \, ][ \, K : F \, ] = [ \, L : F \, ]$, so I think that my answer is correct, but it's so long. Is there a simpler way to do this?

Answer 1

Okay $x^3 - 2=f(x)\in \Bbb Q[x]$ has splitting field $\Bbb Q(\sqrt[3]{2},i\sqrt{3})$ which we can show that $[\Bbb Q(i\sqrt{3}):\Bbb Q] = 2$ and the minimal polynomial argument gives us $[\Bbb Q(\sqrt[3]{2}):\Bbb Q] = 3$ so tower of fields gives us $[\Bbb Q(\sqrt[3]{2},i\sqrt{3}):\Bbb Q]=6$.

$$\{a+bi\sqrt{3} + c\sqrt[3]{2}+ d(\sqrt[3]{2})^2+ e\sqrt[3]{2}\sqrt{3}i + f(\sqrt[3]{2})^2 i\sqrt{3}:a,b,c,d,e,f\in \Bbb Q\}$$

There isn't really a more simple way unfortunately.

Answer 2

The answer is already in the question, so this is a dumbed down version (please edit if there are mistakes).

The splitting field of $x^3 -2$ over the rationals is the smallest field extension containing all the roots. The field extension is necessary because $x^3-2$ is irreducible (e.g. it has no roots that are rational numbers).

The root $2^{1/3}$ will require a field extension $\mathbb Q(2^{1/3})$ containing $2^{1/3}$ but also $2^{2/3}$ because the powers of the root need to be included in the extension to form a field. This extended field can be thought of as a vector space with basis vectors

$$\{\mathbf 1, \mathbf {2^{1/3}}, \mathbf{2^{2/3}} \}$$

such that each element of the field extension can be expressed as a linear combination

$$\mathbb Q(2^{1/3})= \{ a \, \mathbf 1 + b \,\mathbf {2^{1/3}} + c \, \mathbf{2^{2/3}} \}$$

with $a,b,c\in \mathbb Q.$ If $b,c=0$ we get back the base field $\mathbb Q.$

The minimal polynomial of the extension $\mathbb Q(2^{1/3})$ is $x^3-2,$ which, as much as the number of basis vectors, denotes that the extension is of degree $3.$

But there are two additional roots to this degree $x^3-2$ polynomial, given by the product of $2^{1/3}$ by the primitive third root of unity and its complex conjugate, which can be represented as $\zeta_3=e^{\frac{2\pi}{3} i}$ and $\zeta_3^2=e^{\frac{4\pi}{3} i},$ namely $\zeta_3 2^{1/3}$ and $\zeta_3^2 2^{1/3}.$

To adjoin these roots to the field extension $\mathbb Q(2^{1/3})$ we need the minimal polynomial of $\zeta_3.$ Since $\zeta_3$ is a root of $x^3-1,$ but $x^3-1=(x-1)(x^2+x+1),$ and the root of $x-1$ is $1,$ it follows that $\zeta_3$ is root of $(x^2+x+1).$ Now, $(x^2+x+1)$ is the minimal polynomial of $\zeta_3,$ being monic, irreducible, and having $\zeta_3$ as one of its roots. Therefore the field extension of $\mathbb Q(2^{1/3})$ by $\zeta_3$ is of degree $2.$ Correspondingly, the extension of $\mathbb Q$ associated to the minimal polynomial $(x^2+x+1)$ will have as basis the $2$ vectors:

$$\{\mathbf 1, \mathbf{\zeta_3} \}$$

which will be closed under multiplication without need for $\mathbf{ \zeta_3^2}$ because $\{\mathbf 1, \mathbf{\zeta_3} \}=\{\mathbf{\zeta_3^3}, \mathbf{\zeta_3} \}$ renders $\mathbf{\zeta_3^2}$ linearly dependent $-1-e^{\frac{2\pi}{3} i}=\left(e^{\frac{2\pi}{3} i} \right)^2.$

The elements of this extended field can be written as

$$\mathbb Q(\zeta_3,2^{1/3})= \{a \, \mathbf 1 + b \,\mathbf {2^{1/3}} + c \, \mathbf{2^{2/3}} + d \, \mathbf{\zeta_3} + e \, \mathbf{\zeta_3 2^{1/3}} + f \, \mathbf{\zeta_3 2^{2/3}}; a,b,c,d,e,f\in \mathbb Q \}$$

resulting from distributing $\{\mathbf 1, \mathbf{\zeta_3}\}$ over $\{ a \, \mathbf 1 + b \,\mathbf {2^{1/3}} + c \, \mathbf{2^{2/3}}\}.$

The basis vectors of the extension of $\mathbb Q(2^{1/3},\zeta_3)$ over $\mathbb Q,$ i.e. $[\mathbb Q(2^{1/3},\zeta_3):\mathbb Q]$ is the Cartesian product of the basis of $[\mathbb Q(2^{1/3}):\mathbb Q]$ and $[\zeta_3:\mathbb Q],$ which guarantees linearly independent basis:

$$\begin{array}{c|ccc} & \mathbf 1 & \mathbf {2^{1/3}} & \mathbf {2^{2/3}} \\ \hline \mathbf 1 & \mathbf 1 &\mathbf {2^{1/3}} & \mathbf {2^{2/3}} \\ \mathbf {\zeta_3} & \mathbf {\zeta_3} & \mathbf {2^{1/3}}\mathbf {\zeta_3} & \mathbf {2^{2/3}}\mathbf {\zeta_3} \end{array}$$

or $\{\mathbf 1, \mathbf {2^{1/3}}, \mathbf{2^{2/3}}, \mathbf{2^{1/3} \zeta_3} , \mathbf{2^{2/3} \zeta_3} \}$

of degree $$[\mathbb Q(\zeta_3, 2^{1/3}):\mathbb Q] =\underset{\text{min poly } x^2+x+1}{[\mathbb Q(\zeta_3):\mathbb Q(2^{1/3})]} \quad \underset{\text{min poly } x^3-2}{[\mathbb Q(2^{1/3}):\mathbb Q]} =2\times 3=6$$ by the tower law.