Find the splitting field of $x^3-1$ over $\mathbb{Q}$.

Question

Find the splitting field of $x^3-1$ over $\mathbb{Q}$.

My try:

Factoring this to the most I can (in $\mathbb{Q}$), we get that $(x-1)(x^2+x+1)$

So $x=1$ is a root of $f(x)$.

$x^2+x+1$ has no roots in $\mathbb{Q}$ so I conclude it is irreducible (since we have a low degree of $2$)

Can I conclude that the splitting field of $f(x)$ is $\mathbb{Q}(1)=\mathbb{Q}$ because $1\in\mathbb{Q}$?

Or do I have to keep factoring, assuming that the roots I find will be in some extension field?

Like:

$x^3-1=(x-1)(x^2+x+1)$ gives the roots $x=1, x=\frac{-1-\sqrt{3}i}{2}, x=\frac{-1+\sqrt{3}i}{2}$, in which case the splitting field is $\mathbb{Q}(1,\frac{-1-\sqrt{3}i}{2},\frac{-1+\sqrt{3}i}{2})$. But if this is the case, I'm not sure how to simplify this.

Answer 1

Convince yourself, by using the quadratic formula to solve:

$x^2 + x + 1 = 0$

that the splitting field of $x^3 - 1$ is $\Bbb Q(\sqrt{-3})$.

Note: it is customary to denote this field as $\Bbb Q(\omega)$, where:

$\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$

Therefore, convince yourself that $\Bbb Q(\omega) = \Bbb Q(\sqrt{-3})$.

It may be helpful to recall the formula for the discriminant of a quadratic equation.

Answer 2

The splitting field of $x^3 - 1$ is the smallest field containing $\mathbb Q$ and all of its roots in $\mathbb{C}$ i.e. $$\mathbb{Q} \left(1, \frac{-1-\sqrt{3}i}{2}, \frac{-1+\sqrt{3}i}{2}\right) = \mathbb{Q} \left(\frac{-1-\sqrt{3}i}{2}, \frac{-1+\sqrt{3}i}{2}\right)$$ Of course if you want you can always write it as a simple extension of $\mathbb Q$ (using the primitive element theorem)