basis for the splitting field of $x^3 - 3$ over the rationals

Question

Let $F$ be the splitting field of $x^3 - 3$ over the rationals. Find a basis for $F$ as a vector space over $\mathbb{Q},$ and prove your answer is correct.

I think that the roots of the polynomial $x^3 - 3$ are $\sqrt[3]{3}, \sqrt[3]{3}(-1 + i\sqrt{3})/2,$ and $\sqrt[3]{3}(-1 - i\sqrt{3})/2.$ From here I would know that $F = \mathbb{Q}(\sqrt[3]{3}, \sqrt{-3}).$ Would that mean that the basis of $F$ is $\{1, \sqrt[3]{3}, \sqrt{-3}\}?$

Are any of these ideas correct and how far away am I from a proof?

Answer 1

It is a basis as a vector space, not a set of generators as a ring. This means you are allowed only to add them and multiply them with a rational number when making linear combinations: not multiply them with each other.

Actually, what can be proven is that the basis has $6$ elements, and you can take as a basis: $1, \sqrt[3]{3}, \sqrt[3]{9}, \sqrt{-3}, \sqrt{-3}\sqrt[3]{3}, \sqrt{-3}\sqrt[3]{9}$. You need to prove that the splitting field is generated by those elements, and that they are linearly independent.

The first statement is easy, and basically comes from the fact that product of any two elements from the above (purported) basis is a linear combination of elements of the same set.

As for linear independence, if $\alpha+\beta\sqrt[3]{3}+\gamma\sqrt[3]{9}+\delta\sqrt{-3}+\epsilon\sqrt{-3}\sqrt[3]{3}+\zeta\sqrt{-3}\sqrt[3]{9}=0$, $\alpha,\beta,\ldots,\zeta\in\mathbb Q$, then by making the real and the imaginary parts equal to $0$ you have:

$$\alpha+\beta\sqrt[3]{3}+\gamma\sqrt[3]{9}=0$$

and

$$\delta+\epsilon\sqrt[3]{3}+\zeta\sqrt[3]{9}=0$$

Now, the first equality implies $\alpha=\beta=\gamma=0$ because $x^3-3$ is the minimal polynomial of $\sqrt[3]{3}$ over $\mathbb Q$ (it is irreducible by Eisenstein criterion for $p=3$), so no polynomial $\alpha+\beta x+\gamma x^2$ would have a zero in $\sqrt[3]{3}$, unless $\alpha=\beta=\gamma=0$. Similarly, $\delta=\epsilon=\zeta=0$.

Answer 2

Nop. That splitting field has dimension $\;6\;$ over the rationals, not $\;3\;$ . Take

$$\zeta:=e^{2\pi i/3}=\frac12\left(-1+\sqrt3\,i\right)$$

then the roots of the cubic polynomial are $\;\sqrt[3]3,\,\sqrt[3]3\,\zeta\,,\,\,\sqrt[3]3\,\zeta^2\;$ , and the splitting field of $\;x^3-3\in\Bbb Q[x]\;$ is $\;\Bbb Q(\sqrt[3]3,\,\zeta)\;$ , with dimension

$$[\Bbb Q(\sqrt[3]3,\,\zeta):\Bbb Q]=[\Bbb Q(\sqrt[3]3)(\zeta):\Bbb Q(\sqrt[3]3)]\cdot[\Bbb Q(\sqrt[3]3):\Bbb Q]=2\cdot3=6$$

since $\;x^2+x+1\;$ is the minimal polynomial of $\;\zeta\;$ over $\;\Bbb Q(\sqrt[3]3)\;$ . (Check all this!)